# Input impedance of emitter follower

I decided to calculate the input impedance of the following emitter follower circuit as a practice.

I used this well-known small signal model for the BJT:

For voltage gain, I obtained

$$A_\text{v}=\frac{v_\text{out}}{v_\text{in}}=\frac{\beta/r_\pi-1/R_\text{B}}{\beta/r_\pi+1/R_\text{L}}$$

which is somewhat smaller that unity, as expected. For input impedance, I found

$$r_\text{in}=\frac{v_\text{in}}{i_\text{in}}=\frac{v_\text{in}}{v_\text{in}/R_\text{B}+(v_\text{in}-v_\text{out})/r_\pi}=R_\text{B}\;||\; (R_\text{L}||R_\text{B})\left(\beta+\frac{r_\pi}{R_\text{L}}\right)$$

I'd like to know if I've calculated it correctly. The book The Art of Electronics obtains $$\r_\text{in}=R_\text{B}\;||\;(\beta+1)R_\text{L}\$$ by a more or less different method.

• For a small-signal model $v_{in}$ appears directly at the transistor's base, which then means that the resistor $R_B$ has no influence on your voltage gain $A_v$. So your first equation cannot be correct. Commented Jul 20, 2021 at 7:56
• @Mr.Snrub, Thanks. I checked my calculations again and found no mistake. So I guess either the model I've used for BJT is not accurate enough or you're wrong that $R_\text{B}$ should not appear in the relation for small-signal voltage gain. Commented Jul 20, 2021 at 8:54
• @apadana How do you calculate $r_\pi$? And yes, I get similar values for $A_v$. I'm just curious about your methods, just now. What are the steps you used to reach your $A_v$ expression?
– jonk
Commented Jul 20, 2021 at 9:03
• @Mr.Snrub, $r_\pi$ is simply $v_\text{be}/i_\text{b}$. Commented Jul 20, 2021 at 9:08
• @apadana For example, I come up with:$$A_v= \frac1{1+\left[\frac{V_T}{V_{_\text{CC}}-V_{_\text{BE}}}\right]\cdot\left[1+\frac{R_{_\text{B}}}{\left(\beta+1\right)\cdot R_{\text{E}}}\right]}$$which is based entirely upon the DC operating point. As it must be. Your argument appears "circular" to me.
– jonk
Commented Jul 20, 2021 at 9:38

Hamm, the calculation are quite simple.

simulate this circuit – Schematic created using CircuitLab

From the small-signal model we have:

I apply KVL around the loop:

$$v_{in} = i_b r_\pi + i_eR_L$$

Additional we know that $$\I_E = I_B + I_C = I_B + \beta I_B = I_B(\beta + 1) \$$

Therefore

$$v_{in} = i_b r_\pi + i_eR_L = i_b r_\pi + i_b(\beta +1)R_L = i_b(r_\pi +(\beta +1)R_L)$$

And the input current is $$\i_{in} = i_{RB} + i_b = \frac{v_{in}}{R_B} + \frac{v_{in}}{r_\pi + (\beta +1)R_L} \$$

So now we can find the input impedance:

$$r_{in} = \frac{v_{in}}{i_{in}} = R_B||(r_\pi +(\beta +1)R_L)$$

And if we treat the voltage across the RL resistor as output we will get:

$$v_o = i_e*R_L = i_b(\beta +1)R_L$$

Therefore the voltage gain is

$$\frac{v_o}{v_{in}} = \frac{i_b(\beta +1)R_L}{i_b(r_\pi +(\beta +1)R_L)} = \frac{(\beta +1)R_L}{r_\pi +(\beta +1)R_L}$$

Also, notice that if we substitute for $$\r_\pi = (\beta+1)r_e\$$ the gain expresion becomes:

$$\frac{v_o}{v_{in}} = \frac{(\beta +1)R_L}{(\beta+1)r_e +(\beta +1)R_L} = \frac{(\beta +1)R_L}{(\beta+1)(r_e+R_L)} = \frac{R_L}{r_e + R_L}$$

A voltage divider equation.

Where $$\r_e = \frac{V_T}{I_E}\$$

• Thanks. By seeing your diagram, I noticed my mistake. I had used KCL for one of the grounded nodes, but had forgotten to take into account the current that returns to input signal source. Commented Jul 20, 2021 at 19:29

@apadana, your equation for rin is correct, however, the gain expression is not. Wihout any calculation we can see that RB cannot have any influence on Av (Vin ideal voltage source).

Correct: Av=gmRL/(1+gmRL) with emitter transconductance gm=(1+beta)/r_pi

Introducing Av into the formula for rin gives the correct result .

• Thanks. As far as I know, $i_\text{c}=\beta i_\text{b}=g_\text{m} v_\text{be}$, so $r_{\pi}=v_\text{be}/i_\text{b}=\beta/g_\text{m}$. But according to you $r_{\pi}=(\beta+1)/g_\text{m}$. Why? Commented Jul 20, 2021 at 8:27
• apadana, we are speaking about the role of the emitter current which is ie=(beta+1)ib and, hence the emitter-related transconductance gm,e. But we make no distinction between gm,e and gm,c because of the tolerances of BJT parameters . As you know, the difference between beta and (beta+1) is certainly much smaller than the tolerance of the beta value.
– LvW
Commented Jul 20, 2021 at 8:33