enter image description here

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.

enter image description here

  • \$\begingroup\$ For your circuit Ro = Rc||Rout ≈ 1kΩ \$\endgroup\$
    – G36
    Oct 29, 2017 at 16:26

1 Answer 1


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)

enter image description here

\$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\$

enter image description here

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.

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

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Dave Tweed
    Oct 31, 2017 at 13:33
  • \$\begingroup\$ @G36 What value of Vce do you take in "ro" equation? \$\endgroup\$
    – MaxMil
    Feb 4, 2018 at 16:44
  • \$\begingroup\$ @MaxMil The DC operation point value (around 5V for this circuit). \$\endgroup\$
    – G36
    Feb 4, 2018 at 16:49
  • \$\begingroup\$ @G36 Is there "r_o" is only ac analysis quantity and not useful for DC Q-point? \$\endgroup\$
    – MaxMil
    Feb 4, 2018 at 17:44
  • \$\begingroup\$ @MaxMil Yes, r_o is only AC quantity. In DC we usually ignore the Early effect in hand calculations. electronics.stackexchange.com/questions/299672/… \$\endgroup\$
    – G36
    Feb 4, 2018 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.