Microelectronics Questions and Answers - Bipolar Differential Amplifiers - Set 2

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Differential Amplifiers – Set 2”.

1. What is the small signal gain for Q₁ in the following circuit? Assume that both channel length modulation and early effect is present, only in Q₁. The MOSFETs are identical to each other and the Bipolar transistors are identical to each other.

a) –[(g_m1/1+(g_m1*((R₃+R₄+1/g_m2)||r_o1)*(r_o1+(1+g_m1*r_o1) * ((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3)))
b) –[(g_m1/1+(g_m1*(R₃+R₄+1/g_m2)*(r_o1+(1+g_m1*r_o1)||r_o1*((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3)))
c) –[(g_m1/1+(g_m1*(R₃+R₄+1/g_m2)||r_o1*(r_o1+(1+g_m1*r_o1) * ((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3)))
d) –[(g_m1/1+(g_m1*(R₃+R₄+1/g_m2)||r_o1)*(r_o1+(1+g_m1*r_o1) * ((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3)))
View Answer

Answer: a
Explanation: We note that the small signal gain for Q₁ has to be analyzed by turning off all voltage sources and current sources. We note that the emitter of Q₁ is connected to R₃ which is in series with R₄ & the resistance looking into the emitter of Q₂ in presence of early effect. Henceforth, the total resistance comes connected at the emitter which will affect the transconductance, and the output resistance. The transconductance becomes g_m1/1+(g_m1*((R₃+R₄+1/g_m2)||r_o1)) and the output resistance becomes (r_o1+(1+g_m1*r_o1)*((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3))). Noting that the voltage gain will be for a degenerated CE stage, we can write the voltage gain as –[(g_m1/1+(g_m1*(R₃+R₄+1/g_m2)||r_o1)*(r_o1+(1+g_m1*r_o1)*((R₃+R₄+1/g_m2)||r_π1)||(r_o3||1/g_m3))).

2. What is the differential voltage gain of the following circuit? Assume that both channel length modulation and early effect is present, the MOSFETs are identical to each other and the Bipolar transistors are identical to each other.

a) –[(g_m1/1+(g_m1*(R₃+r_o1)*(r_o1+(g_m1*r_o1)*(R₃+r_π1)||(r_o3||1/g_m3)))+(g_m2/1+(g_m2*(R₄||r_o2)*(r_o2+(1-g_m1*r_o1)*(R₃||r_π2)||(r_o4)))]
b) [(g_m1/1+(g_m1*(R₃+r_o1)*(r_o1+(g_m1*r_o1)*(R₃+r_π1)||(r_o3||1/g_m3)))+(g_m2/1+(g_m2*(R₄)*(r_o2+(1-g_m1*r_o1)*(R₃||r_π2)||(r_o4)))]/2
c) -[(g_m1/1+(g_m1*((R₃+R₄+(1/g_m2||r_o2))||r_o1) *(r_o1+(1+g_m1*r_o1) * ((R₃+R₄+(1/g_m2||r_o2))||r_π1)||(r_o3||1/g_m3)))+(g_m2/1+(g_m2*(R₄+(1/g_m1||r_o1)+R₃)||r_o2)*(r_o2+(1+g_m1*r_o1)*((R₄+(1/g_m1||r_o1)+R₃)||r_π2)||(r_o4)))]
d) [(g_m1/1+(g_m1*(R₃+r_o1)*(r_o1+(g_m1*r_o1)*(R₃+r_π1)||(r_o3||1/g_m3)))+(g_m2/1+(g_m2*(r_o2)*(r_o2-(1+g_m1*r_o1)*(R₃||r_π2)||(r_o4)))]/2
View Answer

Answer: c
Explanation: It is evident that the differential voltage gain is the difference between the voltage gain of each transistor. The input to each transistor is different with a phase of 180⁰. The voltage gain for Q₁ is –[(g_m1/1+(g_m1*(R₃+R₄+(1/g_m2||r_o2))||r_o1)*(r_o1+(1+g_m1*r_o1)*((R₃+R₄+(1/g_m2||r_o2))||r_π1)||(r_o3||1/g_m3)). The voltage gain for Q₂ is –(g_m2/1+(g_m2*(R₄+R₃+(1/g_m1||r_o1)||r_o2)*(r_o2+(1+g_m1*r_o1)*(R₄+R₃+(1/g_m1||r_o1)||r_π2)||(r_o4))). The differential voltage gain will be the difference between the voltage gain of \(\frac {(V_{OUT2} – V_{OUT1})}{ (V_{IN2}-V_{IN1})} = \frac { \{ V_{IN}*A_{V2} – (-V_{IN}*A_{v1}) \} }{(V_{IN}-(-V_{IN})}= \frac {(A_{V2}+A_{V1})}{2}\). Note that both the voltage gains are negative and the overall voltage gain will also have a minus sign. We have approximated 1/r_π>>g_m. Note that the transconductance doesn’t get affect by the resistance looking from the drain terminal towards the power supply since the drain resistance typically controls the drain voltage.

3. What is the transconductance of Q₁ if V_A=2V, the collector current is 1mA, R₃ = R₄ = 1kΩ and the temperature is 300K.

a) g_m1/1+(g_m1*((1/g_m2)||r_o1))
b) g_m1/1+(g_m1*((R₃+1/g_m2)||r_o1))
c) g_m1/1+(g_m1*(R₄+1/g_m2||r_o1))
d) g_m1/1+(g_m1*((R₃+R₄+1/g_m2)||r_o1))
View Answer

Answer: d
Explanation: We note that Q₁ is degenerated by R₃ and its transconductance has decreased but the linearity of the circuit will be enhanced. The new transconductance is g_m1/1+(g_m1*((R₃+R₄+1/g_m2)||r_o1)) where g_m1 is the transconductance without degeneration.

4. What is the voltage gain of M₁? Assume that early effect is absent but channel length modulation is present.

a) r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o))/(1/g_m+(r_π+(β+1)*(R₃+R₄(R₃||r_o)/(β+1)))
b) r_π+(β+1)*(R₃+R₄+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+1/g_m+(R₃||r_o)/(β+1)))
c) r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₄+1/g_m+(R₃||r_o)/(β+1)))
d) r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1)))
View Answer

Answer: d
Explanation: M₁ behaves as a source follower so we know that its voltage gain is R_s/(1/g_m+R_s). Now, R_S is the impedance looking into the base of the transistor which is r_π+(β+1)*R_E & R_E is R₃+R₄+1/g_m+R_B(Q2)/(β+1). We note that R_B is simply R₃||r_o and the overall voltage gain of M₁ becomes r_π+(β+1)*(R₃+R₄+1/g_m2+(R₃||r_o5)/(β+1))/(1/g_m+(r_π1+(β+1)*(R₃+R₄+1/g_m2+(R₃||r_o5)/(β+1)))

5. What is the voltage gain of M₅? Assume that early effect is absent but channel length modulation is present.

a) r_π+(β+1)*(R₃+R₄+1/g_m1)||R₃||r_o5
b) r_π+(β+1)*(R₃+R₄+(R₅||r_o1))||R₃||r_o5
c) r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5
d) r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||r_o5
View Answer

Answer: c
Explanation: M₅ is behaving as a CS stage. The voltage gain is R_C || r_o. R_C is connected to the base of Q₂ which has an input resistance of r_π+(β+1)*R_E. Now, R_E is R₃+R₄+1/g_m+R_B(Q1)/(β+1). We note that R_B is simply R₅||r_o which is the resistance connected to the source of the M₁. Moreover, R₃ is also connected to the drain of M₅ & the overall voltage gain becomes r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5.

6. What is the voltage gain from the gate of M₅ to the collector of Q₂? Assume channel length modulation is present but early effect is absent.

a) -(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)
b) -(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5))
c) (r_π+(β+1)*(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃)/(β+1)
d) (r_π+R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(r_o5)/(β+1)
View Answer

Answer: a
Explanation: The voltage gain is a product of the voltage gain of M₅, behaving as a CS stage, and that of Q₂, behaving as a CE stage. The voltage gain of Q₂ is R₂/(R_E+1/g_m+R_B/(β+1). Now, R_E is R₄+R₃+1/g_m1+R₅||r_o/(β+1) while R_B is R₃||r_o5. Hence, the voltage gain of only Q₂ is R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1). The voltage gain of M₅ is -(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5). The overall voltage gain is -(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1).

7. What is the voltage gain from the gate of M₁ to the collector of Q₁? Assume channel length modulation is present but early effect is absent.

a) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂*(1/g_m+(R₅||r_o5)/(β+1))
b) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1) * (R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))
c) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))
d) ((β+1)*(R₃+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))
View Answer

Answer: b
Explanation: We note that the voltage gain is a product of the voltage gain of M₁ & the voltage gain of Q₁. The voltage gain of only Q₁ is R₁/(R_E+1/g_m+R_B/(β+1). Now, R_E is R₃+R₄+1/g_m2+R₃||r_o/(β+1) while R_B is R₅||r_o5. Hence, the voltage gain of only Q₂ is (R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1)). The voltage gain of M₁ is r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+ R₄+1/g_m+(R₃||r_o)/(β+1))). The overall voltage gain is (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+ R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1)).

8. What is the total output impedance of Q₁? Assume that both channel length modulation and early effect is present, the MOSFETs are identical to each other and the Bipolar transistors are identical to each other.

a) r_o1+(1+r_o1)((R₄+R₃+(1/g_m||r_o))||r_π1)||(r_o3+1/g_m3)
b) r_o1+(1+g_m1*r_o1)(r_π1)||(r_o3+1/g_m3)
c) r_o1+(1+g_m1*r_o1)((R₄+R₃+(1/g_m||r_o))||r_π)||(r_o3+1/g_m3)
d) r_01q+(1+g_m1*r_o1)(R₃)||(r_o3+1/g_m3)
View Answer

Answer: c
Explanation: We note that Q₁ is degenerated by R₃ and hence, the output impedance looking into the collector of the NPN transistor is r_o+(1+g_m*r_o)(R₃||r_π). We also note that M₃, behaving as a diode-connected device, offers a resistance of (r_o3+1/g_m3), looking into it’s drain terminal, and the drain is connected to the collector of Q₁. We remember that these resistances are small signal values and the are all connected from the collector terminal to ground, since V_DD is grounded during small signal analysis and the current source is made open. The overall resistance becomes r_o+(1+g_m*r_o)(R₃||r_π)||(r_o3+1/g_m3).

9. What is the total output impedance of Q₂? Assume that both channel length modulation and early effect is present, the MOSFETs are identical to each other and the Bipolar transistors are identical to each other.

a) r_o2+(1+r_o2)(R₃||r_π2)||(r_o3+1/g_m3)
b) r_o2+(1+g_m2+r_o2)(r_π2)||(r_o4)
c) r_o2+(1+g_m2+r_o2)((R₄+R₃+(1/g_m||r_o))||r_π2)||(r_o4)
d) r_o2+(1+g_m2+r_o2)((R₄+R₃+(1/g_m||r_o)))(R₃)||(r_o3+1/g_m3)
View Answer

Answer: c
Explanation: We note that Q₁ is degenerated by R₃ and hence, the output impedance looking into the collector of the NPN transistor is r_o+(1+g_m*r_o)(R₃||r_π). We also note that M₃, behaving as a diode-connected device, offers a resistance of (r_o3||1/g_m3), looking into it’s drain terminal, and the drain is connected to the collector of Q₁. We remember that these resistances are small signal values and the are all connected from the collector terminal to ground, since V_DD is grounded during small signal analysis and the current source is made open. The overall resistance becomes r_o2+(1+g_m2+r_o2)(R₃||r_π2)||(r_o4).

10. What is the output at the collector of Q₁ due to V_IN1 at M₅? Note that early effect is absent but channel length modulation is present.

a) R₃*R_C/(1/g_m+(R₄+1/g_m2+R₃||r_o/(β+1)))
b. [(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+ (R₃+R₄+1/g_m2+R₃||r_o/(β+1)))
c) R_C/(1/g_m+ (R₃+R₃||r_o/(β+1)))
d. [(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+ (R₃+R₃||r_o/(β+1)))
View Answer

Answer: b
Explanation: For V_IN1 at M₁, the output at Q₁ is the gain for a CB stage. A C.B. stage shows a voltage gain of R_C/(1/g_m+R_E) & R_E is the resistance connected to the emitter of Q₁ which is R₃+R₄+1/g_m2+R₃||r_o/(β+1). The overall voltage gain becomes the product of the voltage gain of M₅ as a CS stage and R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1))) i.e. [(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+ (R₃+R₄+1/g_m2+R₃||r_o/(β+1))).

11. What is the output at the collector of Q₂ due to V_IN1 at M₁? Note that early effect is absent but channel length modulation is present.

a) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*R_C/(1/g_m+ (R₃+R₄+1/g_m2+R₃||r_o/(β+1)))
b) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*R_C/( (R₃+R₄+R₃||r_o/(β+1)))
c) (r_π+(R₃+R₄+1/g_m+1/(β+1))/(1/g_m+((R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*R_C/(1/g_m+ (R₃+R₄+1/g_m2+R₃||r_o/(β+1)))
d) (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*R_C/(1/g_m+ (R₃+R₄+1/g_m1+R₅||r_o/(β+1)))
View Answer

Answer: d
Explanation: For V_IN1 at M₁, the output at collector of Q₂ is the gain for a CB stage. A C.B. stage shows a voltage gain of R_C/(1/g_m+R_E) & R_E is the resistance connected to the emitter of Q₂ which is R₄+R₃+1/g_m1+R₅||r_o/(β+1). The overall voltage gain becomes the product of the voltage gain of M₁ as a follower stage and R_C/(1/g_m2+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1))) ie (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1))).

12. What is the output at the collector of Q₁ due to V_IN1 at both M₁ & M₂? Note that early effect is absent but channel length modulation is present. Assume that all are identical and working in the linear region of operation.

a) [(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1) * (R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1)*(R₁||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]
b) [(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1)))) *(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1) * (R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)] * R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]
c) [(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1) * (R₃+R₄+1/g_m+(R₃||r_o)/(β+1)))) *(R₂/((1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1) * (R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]
d) [((β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1)))) *(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1) * R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]
View Answer

Answer: b
Explanation: The output at the collector of Q₁ due to V_IN1 at both M₁ & M₂, is a superposition of the outputs at either MOSFET ie the output due to the voltage gain at Q₁ due to M₁ and M₂ are to be calculated separately and their addition gives the voltage gain at Q₁. The voltage gain due to M₁ is the product of the voltage gain of M₁, behaving as a follower stage, and the voltage gain of Q₁, behaving as a degenerated CE stage ie (r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1)). The voltage gain due to M₅ is [(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1))). The total voltage gain is the sum of the aforementioned gains.

13. What is the output at the collector of Q₂ due to V_IN1 at both M₁ & M₂? Note that early effect is absent but channel length modulation is present. Assume that all are identical and working in the linear region of operation.

a) [-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5) *(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m + (R₃||r_o5)/(β+1)+[r_π+(β+1) * (R₃+R₄+1/g_m + (R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1) * (R₃+R₄+1/g_m +(R₃||r_o)/(β+1))))*-(R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1)))]
b) [-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5) * (R₂/((β+1)+[r_π+(β+1) * (R₃+R₄+1/g_m+(R₃||r_o+R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1)))]
c) [-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5) * (R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))]
d. [-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5) * (R₂/((R₄+R₃+1/g_m1+r_o5)/(β+1)+[r_π+(β+1)*
(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1)))]
View Answer

Answer: a
Explanation: The output at the collector of Q₂ due to V_IN1 at both M₁ & M₂, is a superposition of the outputs at either MOSFET ie the output due to the voltage gain at Q₁ due to M₁ and M₂ are to be calculated separately and their addition gives the voltage gain at Q₁. The voltage gain due to M₂ is the product of the voltage gain of M₅, behaving as a CS stage, and the voltage gain of Q₂, behaving as a degenerated CS stage ie -(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1) . The voltage gain due to M₁ is [r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1))). The total voltage gain is [-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1)))].

14. What is the differential voltage gain of the following circuit? Assume that early voltage is absent and channel length modulation is present.

a) [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))+R₄+1/g_m+(R₃||r_o)/(β+1))))]
b) [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]]/[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]-(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))]
c) [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-(R_C/(1/g_m+(R₄+R₃+1/g_m1+R₅||r_o/(β+1)/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]]/[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]-(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))]
d) [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))+(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-(R_C/(1/g_m+(R₄+R₃+1/g_m1+R₅||r_o/(β+1))))]-[(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]]/[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]-[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))]
View Answer

Answer: d
Explanation: We start from the expression of differential voltage gain which is the ratio of the differential output and the differential input voltage. While, doing so, we remember that the differential change in the node connecting the tail current source is 0 if and only if the difference between the input to the differential pail is equal and opposite. In calculating the small signal differential voltage gain, we will have to assume that the expression really represents the small signal differential voltage gain. Now, the output of Q₁ at its collector is a superposition of the output due to V_IN1 & V_IN2 and the same goes for Q₂.Hence, we will find four different voltage gain and we need to subtract the voltage gain at the output of Q₁ from Q₂ and divide it by the difference in the input to each bipolar transistor to get the expression of differential voltage gain. The total differential output voltage is [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-(R_C/(1/g_m+ (R₄+R₃+1/g_m1+R₅||r_o/(β+1))))]-[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]]. The total differential input voltage is [[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]-[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))||R₃||r_o)]-(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o/(β+1))))]. Note that the differential voltages are calculated as V₂–V₁. The overall voltage gain is now written as [[-(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o5)*(R₂/((R₄+R₃+1/g_m1+R₅||r_o/(β+1))+1/g_m+(R₃||r_o5)/(β+1)+[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))+(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*-(R_C/(1/g_m+(R₄+R₃+1/g_m1+R₅||r_o/(β+1))))]-[(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))*(R₂/((R₃+R₄+1/g_m2+R₃||r_o/(β+1))+1/g_m+(R₅||r_o5)/(β+1))]+[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]*R_C/(1/g_m+(R₃+R₄+1/g_m2+R₃||r_o/(β+1)))]]/[[(r_π+(β+1)*(R₃+R₄+1/g_m1+(R₅||r_o1)/(β+1))||R₃||r_o)]-[r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))/(1/g_m+(r_π+(β+1)*(R₃+R₄+1/g_m+(R₃||r_o)/(β+1))))]. In reality, the expression shall get more complicated if we include body effect but we note that the expression has been derived without using KVL and KCL. The reader will note that each transistor can be represented by their small signal model to find the small signal gain but we can at least derive the expressions of the voltage gain by remembering the basic behavior of the bipolar and MOS amplifier topologies.

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