Output resistance of a multistage BJT amplifier
but how do you get that expression?


In a comment, I wrote an expression for the small-signal resistance looking into the collector of Q7.  The derivation is below.







simulate this circuit &ndash; Schematic created using CircuitLab

In the above, I assume that we can ignore the resistance looking in to the collector of Q5.

The small-signal resistance looking into the collector of Q7 is

$$r_{ic7} = \frac{V_{c7}}{I_{c7}}$$

Let the current \$I_{c7}\$ be due to a test source and write the following equations:

$$V_{c7} = r_{o7} \left( I_{c7} - \beta_7 I_{b7}\right) + V_{e7}$$

$$V_{e7} = I_{c7}\cdot R_4||\left(R_3 + r_{\pi 7}\right) $$
$$I_{b7} = -I_{c7}\cdot \frac{R_4}{R_3 + r_{\pi 7} + R_4}$$

Substituting the final two equations into the first equation yields

$$V_{c7} = r_{o7}\left(I_{c7} + \beta_7 I_{c7}\cdot \frac{R_4}{R_3 + r_{\pi 7} + R_4}  \right) + I_{c7}\cdot \left(R_3 + r_{\pi 7}\right)||R_4 $$

thus

$$r_{ic7} =   \frac{V_{c7}}{I_{c7}} = r_{o7}\left(1 +  \frac{\beta_7 R_4}{R_3 + r_{\pi 7} + R_4}\right) + \left(R_3 + r_{\pi 7}||R_4\right)$$

Note that setting \$R_4 = 0\$ yields

$$r_{ic7} = r_{o7}$$ as expected.  Also, note that for \$R_4\$ large enough

$$r_{ic7} \approx r_{o7}\left(1 + \beta_7 \right) $$