Input impedance of emitter follower

Question

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.

Answer 1

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}\$

Answer 2

@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 .