# Arriving at a wrong output impedance for a BJT Emitter Follower Configuration Circuit

We've to calculate output impedance of the circuit (figure 1). Although i understood the textbook solution (figure 2, figure 3), I took a different approach and arrived at a wrong answer.

step 1: I shorted the input voltage sources. (since, Zout is being calculated)
step 2: since there is no power anywhere in circuit, Ib will definitely be 0.
step 3: the dependent source is opened. (since, Ib = 0)
step 4: also RB is shorted by the source
step 5: thus, when seen b/w emitter terminal and ground, βre and RE are parallel but equation 8.42 says otherwise!

please point where the mistake was in my approach.

• Why would you say "since there is no power anywhere in the circuit"? Mar 29, 2019 at 19:14
• Input voltage is shorted and that's the only external power applied to the circuit.. so Ib has to be zero, no? Mar 29, 2019 at 19:16
• No. Current flows (conventional flow) from Vcc through Rb, through the E-B junction, through Re, to Ground. In other words, the transistor has base bias. The amount of bias current depends on the value of the resistors. Mar 29, 2019 at 19:18
• The small-signal emitter voltage is not zero, so how can Ib be zero? Mar 29, 2019 at 19:20
• @SpehroPefhany: You may be right in that I am confusing small signal analysis with large signal. Been way too many decades since I last touched this stuff. That's why I keep my old textbooks handy. Mar 29, 2019 at 21:14

To be able to see the difference try to analysis these two circuits.

The first one is CE amplifier

And for this circuit

$$\Z_{OUT} = \frac{V_X}{I_X} = R_C\$$

because $$\I_B = 0A\$$

But for the emitter follower, we have a different situation:

And $$\ I_B\$$ is not equal to $$\0A\$$ despite the fact that the Vin = 0

For this circuit $$\I_B = -\frac{V_X}{r_{\pi}}\$$

And $$\Z_{OUT} = \frac{V_X}{I_X}\$$

Let us try to find the $$\Z_{OUT}\$$ for the equivalent circuit:

I hope that you see that $$\R_E\$$ is in parallel with the resistance seen from the emitter terminal into the BJT.

And our test current is

$$I_X = I_B + \beta I_B + I_{RE}$$

But if we ignore $$\R_E\$$ resistance for a moment we can find the transistor resistance seen from the emitter looking into BJT.

$$I_X = I_B + \beta I_B = I_B(\beta +1)$$

Additional we know that $$\I_B = \frac{V_X}{r_\pi}\$$

we can write

$$I_X = I_B(\beta +1)= \frac{V_X}{r_\pi}(\beta +1) = \frac{V_X (\beta +1)}{r_\pi}$$

$$\frac{I_X}{V_X} =\frac{\beta +1}{r_\pi}$$

And finally

$$Z_{OUT} = \frac{V_X}{I_X} = \frac{r_\pi}{\beta +1}||R_E$$

and because

$$r_\pi = (\beta +1)re$$

we have

$$Z_{OUT} = re||R_E$$

• Okay, so do we always have to apply Vtest (which is Vx) at output and only then calculate output impedance? If we HAVE TO, it makes sense but without that, can't we proceed? Mar 29, 2019 at 20:28
• No always. But it is the simplest way. See some examples: electronics.stackexchange.com/questions/295771/… and here electronics.stackexchange.com/questions/342859/… and a good read ittc.ku.edu/~jstiles/412/handouts/…
– G36
Mar 29, 2019 at 20:31
• Because if we do not apply Vtest, then Ib turns out to be zero, right? Mar 29, 2019 at 20:32
• could please answer this one last thing? This was the only sticking point for me through out! Mar 29, 2019 at 20:38
• Yes, because Zout = dVout/dIout
– G36
Mar 29, 2019 at 20:39