# Output resistance of a multistage BJT amplifier

In the amplifier shown: To calculate the output resistance, this equation is used: I understand that the R5/(B+1) comes from resistance reflection, but why do they not take the resistance in the collector of Q7, which I expected to be in parallel with R5, into consideration?

Q7's output resistance will be many times that of R5 and therefore the equation is an "OK" approximation but, sure, if you are able to calculate the output resistance of Q7 then put it in parallel with R5 to get a more accurate formula.

• Thanks. How would I find the output resistance of Q7 though? Set all sources to 0 and apply a test voltage? Or is there an easier way. – Mosho Aug 3 '14 at 20:55
• @Mosho look it up in the data sheet for the device - in common-emitter mode the output impedance is almost a constant current source but there will be a slope that would translate to a highish resistance. – Andy aka Aug 3 '14 at 21:47
• @Mosho, a near exact expression for the resistance looking into the collector of Q7 is: $$r_{o7} \left(1 + \frac{\beta_7 R_4}{R_3 + r_{\pi 7} + R_4}\right) + \left(R_3 + r_{\pi 7} \right)||R_4$$ which, as Andy correctly points out, is much larger than $R_5$ – Alfred Centauri Aug 4 '14 at 3:28
• @AlfredCentauri thanks, but how do you get that expression? – Mosho Aug 4 '14 at 5:40
• Mosho: You even can go further: The resistor R3 in the above equation is in parallel to the output resistance of Q5...and so forth. What I want to say is the following: It makes no sense to be more exact - as far as formulas are concerned - as tolerances or other uncertainties allow. In most cases, it is sufficient to use only re8=1/gm8 as a good and sufficiuent approximation for the output resistance. (Note that the inverse transconductance re8=26mV/Ic8) – LvW Aug 4 '14 at 6:36

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 – 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

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