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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 $$I_E = \dfrac{4,7-0,7}{20k\Omega} = 0,2\,mA$$

$$I_C \approx I_E$$

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

miguel747
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    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 https://electronics.stackexchange.com/questions/342859/bjt-common-base-output-resistance-derivation/342989#342989 – G36 Mar 27 '22 at 15:50
  • @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\$? – jonk Mar 27 '22 at 18:31
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    @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\$ – G36 Mar 27 '22 at 19:01
  • I tried to compute the small signal analysis (pi model) and I find the expression Vx/Ix = $$Rx = ro - (gm\cdot Vbe\cdot ro)/ Ix + RE//r\pi$$ but i cant deal with Ix on denominator. Seems the correct expression is $$Rout = {ro + (1+ gm*ro) (r\pi//RE)}$$ – miguel747 Mar 27 '22 at 19:04
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    @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)\$. – jonk Mar 27 '22 at 19:08
  • i got it. just replace Vbe = Vpi = -V = -Ix.Re//rpi and done. – miguel747 Mar 27 '22 at 23:23
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    @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.) – jonk Mar 28 '22 at 04:29
  • @miguel747 If you want to put things in terms that G36 tried, I find something closer (ignoring confounding factors that are usually small) to: \$ \beta \cdot R_{_\text{E}}\cdot \left[ \frac{V_A}{\beta\cdot V_T +V_{_\text{E}}} \right] \$. The bracketed factor at the end is the modification I'd make to what G36 wrote, if I had to pick something. (It assumes \$V_{_\text{E}}\gg V_T\$, for example.) – jonk Mar 28 '22 at 05:29
  • Ty for the words @jonk, i will write my own answer here. – miguel747 Mar 28 '22 at 20:49
  • @jonk, regarding your last equation: In the brackets you add "1" with a resistance. This can`t be correct. – LvW Mar 29 '22 at 07:32
  • @LvW Yeah, I need to explain that development. I'll do that when I get a chance. Going to sleep now – jonk Mar 29 '22 at 07:35
  • @LvW Okay. Finally got a moment. I completely screwed up. Thanks for the catch. I meant to write: \$r_o+r_o\, g_m\left(R_{_\text{E}}\mid\mid r_\pi\right)+\left(R_{_\text{E}}\mid\mid r_\pi\right)=r_o+\left(R_{_\text{E}}\mid\mid r_\pi\right)\left(1+r_o\,g_m\right)\$ Again, thanks for the catch! – jonk Mar 29 '22 at 20:39

1 Answers1

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

schematic

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):

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

but:

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

Combine (1) and (2):

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

So:

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

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

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

G36
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miguel747
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  • 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 – LvW Mar 29 '22 at 07:22