11.89 The transistor parameters for the circuit in the figure are $ \beta=200 $, $ V_{BE}(on)=0.7V$, and $ V_A=80V$ Determine the differential mode voltage gain $ A_d=v_{o3}/v_d $

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I have done the analysis. I want to know if my analysis is correct or not. What things should be taken care of and other valuable suggestions would be of great help.

DC ANALYSIS(To find quiescent collector currents of the circuit to find \$ g_m \$ for the small signal AC analysis)

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Calculating \$ I_{C5} \$ from the current mirror circuit:


The differential amplifier circuit:


$$ => V_{BE1}=V_{BE2}$$

$$V_{BE1}=V_Tln(\frac{I_{C1}}{I_{S1}})$$ $$V_{BE2}=V_Tln(\frac{I_{C2}}{I_{S2}})$$ If \$I_{S1}=I_{S2}\$ and since \$ V_{BE1}=V_{BE2}[=0.7V\$(given)] $$V_Tln(\frac{I_{C1}}{I_{S1}})=V_Tln(\frac{I_{C2}}{I_{S2}})$$ $$\frac{I_{C1}}{I_S}=\frac{I_{C2}}{I_S}$$ $$I_{C1}=I_{C2}$$ $$\frac{\beta}{\beta +1}I_{C1}=\frac{\beta}{\beta +1}I_{C2}$$ $$=> I_{E1}=I_{E2}$$ Therefore, $$I_{E1}=I_{E2}=\frac{I_{C5}}{2}=\frac{233}{120*2}mA=\frac{223}{240}mA$$ $$I_{C1}=I_{C2}=\frac{\beta}{\beta +1} I_{E1}=\frac{200}{201}*\frac{223}{240}mA$$.

Let us assume, sum of all currents leaving \$ V_{o2} \$ is 0.

$$\frac{V_{o2}-V^+}{R_C}+I_{C2}+I_{B3}=0$$ $$=>\frac{V_{o2}-V^+}{R_C}+I_{C2}+\frac{I_{E3}}{1+\beta}=0$$ $$\frac{V_{o2}-12}{8k}+\frac{200}{201}*\frac{223}{240}m+\frac{I_{E3}}{201} m=0$$ $$\frac{V_{o2}}{8}+\frac{I_{E3}}{201}=\frac{347}{603} \tag1$$


$$V_{o2}-0.7-I_{E3}(3.3)=0 \tag 2$$

From (1) and (2) we get \$I_{E3}=\frac{3317}{2877}mA\$ $$I_{C3}=\frac{200}{201}*\frac{3317}{2877}$$

SMALL SIGNAL AC ANALYSIS enter image description here

At node 1:

$$g_{m1}v_{\pi 1}+\frac{v_{\pi 1}}{r_{\pi 1}}+\frac{v_{\pi 1}}{R_1||r_{o2}}+\frac{v_{\pi 1}}{r_{\pi 1}}=0$$ $$v_{\pi 1}(g_{m1}+\frac{2}{r_{\pi 1}}+\frac{1}{R_1||r_{o2}})=0$$ $$=>v_{\pi 1}=0$$

At node k:

Since \$ g_{m1}v_{\pi 1}=0\$, node k is at ground.

$$\frac{v_{\pi 2}}{r_{\pi 2}}+ g_{m2}v_{\pi 2}+ g_{m2}v_{\pi 3}+\frac{v_{\pi 3} }{r_{\pi 2}}=0$$

$$v_{\pi 2}(\frac{1}{r_{\pi 2}}+g_{m2})=v_{\pi 3}(\frac{1}{r_{\pi 2}}+g_{m2})$$ $$v_{\pi 2}=-v_{\pi 3}$$

Since node k is at ground:

\$v_{\pi 2}=v_1\$


\$v_{\pi 3}=v_2\$

In node 4:

$$\frac{v_4}{R_C||r_{o1}}+g_{m2}v_{\pi 3}+\frac{v_4-v_3}{r_{\pi 4}}=0$$ $$v_4(\frac{1}{R_C||r_{o1}}+\frac{1}{r_{\pi 4}})=\frac{v_3}{r_{\pi 4}}-g_{m2}v_{\pi 3}$$

$$v_4=\frac{\frac{v_3}{r_{\pi 4}}-g_{m2}v_{\pi 3}}{R_C||r_{o1}||r_{\pi 4}} \tag A$$

At node 3:

$$\frac{v_3-v_4}{r_{\pi 4}}+\frac{v_3}{R_E}-g_{m3}v_{\pi 4}+\frac{v_3-v_{o3}}{r_{o3}}=0$$

$$v_3[\frac{1}{r_{\pi 4}}+\frac{1}{R_E}+\frac{1}{r_{o3}}]=\frac{v_{o3}}{r_{o3}}+g_{m3}v_{\pi 4}+\frac{v_4}{r_{\pi 4}}$$

$$v_3[\frac{1}{r_{\pi 4}}+\frac{1}{R_E}+\frac{1}{r_{o3}}]=\frac{v_{o3}}{r_{o3}}+g_{m3}v_{ 4}-g_{m3}v_{ 3}+\frac{v_4}{r_{\pi 4}}$$

$$v_3[r_{\pi 4}||R_E||r_{o3}+g_{m3}]=\frac{v_{o3}}{r_{o3}}+v_4(g_{m3}+\frac{1}{r_{\pi 4}}) \tag {III(b)}$$

Putting \$ v_4 \$ from eq (A) to III(b):

$$v_3[r_{\pi 4}||R_E||r_{o3}+g_{m3}]=\frac{v_{o3}}{r_{o3}}+\frac{g_{m3}+\frac{1}{r_{\pi 4}}}{R_C||r_{o1}||r_{\pi 4}}(\frac{v_3}{r_{\pi 4}}-g_{m2}v_{\pi 3})$$

$$v_3[r_{\pi 4}||R_E||r_{o3}+g_{m3}-\frac{g_{m3}+\frac{1}{r_{\pi 4}}}{r_{\pi 4}(R_C||r_{o1}||r_{\pi 4})}]=\frac{v_{o3}}{r_{o3}}-\frac{g_{m3}+\frac{1}{r_{\pi 4}}}{R_C||r_{o1}||r_{\pi 4}}g_{m2}v_2 \tag {IIIc}$$ (Replaced \$v_{\pi 3}\$ by \$ v_2\$)

or, $$Av_3=Bv_{o3}-Cv_2 \tag {IIId}$$ (for ease of calculation, we have assumed the constants as A,B,C.)


$$v_{o3}=(-g_{m3}v_{\pi 4}+\frac{v_3-v_{o3}}{r_{o3}})R_{C2}$$ $$v_{o3}[1+\frac{R_{C2}}{r_{o3}}]=-g_{m3}v_{\pi 4}R_{C2}+v_3\frac{R_{C2}}{r_{o3}}$$ $$v_{o3}[1+\frac{R_{C2}}{r_{o3}}]=-g_{m3}v_{v_4-v_3}R_{C2}+v_3\frac{R_{C2}}{r_{o3}}$$


$$v_{o3}[1+\frac{R_{C2}}{r_{o3}}]=-g_{m3}R_{C2}[\frac{\frac{v_3}{r_{\pi 4}}-g_{m2}v_{\pi 3}}{R_C||r_{o1}||r_{\pi 4}}]+v_3(\frac{R_{C2}}{r_{o3}}+g_mR_{C2})$$ $$v_{o3}[1+\frac{R_{C2}}{r_{o3}}]=v_3[(\frac{-g_{m3}R_{C2}}{R_C||r_{o1}||r_{\pi 4}})\frac{1}{r_{\pi 4}}+(\frac{R_{C2}}{r_{o3}}+g_mR_{C2})]+\frac{g_{m2}v_2g_{m3}R_{C2}}{R_c||r_{o1}||r_{\pi 4}} \tag {IVa}$$

Replacing the constants in IVa by W,X,Y we get:

$$Wv_{o3}=Xv_3+Yv_2 \tag {IVb}$$

From eq IVb and IIId: $$Wv_{o3}=X(\frac{Bv_{o3}-Cv_2}{A})+Yv_2$$ $$v_{o3}(W-\frac{XB}{A})=v_2(Y-\frac{XC}{A})$$ $$\frac{v_{o3}}{v_2}=\frac{Y-\frac{XC}{A}}{W-\frac{XB}{A}}\tag {10}$$

Differential voltage gain is given by: $$A_d=\frac{v_{o3}}{v_2-v_1}=\frac{v_{o3}}{v_{\pi 3}-v_{\pi 2}}=\frac{v_{o3}}{v_{\pi 3}-(-v_{\pi 3})}=\frac{v_{o3}}{2v_{\pi 3}}=\frac{v_{o3}}{2v_2}=\frac{2}{2}\frac{Y-\frac{XC}{A}}{W-\frac{XB}{A}}$$


closed as unclear what you're asking by Eugene Sh., Bimpelrekkie, uint128_t, winny, PeterJ Aug 11 '17 at 9:53

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    \$\begingroup\$ Do you actually expect anyone (except your TAs, who are paid for this) to go over this and check it for you? \$\endgroup\$ – Eugene Sh. Aug 8 '17 at 20:25
  • \$\begingroup\$ @EugeneSh. Sir, I have edited the question as follows: "I have done the analysis. I want to know if my analysis is correct or not. What things should be taken care of and other valuable suggestions would be of great help." Sir I want to know if my analysis of this differential amplifier circuit is correct or have I done something wrong. I have seen the solution of the book however my solution and the book's solution seem to be different...I shall also post the solution in the book for reference. \$\endgroup\$ – Soumee Aug 8 '17 at 20:28
  • \$\begingroup\$ So what is the final number? If you post a solution to the final equation we can do the short cuts to get an approximate solution (I have done that in my head) to at least indicate whether you are close. \$\endgroup\$ – Kevin White Aug 8 '17 at 20:46
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    \$\begingroup\$ @Soumee The problem with your solution is that it is a complex mess because you're including everything without even considering how the circuit works. For example, including the biasing (R1, Q4, Q5) in the small signal analysis is pointless. Think of how R1, Q4 and Q5 behave small-signal wise and include them as a model. Analog Electronics is not just blindly filling in everything and solving it. You must evaluate the circuit and see how it works and how to find the requested parameters. \$\endgroup\$ – Bimpelrekkie Aug 8 '17 at 20:58
  • \$\begingroup\$ @Bimpelrekkie Thank you sir for your comment. I considered Q4 and Q5 to see if it has any impact on the differential amplifier, and I indeed found out that the node k acts as a 'signal ground' as indicated in the books. I saw that being written on books, but I could't understand why the common point would be the signal ground. Now that I have verified it myself , I feel convinced. \$\endgroup\$ – Soumee Aug 8 '17 at 21:14

The output transistor has gain of ~~ 1.2x.

Each diffpair transistor, running near 1mA each, has gm of 1/26_ohms. However, the total small-signal resistance in the emitters is 26 + 26 or 52 ohms. View that diffpair as a commonbase and commonemitter, hence the 26+26.

The gain is 8,000 ohms / 52 ohms, or 8 * 20 = 160.

Thus, within 10% or 20%, your gain should be near 160x.


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