Common base amplifier output impedance

In Sedra Smith I can found that: Rout = ro + (1 + gm * ro) * Re_s, Re_s = Re || r_pi. Can I see in my circuit using simulation tools (like current probe, voltage meter) that previous equation approximately valid? I think that output impedance I need to meter at collector of Q1 to ground.

• For your circuit Ro = Rc||Rout ≈ 1kΩ – G36 Oct 29 '17 at 16:26

1 Answer

You can always use AC analysis (frequency response) and plot $R_{out} = V_1/I_1$ .

Here you have an example from LTspice (I don't have multisim)

$V_1$ is $5V DC$ source and $1V$ for AC analysis.

And next, I plot $R_{out} = \frac{V_1}{I_1} =\frac{V(n001)}{I(V1)}$

And read from the plot $R_{out} = 2.62\textrm{M}\Omega$

In LTspice I used $2N3904$ with $V_A = 100\textrm{V}$ (Early voltage) and $\beta = 300$

The DC operation point is:

$$I_E = \frac{5\textrm{V} - 0.75\textrm{V}}{15\textrm{k}\Omega}\approx 4.3\textrm{mA}$$

And BJT small-signal parameters:

$$g_m = \frac{I_C}{V_T} = \frac{4.3\textrm{mA}}{26\textrm{mV}} \approx 0.165\:\textrm{S}$$

$$r_O = \frac{V_A + V_{CE}}{I_C}\approx 24.45\textrm{k}\Omega$$

$$r_{\pi} = \frac{\beta}{gm}\approx 1.8\textrm{k}\Omega$$

And finally, we can calculate $R_{out}$ without $R_C$ resistor in the circuit.

$$R_{out}= r_o + (1 +g_m \cdot r_o)\cdot R_E||r_{\pi}\approx 2.61\textrm{M}\Omega$$

• Comments are not for extended discussion; this conversation has been moved to chat. – Dave Tweed Oct 31 '17 at 13:33
• @G36 What value of Vce do you take in "ro" equation? – MaxMil Feb 4 '18 at 16:44
• @MaxMil The DC operation point value (around 5V for this circuit). – G36 Feb 4 '18 at 16:49
• @G36 Is there "r_o" is only ac analysis quantity and not useful for DC Q-point? – MaxMil Feb 4 '18 at 17:44
• @MaxMil Yes, r_o is only AC quantity. In DC we usually ignore the Early effect in hand calculations. electronics.stackexchange.com/questions/299672/… – G36 Feb 4 '18 at 17:54