# 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"? – Dwayne Reid Mar 29 '19 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? – Mohit Singh Chahar Mar 29 '19 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. – Dwayne Reid Mar 29 '19 at 19:18
• The small-signal emitter voltage is not zero, so how can Ib be zero? – Spehro Pefhany Mar 29 '19 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. – Dwayne Reid Mar 29 '19 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

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