I have a doubt respecting how to calculate the output resistance on this question below.

enter image description here

I've searched at 4 different books and find the same way to compute this value but seems its not be right in that situation.

Translated the instruction:

For the circuit on current source as show the picture below, consider that the common emitter current gain (beta) is 160 and the Early voltage is 8V. The Vbe = 0,7V and the V_T = 25mV, the output resistance of circuit is:

Letter D is answer given.

Anyone have some tips?

My attempt. Like Sedra, Razavi and Boyleastad book said que the output resistance has a approximation expression given as: $$R_{out} = > \dfrac{V_A}{I_C}$$ as $$I_E = \dfrac{4,7-0,7}{20k\Omega} = 0,2\,mA$$

$$I_C \approx I_E$$

$$R_{out} = \dfrac{8V}{0,2 mA} = 40k\Omega$$

  • 1
    \$\begingroup\$ Do you know the small-signal analysis? For CB amplifier \$R_{out} \approx R_E*\beta \approx 20k\Omega *160 \approx 3.2M\Omega\$ Look here electronics.stackexchange.com/questions/342859/… \$\endgroup\$
    – G36
    Commented Mar 27, 2022 at 15:50
  • \$\begingroup\$ @G36 That does work out when \$V_A=8\$, which is the OP's case. (Because \$\beta\cdot V_T+V_{_\text{E}}=8.16\$ and is close to \$V_A\$.) But what if \$V_A=80\$? Or \$V_A=800\$? \$\endgroup\$
    – jonk
    Commented Mar 27, 2022 at 18:31
  • 1
    \$\begingroup\$ @jonk It seems that I made a mistake. For the CB stage the maximum value of a Rour you can get is equal to \$R_{out_{max}} = ro*\beta\$ not \$R_E*\beta\$ \$\endgroup\$
    – G36
    Commented Mar 27, 2022 at 19:01
  • 1
    \$\begingroup\$ @miguel747 Yeah. That's about what I got: \$r_o+r_o \,g_m\left(1+R_{_\text{E}}\mid\mid r_\pi\right)\$ or \$r_o\left(1+g_m\left(1+R_{_\text{E}}\mid\mid r_\pi\right)\right)\$. \$\endgroup\$
    – jonk
    Commented Mar 27, 2022 at 19:08
  • 2
    \$\begingroup\$ @miguel747 You can write your own answer and then select it!! Might be a good thing because you know your question better than anyone here. I'd love to upvote it, myself. If you have the pieces, why not go ahead and write an answer here? It sounds as though you are ready! Go for it! (Also, providing an answer and closing the question provides a resource for others, later on. It's the right thing to do if you feel able.) \$\endgroup\$
    – jonk
    Commented Mar 28, 2022 at 4:29

1 Answer 1


For regarding several tips and good insights coming from @jonk and @G36, I find the solution using small signal analysis.


simulate this circuit – Schematic created using CircuitLab

DC bias:

$$I_E = \dfrac{4.7-0.7}{20k\Omega} = 0.2\, mA$$

As follow (small signal):

$$V_X = ro\cdot (i_x - v_{be}\cdot gm)+V\\ = ro\cdot (i_x - v_{be}\cdot gm)+ix\cdot (R_E||r_{\pi}) \tag{1}$$


$$V = i_x\cdot R_E||r_{\pi} = -v_{\pi} = -v_{be}\tag{2}$$

Combine (1) and (2):

$$\require{cancel}\dfrac{V_X}{i_x} = ro\cdot (\cancelto{1}{i_x} +\cancel{i_x}\cdot R_E||r_{\pi}\cdot gm)+\cancel{i_x}\cdot (R_E||r_{\pi})\\ \boxed{R_{out} = ro + (1+ro\cdot gm)\cdot R_E||r_{\pi}} $$


$$r_{\pi} = \dfrac{V_T}{I_B} = \dfrac{V_T\cdot \beta}{I_C};\quad\beta = 160;\quad ro = \dfrac{V_A}{I_C} = 40k\Omega\quad \text{(My initial mistake)}$$

$$\require{cancel} R_{out} = 0,04M\Omega + 20k\Omega||\cancelto{20k\Omega}{\dfrac{25mV\cdot 160}{0,2mA}} \cdot \left (1+\dfrac{40k\Omega}{125\Omega}\right )\\ = 0.04M\Omega + 0,01M\Omega + \dfrac{400M\Omega}{125}$$

$$R_{out} = 0.04 + 0.01 + 3.2 = \boxed{3.25M\Omega}\tag{3}$$

  • 1
    \$\begingroup\$ Just one small comment: It is correct that the differential (dynamic) resistances r_o and r_pi are written with small letters. To be consistent, the same should apply also to r_out . In general, one should clearly distinguish between static and dynamic resistances - this helps to understand the working principles of electronic circuits \$\endgroup\$
    – LvW
    Commented Mar 29, 2022 at 7:22

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