# ECE 271 Electronic Circuits I Topic 6 Analog ECE 271 Electronic Circuits I Topic 6 Analog Circuits Ideal Operational Amplifiers One Transistor Amplifier NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 1 Example of Analog Electronic System: FM Stereo Receiver Much information in the world (temperature, pressure, light intensity, sound, etc.) is analog in nature. In electric form those signals are transformed by different linear and nonlinear functions. Linear functions: Radio and audio frequency amplification, frequency selection (tuning), impedance matching, local oscillator Nonlinear functions: DC power supply(rectification), frequency conversion

(mixing), detection/demodulation The characteristics of signals are most often manipulated with linear amplifiers. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 2 Amplification: Introduction Example: Audio amplifier. The input is a form of complicated periodic signal. i i v i R i A complex periodic signal can be represented as the sum of many individual sine waves, one component of which has amplitude Vi = 1 mV and frequency ws with 0 phase (signal is used as reference): vi Vi sin wst After amplification, linear amplifier output is sinusoidal with same frequency but different amplitude VO and phase : vo Vo (sin wst ) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 3

Amplification: Introduction (cont.) Amplifier output power is: PO Vo 2 1 2 RL If we need to deliver PO = 100 W and have RL = 8 W, , then Vo 2PO RL 21008 40V

This power results in output current: io Io (sin wst ) where Io Vo 40V 5A RL 8W, Input current is given by (Vi = 1 mV ) -3 V Vi 10 Ii 1.8210 8 A Ri Rin 5kW, 50kW, and the phase is zero because circuit is purely resistive. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 4 Amplification: Gain The main parameter of an amplifier is the gain. Using phasor representation v V , we can introduce three types of gain. Voltage Gain (complex number): A vo Vo Vo v vi Vi 0 Vi Magnitude and phase of voltage gain are given by Vo

Av and Av Vi Vo 40V For our example, Av 3 4104 Vi 10 V Current Gain: i I Io Ai o o ii Ii 0 Ii Magnitude of current gain is given by I 5A 8 Ai o 2.75 10 Ii 1.8210-8 A NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 5 Amplification: Gain (cont.) Vo

P Power Gain: AP O 2 Pi Vi 2 For our example, Io 2 Vo I o A A v i I i Vi I i 2 405 13 AP 3 1.10 10 10 1.8210 8 The gain is often expressed in decibel scale: AvdB 20logAv AidB 20logAi APdB 10logAP NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 6 Two-port Model for Amplifier Type of input-output model, black box that relates outputs to inputs. input x1 Amplifier output x2 Simplifies amplifier-behavior modeling in complex systems. Two-port (four terminal) models are linear network models, valid only under small-signal conditions. Represented by g-, h-, y- and z-parameters. (v1, i1) and (v2, i2) represent signal components of voltages and currents at the network ports. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 7 Two-port Model Parameters Theoretically, two-port model could be described fully by a 4x4 matrix or 16 parameters. In practice, several sets of 4 parameters are used.

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 8 Two-port Model Parameters Theoretically, two-port model could be described fully by a 4x4 matrix or 16 parameters. In practice, several sets of 4 parameters are used. Impedance (z)-parameters NJIT ECE 271 Dr. Serhiy Levkov Admittance (y)-parameters Topic 6 - 9 Two-port Model Parameters Theoretically, two-port model could be described fully by a 4x4 matrix or 16 parameters. In practice, several sets of 4 parameters are used. Impedance (z)-parameters

Hybrid (h)-parameters NJIT ECE 271 Dr. Serhiy Levkov Admittance (y)-parameters Inverse hybrid (g)-parameters Topic 6 - 10 Two-port Model Parameters Theoretically, two-port model could be described fully by a 4x4 matrix or 16 parameters. In practice, several sets of 4 parameters are used. Impedance (z)-parameters Hybrid (h)-parameters Admittance (y)-parameters Inverse hybrid (g)-parameters Transfer (A,B,C,D) -parameters NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 11 g-parameters i1g11 v1 g12i 2 v2 g21v1 g22i 2 i g11 v1 1 i20 i g12 1 i2 v 0 1 NJIT ECE 271 Dr. Serhiy Levkov Use open-circuit (i = 0) and short-circuit (v = 0) terminal conditions Open-circuit input conductance v g21 v2 Reverse short-circuit current

gain v g22 2 i2 v 0 1 i2 0 1 Forward open-circuit voltage gain Short-circuit output resistance Topic 6 - 12 g-parameters i1g11 v1 g12i 2 v2 g21v1 g22i 2 i g11 v1 1 i20 i

g12 1 i2 v 0 1 If g12 = 0 NJIT ECE 271 Dr. Serhiy Levkov Use open-circuit (i = 0) and short-circuit (v = 0) terminal conditions Open-circuit input conductance v g21 v2 Reverse short-circuit current gain v g22 2 i2 v 0 1 i2 0 1

Forward open-circuit voltage gain Short-circuit output resistance Norton transformation Topic 6 - 13 g-parameters: Example Problem: Find g-parameters. Approach: Apply specified boundary conditions for each g-parameter, use circuit analysis. For g11 and g21: apply voltage v1 to input port and open circuit output port. For g12 and g22: apply current i2 to output port and short circuit input port. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 14 g-parameters: Example i g11 v1 1i

2 1 210 4 W, 51(200kW, ) 0 g11 9.7910 8S v g21 v2 g11(51)(200kW, )0.998 1 i 0 2 Problem: Find g-parameters. Approach: Apply specified boundary conditions for each g-parameter, use circuit analysis. For g11 and g21: apply voltage v1 to input port and open circuit output port. For g12 and g22: apply current i2 to output port and short circuit input port. NJIT ECE 271 Dr. Serhiy Levkov v

g22 2 i2 v 1 1 391W, 1 51 200kW, 20kW, 391W, 0.0196 20kW, 0 1 i g12 1 i2 v 0

i 19.7910 8 v1 1.9610 2 i2 v2 0.998v1 3.91102 i2 Topic 6 - 15 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. RL Rout RL Rin v 1vs RS Rin V Rin RL Av o A Vs RS Rin Rout RL v o Av1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 16 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load

voltage. RL Rout RL Rin v 1vs RS Rin V Rin RL Av o A Vs RS Rin Rout RL v o Av1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 17 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. RL Rout RL Rin v 1vs RS Rin

v Rin RL Av o A vs RS Rin Rout RL v o Av1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 18 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. RL Rout RL Rin v 1vs RS Rin v Rin RL Av o A vs RS Rin Rout RL v o Av1

Conclusion: full gain depends on source and load parameters. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 19 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. How to make it non- dependent? RL Rout RL Rin v 1vs RS Rin v Rin RL Av o A vs RS Rin Rout RL v o Av1 Conclusion: full gain depends on source and load parameters. NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 20 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. How to make it non- dependent? Rin >> Rs and Rout<< RL, Av A RL Rout RL Rin v 1vs RS Rin V Rin RL Av o A Vs RS Rin Rout RL v o Av1 Ideal voltage amplifier: Rout = 0, Rin

Conclusion: full gain depends on source and load parameters. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 21 Ideal Voltage Amplifier Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage. How to make it non- dependent? Rin >> Rs and Rout<< RL, Av A RL Rout RL Rin v 1vs RS Rin V Rin RL Av o A Vs RS Rin Rout RL v o Av1

Conclusion: full gain depends on source and load parameters. NJIT ECE 271 Dr. Serhiy Levkov Ideal voltage amplifier: Rout = 0, Rin For current: Vo V R R R R I R Ai o L o S in Av S in Vs RL RL Vs I1 RS Rin Topic 6 - 22 Other Amplifier Types Voltage amplifier Ri , Ro 0 v

Avo vo (V/V) Ri 0, Ro i i 0 o NJIT ECE 271 Dr. Serhiy Levkov i Ais o ii v (A/A) o 0 Transresistance amplifier Transconductance amplifier Ri , Ro Current amplifier

i Gm vo i v 0 o (A/V) Ri 0, Ro 0 v Rm o ii i o (V/A) 0 Topic 6 - 23 Other Amplifier Types Voltage amplifier Ri , Ro 0 v Avo vo

(V/V) Ri 0, Ro i i 0 o NJIT ECE 271 Dr. Serhiy Levkov i Ais o ii v (A/A) o 0 Transresistance amplifier Transconductance amplifier Ri , Ro Current amplifier i

Gm vo i v 0 o (A/V) Ri 0, Ro 0 v Rm o ii i o (V/A) 0 Topic 6 - 24 Other Amplifier Types Voltage amplifier Ri , Ro 0 v Avo vo

(V/V) Ri 0, Ro i i 0 o NJIT ECE 271 Dr. Serhiy Levkov i Ais o ii v (A/A) o 0 Transresistance amplifier Transconductance amplifier Ri , Ro Current amplifier i Gm vo

i v 0 o (A/V) Ri 0, Ro 0 v Rm o ii i o (V/A) 0 Topic 6 - 25 Other Amplifier Types Voltage amplifier Ri , Ro 0 v Avo vo (V/V)

Ri 0, Ro i i 0 o NJIT ECE 271 Dr. Serhiy Levkov i Ais o ii v (A/A) o 0 Transresistance amplifier Transconductance amplifier Ri , Ro Current amplifier i Gm vo

i v 0 o (A/V) Ri 0, Ro 0 v Rm o ii i o (V/A) 0 Topic 6 - 26 Other Amplifier Types Voltage amplifier Ri , Ro 0 v Avo vo (V/V)

Ri 0, Ro i i 0 o NJIT ECE 271 Dr. Serhiy Levkov i Ais o ii v (A/A) o 0 Transresistance amplifier Transconductance amplifier Ri , Ro Current amplifier i Gm vo i v 0

o (A/V) Ri 0, Ro 0 v Rm o ii i o (V/A) 0 Topic 6 - 27 Operational Amplifier (Op Amp) Op amp is a fundamental building block in electronic design. They are cheap mass produced IC and can be used to build many different types of electronic devices like instrumentation amplifiers, active filters, rectifiers, A/D and D/A converters, and others. Typical integrated circuit amplifier: LM386N (~ \$1.00 each) The practical op amp is a form of differential amplifier: responds to a difference of two input signals. Typically: VCC >0, VEE <0 so the voltages

are symmetric 5V, 10V, 18V. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 28 Differential Amplifier Signal Amplification The typical voltage transfer characteristic for a differential amplifier biased by two symmetric power supplies: A - open-circuit voltage gain, A =10, Adb =20log(10) = 20 db vID = (v+-v--) - differential input signal vID = VID + vid = 1 + 0.25sin t, where VID dc value, vid signal component vO = VO + vo = 5 + 2.5sin t NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 29 Distortion in Amplifiers If the ac input signal exceeds 0.5V, the output signal will be clipped off (see vO2 ) and get distorted. Different gains for positive and negative values of input also cause distortion in output. Total Harmonic Distortion (THD) is a measure of signal distortion that compares

undesired harmonic content of a signal to the desired component. v(t)VO V1(sinwot 1)V2(sin2wot 2 )V3(sin3wot 3)... dc desired output 2nd harmonic distortion THD 100% 3rd harmonic distortion NJIT ECE 271 Dr. Serhiy Levkov 2 Vi i 2 V1 Topic 6 - 30

Differential Amplifier Model: Basic The basic model of differential amplifier is represented by: A - open-circuit voltage gain vid = (v+-v-) = differential input signal voltage Rid - amplifier input resistance Ro - amplifier output resistance Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 180o out of phase with that applied at - input (inverting) terminal. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 31 Differential Amplifier Model: With Source and Load RL = load resistance RS = Thevenin equivalent resistance of signal source vs = Thevenin equivalent voltage of signal source RL v Av o id Ro RL

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 32 Differential Amplifier Model: With Source and Load RL = load resistance RS = Thevenin equivalent resistance of signal source vs = Thevenin equivalent voltage of signal source RL v Av o id Ro RL NJIT ECE 271 Dr. Serhiy Levkov and R id v v id s R R id s Topic 6 - 33

Differential Amplifier Model: With Source and Load RL = load resistance RS = Thevenin equivalent resistance of signal source vs = Thevenin equivalent voltage of signal source RL v Av o id Ro RL and R id v v id s R R id s v R R o id

L A A v v R R R R s id S o L NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 34 Differential Amplifier Model: With Source and Load (Example) Problem: Calculate voltage gain Given Data: A=100, Rid =100kW, , Ro = 100W, , RS =10kW, , RL =1000W, v R R id L A o A v v R R R R

s id S o L 100kW, 1000W, 100 82.6 38.3dB 10k W, 100k W, 100 W, 1000 W, NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 35

Differential Amplifier Model: With Source and Load (Example) Problem: Calculate voltage gain Given Data: A=100, Rid =100kW, , Ro = 100W, , RS =10kW, , RL =1000W, v R R id L A o A v v R R R R s id S o L 100kW, 1000W, 100 82.6 38.3dB 10k

W, 100k W, 100 W, 1000 W, Ideal amplifiers output depends only on input voltage difference and not on source and load resistances. This can be achieved by using resistance condition (Rid >> RS or infinite Rid and Ro << RL or zero Ro ): v A o A v v id A - open-loop gain (maximum voltage gain available from the device) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 36

Ideal Operational Amplifier Ideal op amp is a special case of ideal differential amplifier with: A - infinite gain, Rid - infinite Rid , Ro 0 NJIT ECE 271 Dr. Serhiy Levkov - zero Ro . Topic 6 - 37 Ideal Operational Amplifier Ideal op amp is a special case of ideal differential amplifier with: A - infinite gain, v Rid - infinite Rid , v o, lim v 0 Ro 0 - zero Ro .

id A A id Thus: 1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite). 2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite). NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 38 Ideal Operational Amplifier Ideal op amp is a special case of ideal differential amplifier with: A - infinite gain, v Rid - infinite Rid , v o, lim v 0 Ro 0

- zero Ro . id A A id Thus: 1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite). 2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite). Additional properties: infinite common-mode rejection (rejects input signals common to both inputs) infinite open-loop bandwidth (frequency magnitude response is flat everywhere with zero phase shift) infinite output voltage range ( in practice limited by VS+ and VS- )

infinite output current capability infinite slew rate (rate of change of output voltage) zero input-offset voltage (when the input terminals are shorted, vout = 0) The major benefit of ideal op-amp: the same standard op-amp allows to build various circuits with very specific properties by using external to op-amp circuit elements. The classic op-amps circuits are considered further. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 39 Inverting Amplifier: Configuration Non-inverting input is grounded. Feedback network (resistor R2 ) is connected between inverting input and output Input network (resistor R1 ) is connected between inverting input and signal source. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 40 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 41 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 42 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is NJIT ECE 271 Dr. Serhiy Levkov since i 0 by 2 Topic 6 - 43 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is vs v vs iS

R1 R1 NJIT ECE 271 Dr. Serhiy Levkov since i 0 by 2 Topic 6 - 44 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is since i 0 by 2 vs v vs since v 0 and vid 0 by 1 iS R1 R1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 45 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is since i 0 by 2 vs v vs

since v 0 and vid 0 by 1 iS R1 R1 Substituting iS and i2 into KVL: vo R2 R vS vS ( R1 R2 ) vO 0 vS R1 vS (R1 R2 ) vO R1 0 vS vO or A 2 v v R1 R1 R1 s NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 46 Inverting Amplifier:Voltage Gain KVL: - vS is R1 i2 R2 vO 0 KCL: is i2 i 0 i2 is since i 0 by 2 vs v vs since v 0 and vid 0 by 1 iS

R1 R1 Substituting iS and i2 into KVL: vo R2 R vS vS ( R1 R2 ) vO 0 vS R1 vS (R1 R2 ) vO R1 0 vS vO or A 2 v v R1 R1 R1 s Conclusions: The gain of the inverting op-amp circuit is determined by R1 and R2 . The gain is a result of imposing the negative feedback. Gain is greater than 1 if R2 > R1. Gain is less than 1 if R1 > R2 Negative voltage gain implies 1800 phase shift between input and output signals. Inverting input of op amp is at ground potential (not connected directly to ground) and is said to be at virtual ground. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 47 Inverting Amplifier: Input and Output Resistances

vs Rin Rin R1 is NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 48 Inverting Amplifier: Input and Output Resistances vs Rin Rin R1 is Rout is found by applying a test current (or voltage) source to amplifier output, turning off all independent sources and determining the voltage (or current): NJIT ECE 271 Dr. Serhiy Levkov KVL: vx i2 R2 i1R1 0 vx i1(R2 R1) But i1=i2 (i- =0) Since v- = 0, i1=0 and vx = 0 irrespective of the value of ix ,

Rout vx 0 i x Topic 6 - 49 Inverting Amplifier: Example Problem: Design an inverting amplifier with Av=40 dB, Rin =20kW, , Assumptions: Ideal op amp Convert the gain from db: Av[db] 20log A 40 A 1040/ 20 100 v v Input resistance is controlled by R1 and voltage gain is set by R2 / R1. Thus and R1 Rin 20kW, R A 2 R2 100R1 2M W, v R 1

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 50 Summing Amplifier Since i-=0, i3= i1 + i2, vid Since negative amplifier input is at virtual ground (vid = 0): v i1 1 R1 NJIT ECE 271 Dr. Serhiy Levkov v i2 2 R2 vO i3 R3 vo R3 R v1 3 v2

R1 R2 Scale factors for the 2 inputs can be independently adjusted by proper choice of R2 and R1. Any number of inputs can be connected to summing junction through extra resistors. This is an example of a simple digital-to-analog converter. Topic 6 - 51 Non-inverting Amplifier: Configuration Input signal is applied to the non-inverting input terminal. Portion of the output signal is fed back to the negative input terminal. Analysis is done by relating voltage at v1 to input voltage vs and output voltage vo . NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 52 Non-inverting Amplifier: Analysis R Since i-=0, v1 vo R 1R (voltage divider)

i- + 1 2 vid i+ NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 53 Non-inverting Amplifier: Analysis R Since i-=0, v1 vo R 1R i- + 1 vid

2 From KVL vS vid v1 0 i+ NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 54 Non-inverting Amplifier: Analysis R Since i-=0, v1 vo R 1R i- + 1 vid i+ NJIT ECE 271 Dr. Serhiy Levkov 2 From KVL vS vid v1 0 , since vid =0

vs v1 and vo vs R1 R2 R1 Topic 6 - 55 Non-inverting Amplifier: Analysis R Since i-=0, v1 vo R 1R i- + 1 vid i+ 2

From KVL vS vid v1 0 , since vid =0 vs v1 and vo vs R1 R2 R1 vo R R R A 1 2 1 2 . v v R1 R1 s NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 56 Non-inverting Amplifier: Analysis R Since i-=0, v1 vo R 1R

i- + 1 vid From KVL vS vid v1 0 , since vid =0 i+ Since i+=0, NJIT ECE 271 Dr. Serhiy Levkov 2 vs v1 vs Rin . i and vo vs R1 R2

R1 vo R R R A 1 2 1 2 . v v R1 R1 s Topic 6 - 57 Non-inverting Amplifier: Analysis vo R R 1 2 1 R2 A R R1 R1 Since i-=0, v1 vo R 1R v vs 1 2 i-

+ vid From KVL vS vid v1 0 , since vid =0 i+ Since i+=0, vs v1 vs Rin . i and vo vs R1 R2 R1 vo R R R A 1 2 1 2 .

v v R1 R1 s Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 58 Non-inverting Amplifier: Analysis vo R R 1 2 1 R2 A R R1 R1 Since i-=0, v1 vo R 1R v vs 1 2 i-

+ vid From KVL vS vid v1 0 , since vid =0 i+ Since i+=0, vs v1 and vo vs R1 R2 R1 vo R R R A 1 2 1 2 . v v R1 R1 s

vs Rin . i Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0. R A 1 2 , Rin , Rout 0 v R 1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 59 Non-inverting Amplifier: Example i- + vid i+ io

Problem: Determine the characteristics of given non-inverting amplifier. Given Data: R1= 3kW, , R2 =43kW, , vs=+0.1 V Assumptions: Ideal op amp. Analysis: R 43kW, A 1 2 1 15.3 v R 3kW, 1 vo A vs (15.3)(0.1V ) 1.53V v Since i-=0, v 1.53V iO O 33.3 A R2 R1 43k W, 3k W, NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 60 Voltage Follower (Unity-gain Buffer) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 61 Voltage Follower (Unity-gain Buffer) A special case of non-inverting amplifier, also called voltage follower with infinite R1 and zero R2. Hence Av 1 R2 / R1 1 Provides excellent impedance-level transformation while maintaining signal voltage level. Ideal voltage buffer does not require any input current and can drive any desired load resistance without loss of signal voltage. Unity-gain buffer is used in many sensor and data acquisition systems. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 62 Difference Amplifier A combination of inverting and non-inverting amplifier. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 63

Difference Amplifier A combination of inverting and non-inverting amplifier. vo v- i2 R2 v- i1R2 (since i 0) R R (v1 v- ) R v- R2 1 2 v- 2 v1 R1 R1 R1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 64 Difference Amplifier A combination of inverting and non-inverting amplifier. vo v- i2 R2 v- i1R2 (since i 0) R R (v1 v- ) R v- R2 1 2 v- 2 v1

R1 R1 R1 R Also, v 2 v (voltage division) R1 R2 2 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 65 Difference Amplifier A combination of inverting and non-inverting amplifier. Since v-= v+, R vo 2 (v v ) R 1 2 1 For R2= R1, vo (v v ) 1 2 vo v- i2 R2 v- i1R2 (since i 0) R R (v1 v- )

R v- R2 1 2 v- 2 v1 R1 R1 R1 R Also, v 2 v (voltage division) R1 R2 2 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 66 Difference Amplifier A combination of inverting and non-inverting amplifier. Since v-= v+, R vo 2 (v v ) R 1 2 1 For R2= R1,

vo (v v ) 1 2 The circuit amplifies difference between input signals (differential subtractor). vo v- i2 R2 v- i1R2 (since i 0) R R (v1 v- ) R v- R2 1 2 v- 2 v1 R1 R1 R1 R Also, v 2 v (voltage division) R1 R2 2 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 67 Difference Amplifier A combination of inverting and non-inverting amplifier.

vo v- i2 R2 v- i1R2 (since i 0) R R (v1 v- ) R v- R2 1 2 v- 2 v1 R1 R1 R1 R Also, v 2 v (voltage division) R1 R2 2 NJIT ECE 271 Dr. Serhiy Levkov Since v-= v+, R vo 2 (v v ) R 1 2 1 For R2= R1, vo (v v ) 1 2

The circuit amplifies difference between input signals (differential subtractor). To understand better, use superposition: For v2=0, Rin1= R1, the circuit reduces to an inverting amplifier with the gain R2 /R1 : vO1= ( R2 /R1 ) v1 For v1=0, v2 is attenuated by the voltage divider and then amplified by non-inverting gain 1+R2 /R1: vO2= (R2 /(R1 +R1 ))(1+ R2 /R1 ) v2 Then vO = vO1 + vO2, R vo 2 (v v ) R1 1 2 Topic 6 - 68 Difference Amplifier: Example Do example on the board NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 69 Difference Amplifier: Example Problem: Determine Vo, V+, V-, Io, I1, I2, I3 .

Given Data: R1= 10kW, , R2 =100kW, , V1= 5V, V2=3V Assumptions: Ideal op amp. Hence, V-= V+ and I-= I+= 0. Analysis: R2 100k W, V V V 3V 2.73V - R R 2 10 k W, 100 k W, 1 2 Io I 227 A

2 V V 5V -2.73V I I 1 - 227 A 1 2 R1 10k W, V2 0 3V I 27 A 3 R R 110k W, 1 2 Vo V I R I R 5V (227 A)(110k W, ) 20.0V 1 11 2 2 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 70 Integrator NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 71

Integrator Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. dvo vs i C is c dt R NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 72 Integrator Since ic= is , dvo 1 vs RC

dt Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. dvo vs i C is c dt R NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 73 Integrator Since ic= is , dvo 1 vs RC

dt 1 dv vsd o RC t vo (t ) 1 vs ( )d vo (0) RC 0 vo (0) V (0) c Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. dvo vs i C is c dt R

NJIT ECE 271 Dr. Serhiy Levkov Voltage at the circuits output at time t is given by the initial capacitor voltage plus integral of the input signal from start of integration interval, here, t=0. Topic 6 - 74 Integrator Since ic= is , dvo 1 vs RC dt 1 dv vsd o RC t vo (t ) 1 vs ( )d vo (0)

RC 0 vo (0) V (0) c Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. dvo vs i C is c dt R NJIT ECE 271 Dr. Serhiy Levkov Voltage at the circuits output at time t is given by the initial capacitor voltage plus integral of the input signal from start of integration interval, here, t=0. Integration of an input step signal results in a ramp at the output. Topic 6 - 75

Differentiator NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 76 Differentiator Input resistor R1 in the inverting amplifier is replaced by capacitor C. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 77 Differentiator v i o R R is C dvs dt Since iR= is vo RC

dvs dt Input resistor R1 in the inverting amplifier is replaced by capacitor C. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 78 Differentiator v i o R R is C dvs dt Since iR= is vo RC Input resistor R1 in the inverting amplifier is replaced by capacitor C. NJIT ECE 271 Dr. Serhiy Levkov

dvs dt Output is scaled version of derivative of input voltage. Topic 6 - 79 Differentiator v i o R R is C dvs dt Since iR= is vo RC Input resistor R1 in the inverting amplifier is replaced by capacitor C. dvs dt Output is scaled version of

derivative of input voltage. Derivative operation emphasizes high-frequency components of input signal, hence is less often used than the integrator. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 80 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 81 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators

Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 82 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) Op-amps applications (ch. 12)

instrumentation amplifiers active filters D/A and A/D converters Oscillators Nonlinear circuits (precision rectifiers) Circuits with positive feedback (comparators, triggers) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 83 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) Op-amps applications (ch. 12)

instrumentation amplifiers active filters D/A and A/D converters Oscillators Nonlinear circuits (precision rectifiers) Circuits with positive feedback (comparators, triggers) Single transistor amplifiers linear amplification and small signal model (ch. 13) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 84 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators

Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) Op-amps applications (ch. 12) instrumentation amplifiers active filters D/A and A/D converters Oscillators Nonlinear circuits (precision rectifiers) Circuits with positive feedback (comparators, triggers) Single transistor amplifiers linear amplification and small signal model (ch. 13) Single transistor amplifiers detailed design (ch. 14) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 85 Op-Amps Whats Next

Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) Op-amps applications (ch. 12) instrumentation amplifiers active filters D/A and A/D converters Oscillators Nonlinear circuits (precision rectifiers)

Circuits with positive feedback (comparators, triggers) Single transistor amplifiers linear amplification and small signal model (ch. 13) Single transistor amplifiers detailed design (ch. 14) Op-amps design (multistage amplifiers) (ch. 15) What do we study in this course? NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 86 Op-Amps Whats Next Frequency dependent feedback (cont. ch 10): high, low and band pass amplifiers (filters) integrators and differentiators Analysis of non-ideal amplifiers (ch. 11) Amplifier frequency response and stability (ch. 11) Op-amps applications (ch. 12)

instrumentation amplifiers active filters D/A and A/D converters Oscillators Nonlinear circuits (precision rectifiers) Circuits with positive feedback (comparators, triggers) Single transistor amplifiers linear amplification and small signal model (ch. 13) Single transistor amplifiers detailed design (ch. 14) Op-amps design (multistage amplifiers) (ch. 15) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 87 Transistors as Amplifiers NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 88 Transistors as Amplifiers

In this section we will study The general techniques for employing individual transistors as amplifiers NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 89 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 90 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 91 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers

Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 92 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) BJT can be used as an amplifier when biased in the forward-active region (FAR) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 93 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier

What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) BJT can be used as an amplifier when biased in the forward-active region (FAR) We will refer to the FAR and SR as to the active region NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 94 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) BJT can be used as an amplifier when biased in the forward-active region (FAR) We will refer to the FAR and SR as to the active region In these regions, transistors can provide high voltage, current and power gains NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 95 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers

Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) BJT can be used as an amplifier when biased in the forward-active region (FAR) We will refer to the FAR and SR as to the active region In these regions, transistors can provide high voltage, current and power gains Bias is provided to stabilize the operating point in a desired operation region NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 96 Transistors as Amplifiers In this section we will study The general techniques for employing individual transistors as amplifiers Operation of common source MOSFET amplifier Operation of common emitter BJT amplifier What do we know: MOSFET (FET) can be used as amplifier if operated in saturation region (SR) BJT can be used as an amplifier when biased in the forward-active region (FAR) We will refer to the FAR and SR as to the active region In these regions, transistors can provide high voltage, current and power gains Bias is provided to stabilize the operating point in a desired operation region The Q-point also determines

Small-signal parameters of transistor Voltage gain, input resistance, output resistance Maximum input and output signal amplitudes Power consumption NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 97 BJT Amplifier Consider a BJT (I s 10 15, F 100, VT 0.025) , biased in active region by dc voltage source VBE = 0.7V. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 98 BJT Amplifier Consider a BJT (I s 10 15, F 100, VT 0.025) , biased in active region by dc voltage source VBE = 0.7V. From the simplified FAR model we get: VBE VT

I IB S e F 15A1.5mA103.3103.31.55A NJIT ECE 271 Dr. Serhiy Levkov IC F I 1.5mA VCE 10 3.3K W, IC 10 3.31.5 5V B Topic 6 - 99 BJT Amplifier Consider a BJT (I s 10 15, F 100, VT 0.025) , biased in active region by dc voltage source VBE = 0.7V. From the simplified FAR model we get: VBE VT I IB S e F 15A1.5mA103.3103.31.55A NJIT ECE 271 Dr. Serhiy Levkov

IC F I 1.5mA VCE 10 3.3K W, IC 10 3.31.5 5V B Topic 6 - 100 BJT Amplifier Consider a BJT (I s 10 15, F 100, VT 0.025) , biased in active region by dc voltage source VBE = 0.7V. From the simplified FAR model we get: VBE VT I IB S e F 15A1.5mA103.3103.31.55A NJIT ECE 271 Dr. Serhiy Levkov IC F I 1.5mA VCE 10 3.3K W, IC 10 3.31.5 5V B Topic 6 - 101 BJT Amplifier Consider a BJT (I s 10 15, F 100, VT 0.025) , biased in active region by dc

voltage source VBE = 0.7V. From the simplified FAR model we get: VBE VT I IB S e F 15A1.5mA103.3103.31.55A IC F I 1.5mA VCE 10 3.3K W, IC 10 3.31.5 5V B Thus the Q-point is set at (IC, VCE) = (1.5 mA, 5 V) NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 102 BJT Amplifier Now lets inject the sin signal with the amplitude 8 mV by adding the voltage source vbe The total base-emitter voltage becomes: vBE VBE vbe NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 103 BJT Amplifier Now lets inject the sin signal with the amplitude 8 mV by adding the voltage source vbe The total base-emitter voltage becomes: vBE VBE vbe From the FAR model for VBEh = 0.708V and VBEl = 0.692V we will get I Bh 20A1.5mA103.3103.31.55A and I Bl 10A1.5mA103.3103.31.55A NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 104 BJT Amplifier Now lets inject the sin signal with the amplitude 8 mV by adding the voltage source vbe The total base-emitter voltage becomes: vBE VBE vbe From the FAR model for VBEh = 0.708V and VBEl = 0.692V we will get I Bh 20A1.5mA103.3103.31.55A and I Bl 10A1.5mA103.3103.31.55A Thus 8 mV peak change in vBE : 5 A change in iB and 0.5 mA change in iC. NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 105 BJT Amplifier Now lets inject the sin signal with the amplitude 8 mV by adding the voltage source vbe The total base-emitter voltage becomes: vBE VBE vbe From the FAR model for VBEh = 0.708V and VBEl = 0.692V we will get I Bh 20A1.5mA103.3103.31.55A and I Bl 10A1.5mA103.3103.31.55A Thus 8 mV peak change in vBE : 5 A change in iB and 0.5 mA change in iC. From the load line equation vCE 10 iC 3300 this will produce the 1.65V change in vCE . NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 106 BJT Amplifier If changes in operating currents and voltages are small enough, then iC and vCE waveforms are undistorted replicas of input signal. Small voltage change at base causes large voltage change at collector. Voltage gain is given by: V 1.65180 Av ce 206180 206 V

0.0080 be Minus sign indicates 1800 phase shift between input and output signals. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 107 MOSFET Amplifier Consider MOSFET biased in active region by dc voltage source VGS = 3.5 V, which Will set the Q-point at (ID, VDS) = (1.56 mA, 4.8 V). Then we introduce the signal vGS and total gate-source voltage becomes vGS VGS vgs Doing similar analisis as for BJT, we get : 1 V change in vGS 1.25 mA change in iD 4 V change in vDS. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 108 Coupling and Bypass Capacitors The constant voltage biasing is not a good method, 4 resistor is better.

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 109 Coupling and Bypass Capacitors The constant voltage biasing is not a good method, 4 resistor is better. We need to introduce the input signal without disturbing the bias point. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 110 Coupling and Bypass Capacitors The constant voltage biasing is not a good method, 4 resistor is better. We need to introduce the input signal without disturbing the bias point. AC coupling through capacitors is used to inject ac input signal and extract output

signal Capacitors provide negligible impedance at frequencies of interest and provide open circuits at dc. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 111 Coupling and Bypass Capacitors The constant voltage biasing is not a good method, 4 resistor is better. We need to introduce the input signal without disturbing the bias point. AC coupling through capacitors is used to inject ac input signal and extract output signal Capacitors provide negligible impedance at frequencies of interest and provide open circuits at dc. C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible. NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 112 Coupling and Bypass Capacitors The constant voltage biasing is not a good method, 4 resistor is better. We need to introduce the input signal without disturbing the bias point. AC coupling through capacitors is used to inject ac input signal and extract output signal Capacitors provide negligible impedance at frequencies of interest and provide open circuits at dc. C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible. C2 is a bypass capacitor that provides a low impedance path for ac current from emitter to ground thereby removing RE from the circuit when ac signals are considered. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 113 DC and AC Analysis DC analysis: Make dc equivalent circuit by replacing all capacitors by open circuits

and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 114 DC and AC Analysis DC analysis: Make dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. AC analysis: Make ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 115 DC and AC Analysis DC analysis: Make dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. AC analysis: Make ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections

and dc current sources by open circuits. Replace transistor by small-signal model NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 116 DC and AC Analysis DC analysis: Make dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. AC analysis: Make ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. Replace transistor by small-signal model Use small-signal ac equivalent to analyze ac characteristics of amplifier. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 117 DC and AC Analysis DC analysis: Make dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. AC analysis:

Make ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. Replace transistor by small-signal model Use small-signal ac equivalent to analyze ac characteristics of amplifier. Combine end results of dc and ac analysis to yield total voltages and currents in the network. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 118 DC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make dc equivalent circuit by replacing all ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 119 DC Equivalent for MOSFET Amplifier

Full circuit dc equivalent Make dc equivalent circuit by replacing all - capacitors by open circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 120 DC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make dc equivalent circuit by replacing all - capacitors by open circuits - inductors by short circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov

Simplified ac equivalent Topic 6 - 121 DC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make dc equivalent circuit by replacing all - capacitors by open circuits - inductors by short circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 122 AC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make ac equivalent circuit by replacing all

- capacitors by short circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 123 AC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 124 AC Equivalent for MOSFET Amplifier

Full circuit dc equivalent Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 125 AC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections - dc current sources by open circuits ac equivalent

NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 126 AC Equivalent for MOSFET Amplifier Full circuit dc equivalent Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections - dc current sources by open circuits ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 127 AC Equivalent for MOSFET Amplifier Full circuit

dc equivalent Simplify the ac equivalent circuit ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 128 DC and AC Equivalents for MOSFET Amplifier Full circuit dc equivalent ac equivalent NJIT ECE 271 Dr. Serhiy Levkov Simplified ac equivalent Topic 6 - 129 DC Equivalent for BJT Amplifier All capacitors in original amplifier circuits need to be replaced by

open circuits, disconnecting vI , RI , and R3 from circuit. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 130 DC Equivalent for BJT Amplifier All capacitors in original amplifier circuits need to be replaced by open circuits, disconnecting vI , RI , and R3 from circuit. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 131 DC Equivalent for BJT Amplifier All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vI , RI , and R3 from circuit. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 132 AC Equivalent for BJT Amplifier Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits

- dc voltage sources by ground connections - dc current sources by open circuits NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 133 AC Equivalent for BJT Amplifier Simplify the ac equivalent circuit NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 134 AC Equivalent for BJT Amplifier Simplify the ac equivalent circuit NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 135 AC Equivalent for BJT Amplifier RB R1 R2 160kW, 300kW, RL RC R3 22kW, 100kW, NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 136 Small-Signal Modeling For AC analysis (or general time varying analysis) we would need to use phasor , time domain, Laplace Transform (or similar) methods. Those are working better with linear systems. Need to make the linear model small signal model. Assume that the time varying components are small signals and construct the two port model, which is linear. The concept of small signal is device dependent. We start from developing small signal model for a diode and then switch to transistor NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 137 Small-Signal Operation of Diode The slope of the diode characteristic at the Qpoint is called the diode conductance and is given by: ID and VD represent the DC bias point Q-point, vD and iD are small changes around Q-point NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 138 Small-Signal Operation of Diode The slope of the diode characteristic at the Qpoint is called the diode conductance and is given by: i gd D vD v v I I S exp D 1 S exp D V V vD VT i I I v S exp D 11 D S VT VT VT Qp

T T i i gd D D 40I D for iD I S VT 0.025V gd is small but non-zero for ID = 0 because slope of diode equation is nonzero at the origin. ID and VD represent the DC bias point Q-point, vD and iD are small changes around Q-point NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 139 Small-Signal Operation of Diode

The slope of the diode characteristic at the Qpoint is called the diode conductance and is given by: i gd D vD v v I I S exp D 1 S exp D V V vD VT i I I v S exp D 11 D S VT VT VT Qp T

T i i gd D D 40I D for iD I S VT 0.025V gd is small but non-zero for ID = 0 because slope of diode equation is nonzero at the origin. Diode resistance is given by: r 1 d gd ID and VD represent the DC bias point Q-point, vD and iD are small changes around Q-point NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 140 Small-Signal Operation of Diode

The slope of the diode characteristic at the Qpoint is called the diode conductance and is given by: i gd D vD v v I I S exp D 1 S exp D V V vD VT i I I v S exp D 11 D S VT VT VT Qp

T T i i gd D D 40I D for iD I S VT 0.025V gd is small but non-zero for ID = 0 because slope of diode equation is nonzero at the origin. 1 Diode resistance is given by: rd Thus, linear model in the vicinity of ID : ID and VD represent the DC bias point Q-point, vD and iD are small changes around Q-point

NJIT ECE 271 Dr. Serhiy Levkov gd iD I D id , id gd vd small signal model when vd 2VT 0.05V or vd 5 mV Topic 6 - 141 Small-Signal Model of BJT BJT is a three terminal device and to build similar model to a diode, we would need to use 2-port y-parameter network. i y21 c v be Using 2-port y-parameter network, ib y11vbe y12vce ic y vbe y22vce v 0

be v 0 ce ic 21 The port variables can represent either time part of total voltages and currents or varying small changes in them away from Q-point values. NJIT ECE 271 Dr. Serhiy Levkov i y12 b v ce y22 v ce i C v BE

0 Q point Q point iC v 0 be ib y11 v be i B vCE vCE Q point v 0 ce IC

VA VCE IC i B v BE I C VT Q point oVT Topic 6 - 142 Small-Signal Model of BJT BJT is a three terminal device and to build similar model to a diode, we would need to use 2-port y-parameter network. i y21 c v be

Using 2-port y-parameter network, ib y11vbe y12vce ic y vbe y22vce v 0 be v 0 ce ic 21 The port variables can represent either time part of total voltages and currents or varying small changes in them away from Q-point values. NJIT ECE 271 Dr. Serhiy Levkov i y12 b v ce

y22 v ce i C v BE 0 Q point Q point iC v 0 be ib y11 v be i B vCE vCE

Q point v 0 ce IC VA VCE IC i B v BE I C VT Q point oVT Topic 6 - 143 Small Signal Hybrid-Pi Model of BJT

The hybrid-pi small-signal model is most widely accepted model for BJT amplifier. Small-signal parameters are controlled by the Q-point NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 144 Small Signal Hybrid-Pi Model of BJT The hybrid-pi small-signal model is most widely accepted model for BJT amplifier. Small-signal parameters are controlled by the Q-point VA - Early Voltage Trans-conductance: I gm 1 C 40IC y21 VT Input resistance: V

r 1 o o T y11 g m IC Output resistance: V 1 V V ro A CE A y22 IC IC NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 145 Small Signal Hybrid-Pi Model of BJT The hybrid-pi small-signal model is most widely accepted model for BJT amplifier. Small-signal parameters are controlled by the Q-point VA - Early Voltage o is the small-signal commonemitter current gain of the BJT. Trans-conductance:

I gm iC |Q C 40IC vBE VT Input resistance: V r o o T gm IC Output resistance: 1 VA VCE VA ro y22 IC IC NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 146 Small Signal Hybrid-Pi Model of BJT The hybrid-pi small-signal model is most

widely accepted model for BJT amplifier. Small-signal parameters are controlled by the Q-point VA - Early Voltage Trans-conductance: I gm iC |Q C 40IC vBE VT Input resistance: o is the small-signal commonemitter current gain of the BJT. V r o o T gm IC Definition of small signal for BJT: Output resistance: 1 VA VCE VA ro y22 IC IC

vbe 2VT or vbe 0.005V NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 147 Small-Signal Current Gain and Voltage Gain of BJT The important small signal parameter is trans-conductance, or current-voltage (iC/ vBE ) gain. It shows how the collector current changes in response to baseemitter voltage. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 148 Small-Signal Current Gain and Voltage Gain of BJT The important small signal parameter is trans-conductance, or current-voltage (iC/ vBE ) gain. It shows how the collector current changes in response to baseemitter voltage. Other two important parameters are: - small signal current gain Small signal current gain (iC/iB = iC/(vBE /r) - gain) : o gmr F

C 1 F 1 I F iC Q point In practice, however, F and o are often assumed to be equal.

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 149 Small-Signal Current Gain and Voltage Gain of BJT The important small signal parameter is trans-conductance, or current-voltage (iC/ vBE ) gain. It shows how the collector current changes in response to baseemitter voltage. Other two important parameters are: - small signal current gain - intrinsic voltage gain Small signal current gain (iC/iB = iC/(vBE /r) - gain) : o gmr F C 1 F 1 I

F iC Q point In practice, however, F and o are often assumed to be equal. NJIT ECE 271 Dr. Serhiy Levkov Intrinsic voltage gain (vCE/vBE gain) : C T

I VA VCE VA VCE F g ro V I VT C For V << V , m CE A V F A 40VA VT F represents maximum voltage gain individual BJT can provide, doesnt change with operating point, and ranges from 1000 to 4000.

Small signal model for pnp-BJT is similar Topic 6 - 150 Small-Signal Model of MOSFET (nmos) ig Using 2-port y-parameter network, ig y11vgs y12vds id y21vgs y22vds The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. NJIT ECE 271 Dr. Serhiy Levkov y11 v gs i y12 g v ds i y21 d v gs

v ds v v gs ds 0 0 vGS 0 i G v DS 0 i D

vGS id y22 v ds iG Q point 0 Q point Q point iD v gs 0 v DS 2ID VGS VTN

Q point ID 1 VDS Topic 6 - 151 Small-Signal Model of MOSFET (nmos) ig Using 2-port y-parameter network, ig y11vgs y12vds id y21vgs y22vds The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. NJIT ECE 271 Dr. Serhiy Levkov y11 v gs

i y12 g v ds i y21 d v gs v ds v v gs ds 0 0 vGS 0 i

G v DS 0 i D vGS id y22 v ds iG Q point 0 Q point Q point iD v gs 0

v DS 2ID VGS VTN Q point ID 1 VDS Topic 6 - 152 Small-Signal Model of MOSFET (nmos) Trans-conductance: Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite.

Small-signal parameters are controlled by the Q-point. For same operating point, MOSFET has higher transconductance and lower output resistance that BJT. Definition of small signal for MOSFET: vgs 0.2 VGS VTN NJIT ECE 271 Dr. Serhiy Levkov 2I D gm y21 2K n I D VGS VTN Output resistance: ro 1 1/ VDS 1 y22 ID ID Amplification factor (intrinsic voltage gain) for VDS<<1: 2K n

1/ VDS f gmro 1 (VGS VTN )/2 I D Where channel-length modulation parameter. Topic 6 - 153 Small-Signal Model of JFET iG 1 y11 rg vGS i gm y21 D vGS iD I DSS

1 v gm 2 2 GS P V 1 v DS vDS vGS VP v iG ISG exp GS 1 V T I D VGS VP

Q point 2 IDSS (V V ) VP2 GS P iD 1 y22 ro v DS for Q point IG I SG 0 for IG 0 VT Q point ID 1 VDS

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 154 Small-Signal Model of JFET For small-signal operation, the input signal limit is: vgs 0.2VGS VP The amplification factor is given by: Since JFET is normally operated with gate junction reversebiased, 1 VDS f gmro 2 2 VGS VP VP I ISG rg G

IDSS ID NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 155 Summary of FET and BJT Small-Signal Models NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 156 Common-Emitter (C-E) Amplifier Construct AC equivalent circuit assuming Q-point is known. Input is applied to Base Output appears at Collector Emitter is common (through RE) to both input and output signal Common-Emitter (CE) Amplifier. Developing a small signal model:

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 157 Common-Emitter (C-E) Amplifier Construct AC equivalent circuit assuming Q-point is known. NJIT ECE 271 Dr. Serhiy Levkov Use the small signal hybrid model for BJT Topic 6 - 158 Common-Emitter (C-E) Amplifier Construct AC equivalent circuit assuming Q-point is known. Simplify: RL ro || RC || R3 NJIT ECE 271 Dr. Serhiy Levkov

Use the small signal hybrid model for BJT Topic 6 - 159 Common-Emitter (C-E) Amplifier Construct AC equivalent circuit assuming Q-point is known. Simplify: RL ro || RC || R3 NJIT ECE 271 Dr. Serhiy Levkov Use the small signal hybrid model for BJT Topic 6 - 160 C-E Amplifier Voltage Gain

Goal develop the overall voltage gain from vi to vo - AvCE . NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 161 C-E Amplifier Voltage Gain Goal develop the overall voltage gain from vi to vo - AvCE . v v v CE o A o b v v v v i b i v ACE o v where vt

b v CE A b vt v i - terminal gain - gain between transistor terminals. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 162 C-E Amplifier Voltage Gain

Goal develop the overall voltage gain from vi to vo - AvCE . v v v CE o A o b v v v v i b i v ACE o v where vt b v CE A b vt v

i - terminal gain - gain between transistor terminals. To find AvtCE , we connect a test source vb to the base terminal (right circuit): v g R v or ACE g R o m L b vt m L NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 163 C-E Amplifier Voltage Gain

Goal develop the overall voltage gain from vi to vo - AvCE . v v v CE o A o b v v v v i b i v ACE o v where vt b v CE A b vt v

i For the input resistance RiB : RiB vb r ib - terminal gain - gain between transistor terminals. To find AvtCE , we connect a test source vb to the base terminal (right circuit): v g R v or ACE g R o m L b vt m L

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 164 C-E Amplifier Voltage Gain Goal develop the overall voltage gain from vi to vo - AvCE . v v v CE o A o b v v v v i b i v ACE o v where vt b

v CE A b vt v i - terminal gain - For the input resistance RiB : RiB Now, since vb vi RB || Rib , RI RB || Rib

vb r ib we have v AvCE AvtCE b vi RB || r g m RL RI RB || r gain between transistor terminals. To find AvtCE , we connect a test source vb to the base terminal (right circuit): v g R v or ACE g R o m L b vt m L NJIT ECE 271 Dr. Serhiy Levkov

Topic 6 - 165 C-E Amplifier Voltage Gain Goal develop the overall voltage gain from vi to vo - AvCE . v v v CE o A o b v v v v i b i v ACE o v where vt b v

CE A b vt v i Now, since vb vi RB || Rib , RI RB || Rib vb r ib we have v AvCE AvtCE b vi - terminal gain -

To find AvtCE , we connect a test source vb to the base terminal (right circuit): v g R v or ACE g R o m L b vt m L NJIT ECE 271 Dr. Serhiy Levkov For the input resistance RiB : RiB gain between transistor terminals. RB || r g m RL

RI RB || r We see that the overall voltage gain is the terminal gain reduced by the voltage division between R1 and equivalent resistance at the base. Thus AvtCE is the upper limit of voltage gain. Topic 6 - 166 C-E Amplifier - Simplifications and Limits If RI RB r we obtain the max gain: AvCE gm RL gm ro RC R3 v RB || r

AvCE AvtCE b g m RL vi RI RB || r NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 167 C-E Amplifier - Simplifications and Limits we obtain the max gain: AvCE gm RL gm ro RC R3 If RI RB r For maximum gain we set R3 >> RC and RC << rO ro RC. R3 Rc. If we assume IC RC = VCC with 0 < < 1: I R Av Avt gm RC C C 40 VCC

VT v AvCE AvtCE b vi NJIT ECE 271 Dr. Serhiy Levkov RB || r g m RL RI RB || r Topic 6 - 168 C-E Amplifier - Simplifications and Limits we obtain the max gain: AvCE gm RL gm ro RC R3 If RI RB r

For maximum gain we set R3 >> RC and RC << ro assume IC RC = VCC with 0 < < 1: Av Avt gm RC ro R.C R3 Rc. If we I C RC 40 VCC VT Typically, = 1/3, since common design allocates one-third power supply across RC. To further account for other approximations leading to this result, we use: Av 10VCC NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 169 C-E Amplifier - Simplifications and Limits we obtain the max gain:

AvCE gm RL gm ro RC R3 If RI RB r For maximum gain we set R3 >> RC and RC << ro assume IC RC = VCC with 0 < < 1: Av Avt gm RC ro R.C R3 Rc. If we I C RC 40 VCC VT Typically, = 1/3, since common design allocates one-third power supply across RC. To further account for other approximations leading to this result, we use: Av 10VCC

Also, if the load resistor approaches ro (RC and R3 are very large), voltage gain is limited by amplification factor, f , of BJT itself. gm R R CE L Avt L 1g m R R E E For large R , voltage gain can be approximated as: E NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 170 C-E Amplifier Example

Problem: Find voltage gain, input and output resistances. Given data: F = 65, VA = 50 V Assumptions: Active-region operation, VBE = 0.7 V, small signal operating conditions. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 171 C-E Amplifier Example We start from finding the Q-point. DC equivalent circuit: The KVL for the input loop : 105 I B VBE (F 1)I B (1.6104 ) 5 0 I B 3.71 A IC 65I B 241 A Problem: Find voltage gain, input and output resistances. Given data: F = 65, VA = 50 V

Assumptions: Active-region operation, VBE = 0.7 V, small signal operating conditions. I E 66I B 245 A The KVL for the output loop (blue): 5 104 IC VCE (1.6104 )I E 5 0 VCE 3.67 V NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 172 C-E Amplifier Example Next we construct the ac equivalent and simplify it. Rin RB r 6.23 kW, gm 40IC 9.6410 3 S oVT r 6.64 kW, IC V A VCE ro

223 kW, IC Rout RC ro 9.57 kW, v Rin 84.0 Av o gm (Rout R3) vi RI Rin NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 173 Common-Source Amplifier NJIT ECE 271 Dr. Serhiy Levkov

Construct AC equivalent circuit. Assume that Q-point is known. Topic 6 - 174 Common-Source Amplifier NJIT ECE 271 Dr. Serhiy Levkov Construct AC equivalent circuit. Assume that Q-point is known. Use the small signal hybrid model for BJT Topic 6 - 175 Common-Source Amplifier Simplify: RL ro || RC || R3

NJIT ECE 271 Dr. Serhiy Levkov Construct AC equivalent circuit. Assume that Q-point is known. Use the small signal hybrid model for MOSFET Topic 6 - 176 Common-Source Amplifier Simplify: RL ro || RC || R3 NJIT ECE 271 Dr. Serhiy Levkov Construct AC equivalent circuit. Assume that Q-point is known. Use the small signal model Topic 6 - 177

C-S Amplifier - Terminal Voltage Gain Terminal voltage gain between gate and drain: v v Avt vd v o gm RL g gs Overall voltage gain from source vi to output voltage across R3 v vo vo vgs Av v v v Avt vgs gs i i i Voltage division NJIT ECE 271 Dr. Serhiy Levkov

L Av gm R RG RI RG Topic 6 - 178 C-S Amplifier - Simplifications If we assume RI << RG Av Avt gm RL gm ro RD R3 This implies that total signal voltage at input appears across gate-source terminals.

Generally R3 >> RD and RD << ro. Hence, total load resistance on drain is RD. For this case, common design allocates half the power supply for voltage drop across RD and (VGS - VTN ) = 1V Av gm RD I D RD VDD VGS VTN 2 Also, if load resistor approaches ro, (RD and R3 are very large), voltage gain is limited by amplification factor, f of the MOSFET itself. NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 179 C-S Amplifier - Input and Output Resistance Input resistance of C-S amplifier is much larger than that of corresponding C-E amplifier.

vx ix RG Rin RG For comparable bias points, output resistances of C-S and C-E amplifiers are similar. In this case, vgs= 0. v Rout x RD ro RD ix NJIT ECE 271 Dr. Serhiy Levkov for ro RD Topic 6 - 180 C-S Amplifier - Example Do example on the board Problem: Find voltage gain, input

and output resistances. Given data: Kn = 500 A/V2, VTN = 1V, = 0.0167 V-1 NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 181 C-S Amplifier - Example Construct dc equivalent circuit Since IG 0, V I1 DS 6 , 510 KVL: Problem: Find voltage gain, input and output resistances. Given data: Kn = 500 A/V2, VTN = 1V, = 0.0167 V-1 NJIT ECE 271 Dr. Serhiy Levkov 10 VDS 2104 (I D I1) 0

K I D n (0.4VDS VTN )2 2 VDS 5 V , VGS 2 V , I D 250 A Topic 6 - 182 C-S Amplifier - Example Next we construct the ac equivalent and simplify it. gm 2K n I DS (1 VDS ) 5.20104 S Rin RG1 RG2 1 MW, 1 VDS ro 260 kW, I D RI RG Rout ro RD RG3 18.2 k W,

Rin vo Av gm (Rout R3 ) 7.93 RI Rin vi NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 183 Summary of CE / CS Amplifiers NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 184