next up previous contents
Next: Input and Output Resistance Up: Differential Amplifiers Previous: Differential Pair DC Bias

Differential Pair Small Signal Voltage Gain

For the sake of understanding, let's assume we take the small signal output from the collector of Q2, and input the small signal voltage to the base of Q1. This is shown in Fig. 5.3.
  
Figure 5.3: Differential Amplifier Single-Ended Differential Mode Gain Configuration
\begin{figure}
\centering{
\fbox {\psfig{file=./413_figs/fig5_3b.ps,width=4.0in}}
}\end{figure}

The small signal voltage gain will then be defined as the change in the collector voltage at Q2 divided by the change in the base voltage of Q1, or $A_v=\frac{V_o}{V_i}=\frac{\Delta V_{C2}}{\Delta V_{B1}}$.Taking the small signal change of (5.5), we find the small signal output voltage Vo.  
 \begin{displaymath}
V_o=\Delta V_c = -\Delta I_{c2}R_C\end{displaymath} (106)
To find the small signal change in the input, we start with the large signal KVL equation

 
Vb1 -Vb2= Vbe1-Vbe2 (107)

Now, if we ground Vb2, and make a small signal change in Vb1 we obtain  
 \begin{displaymath}
\Delta V_{b1} = \Delta V_{be1}-\Delta V_{be2}\end{displaymath} (108)
Recall from previous labs that a first order Taylor series gives $\Delta V_{be}=\frac{\Delta I_c}{g_m}$.We then find that
\begin{displaymath}
\Delta V_{b1} = \frac{\Delta I_{c1}}{g_{m1}}-\frac{\Delta I_{c2}}{g_{m2}}\end{displaymath} (109)
Since $g_m=\frac{I_C}{V_t}$ and, under DC conditions IC1=IC2, then

gm1= gm2= gm

(110)

Now, recall that the total current in the diff-amp is determined by the DC bias current IEE which is fixed. Thus, if we make a small increase in Ic1 then there will be a corresponding small decrease in Ic2. We express this mathematically by taking the small signal change of equation (5.2), which gives
\begin{displaymath}
\Delta I_{c1}+\Delta I_{c2}=\Delta I_{EE}=0\end{displaymath} (111)
which leads to
\begin{displaymath}
\Delta I_{c1}~=~-\Delta I_{c2}\end{displaymath} (112)
Substituting the previous expressions into equation (5.9) leads to  
 \begin{displaymath}
v_{in1}\Delta V_{b1} = -2\frac{\Delta I_{c2}}{g_{m}}\end{displaymath} (113)
Taking the ratio of equations (5.7) and (5.14) leads to the following expression for voltage gain Av12 of a differential amplifier when the output is take from the opposite transistor as the input.
\begin{displaymath}
A_{v12}=\frac {V_{o2}}{V_{in1}} = \frac{\Delta V_{c2}}{\Delta V_{b1}}=\frac{g_m R_C}{2}\end{displaymath} (114)

If the output were to be taken from the collector of Q1 instead of Q2, we could follow virtually the same analysis and would find that the voltage gain Av11 would differ only by a minus sign.
\begin{displaymath}
A_{v11}= \frac {V_{o1}}{V_{in1}}=\frac{\Delta V_{c1}}{\Delta V_{b1}}=\frac{- g_m R_C}{2}\end{displaymath} (115)
If we apply a signal to both inputs, and take the output from the collector of Q2, then we can define the differential gain to be
\begin{displaymath}
A_{dm}= \frac{\Delta V_{c2}}{\Delta V_{b1}-\Delta V_{b2}}=\frac{ g_m R_C}{2}
= \frac{v_{o2}}{v_{in1}-v_{in2}}\end{displaymath} (116)


 
next up previous contents
Next: Input and Output Resistance Up: Differential Amplifiers Previous: Differential Pair DC Bias
Neil Goldsman
10/23/1998