Small Signal Open Loop Gain

To determine the small signal open loop gain we break the circuit up into stages. Looking at Fig. 5.6, we realize that the circuit is composed of a gain stage that is composed of the differential amplifier, and the emitter follower output stage, which acts to lower the output resistance of the circuit. The overall gain of the circuit is then the product of the gain of the two stages:

$$\frac{v_o}{v_{in}}= \left(\frac{\Delta V_{C2}}{v_{in}}\right)\left( \frac{v_o}{\Delta V_{C2}}\right)$$ (123)

where ${\Delta V_{C2}}={\Delta V_{B4}}$The approximate result of this calculation can be obtained almost immediately by inspection to be $\frac{v_o}{v_{in}}\approx g_{m2}R_{C2}/2$.Now, let's perform a more detailed analysis to show that this is indeed the case. First of all the gain of the first stage is:  

$$\frac{\Delta V_{C2}}{v_{in}}= \frac{g_{m2}R_{C2}\vert\vert(r_\pi+\beta R_{E4})}{2}$$ (124)

Where $(r_\pi+\beta R_{E4})$ is the input resistance looking into the base of Q4.

Now, the gain of the second stage is given by  

$$\frac{v_o}{\Delta V_{C2}}= \frac{R_{E4}}{R_{E4}+1/g_{m4}}$$ (125)

If we multiply the gain of the two stages together, we arrive at the following expression for the total differential mode gain of the circuit A_dm.

$$\frac{v_o}{v_{in}}=A_{dm} = \left(\frac{g_{m2}R_{C2}\vert\ve... ...R_{E4})}{2}\right) \left(\frac{R_{E4}}{R_{E4}+1/g_{m4}}\right)$$ (126)

Since R_E4 is usually much greater than 1/g_m4, in most cases the gain of the second stage can often be approximated by $\frac{\Delta V_{C2}}{v_{in}} \approx 1$.So, the gain of the overall amp is usually given by the gain of the first stage alone described by equation (5.25). Finally, we it is often the case that we can neglect $(r_\pi+\beta R_{E4})$ because it is usually much greater than R_C2. Under these conditions, the total gain of the circuit can be given by $\frac{v_o}{v_{in}}=A_{dm}\approx g_{m2}R_{C2}/2$, which was mentioned above.