Stuck in Ri, Rout and Av calculation for BJT circuit!

I have been asked to calculate Ri, Rout and Av for the following Emitter Follower circuit:

I s=7*10-16A, β=100, VA=5V , C=very large I analysed the circuit in DC and then I drew the small signal model as below: I stuck in this point! I have no idea how to calculate Ri and Av! Can you please shed any light for me in this problem?

• "calculate Ri, Rout and Av". But you haven't defined them! Closing as unclear. – Olin Lathrop Aug 4 '17 at 11:51
• Ri is input resistance, Ro is output resistance. (from vi and vo points). Av is voltage gain. – Abraham Aug 4 '17 at 11:53
• Inject a current (in your small signal model) into the output. There will be some output voltage caused by that current. Now use Ohms Law to compute the "resistance". Do the same at the input: inject a current, compute the resultant voltage, and divide. – analogsystemsrf Aug 4 '17 at 16:19

The gain of the emitter follower is a bit less than one. To find a relation for Av you just need to find vo in terms of vi. Based on your small signal model, we have using kcl at node vo:

$$\frac{v_o}{R_E||R_L||r_0} - g_m V_\pi - \frac{V_\pi}{r_\pi} = 0$$

Now you need to find Vpi in terms of vi:

$$V_\pi=v_i-v_o$$ Now substituting this back into the first equation you simply find for Av: $$A_v=\frac{v_o}{v_i}=\frac{R_E||R_L||r_0}{R_E||R_L||r_0+\frac{1}{g_m}}$$

Just follow the same procedure to find input/output impedances. For example, to calculate the input impedance put a test voltage source Vt at the input and find Ri=Vt/It. For now just neglect RB we will get back to it later.

$$V_t=V_\pi+v_o$$ $$I_t=\frac{V_\pi}{r_\pi}$$ $$v_o=R_E||R_L||r_o\times V_\pi(\frac{1}{r_\pi}+g_m)$$ Now Vt/It becomes $$R_i=[r_\pi+R_E||R_L||r_o(1+\beta)]$$

This is just the impedance looking into the base. To find the overall input impedance including RB, you need to consider Ri in parallel with RB. Therefore the input impedance becomes:

$$R_{in}=R_i||R_B$$ The ouput impedance can also be solved in the same manner. Just put a test voltage source Vt at the output and solve for Vt/It.

• Thank you. For the Rout, I must grounded the Vi, right? That means the 10K resistor will remove from the circuit. Am I right? – Abraham Aug 4 '17 at 11:34
• No, RB is still there. In SSM you still have RB connected to base and ground. Just remove the input signal frequency Vi and solve for Rout. – dirac16 Aug 4 '17 at 12:14