I have the following BJT based current mirror

enter image description here

for part a, I used an approximate analysis illustrated below: \begin{equation} V_{B2}=0.7V=V_{C2}\\V_{BE1}=V_{B1}-V_{C2\:}\rightarrow V_{B1}=0.7+0.7=1.4=V_{C1}\\V_{BE3}=V_{B1}-V_{E3\:}\rightarrow V_{E3}=1.4-0.7=0.7V\\R=\frac{V_{E3}}{I_{E3}}=\frac{0.7V}{10\mu A}=70K\Omega \end{equation} Now for part b, to find the output resistance, I must use the small signal equivalent circuit. That is done by opening the current source and attaching a test voltage vo to the output. The equivalent circuit is displayed below:


simulate this circuit – Schematic created using CircuitLab

Here is my mathematical analysis for the small signal circuit: \begin{equation} v_o=r_o3\left(i_o-g_{m3}v_{\pi 3}\right)+R\left(i_o-g_{m3}v_{\pi 3}+\frac{v_{\pi 3}}{r_{e3}}\right)\\re-arranging:\\v_o=i_o\left(r_{o3}+R\right)-\left(g_{m3}\left(r_{o3}+R\right)-\frac{R}{r_{e3}}\right)v_{\pi 3} \end{equation} Now using the v_pi3 loop of the emitter of Q1 and R we get: \begin{equation} v_{\pi 3}+R\left(i_o-g_{m3}v_{\pi 3}+\frac{v_{\pi 3}}{r_{e3}}\right)=0\\re-arranging:\\v_{\pi 3}=\frac{R}{Rg_{m3}-\frac{R}{r_{e3}}-1}i_o \end{equation} Now substituting we get the final Ro as: \begin{equation} \frac{v_o}{i_o}=R_o=r_{o3}+R-\left[g_{m3}\left(r_{o3}+R\right)-\frac{R}{r_{e3}}\right]\left(\frac{R}{Rg_{m3}-\frac{R}{r_{e3}}-1}\right)\\now:\\r_{o3}=r_o=\frac{V_A}{I_o}=\frac{40V}{10\mu A}=4M\Omega\\r_{e3}=r_e=\frac{\alpha V_T}{I_o}=\frac{\left(0.99\right)\left(25mV\right)}{10\mu A}=2.5K\Omega\\g_{m3}=g_m=\frac{I_o}{V_T}=\frac{10\mu A}{25mV}=0.4mS \end{equation} Substituting the corresponding values, I get Ro to equal: \begin{equation} R_o=116M\Omega \end{equation} Now I tried to verify my results, so I decided to simulate my circuit using LTSpice. Here is my schematic:

enter image description here

Below are my DC operating point results, they do seem close to my hand calculated values:

enter image description here

Now utilizing the AC analysis to find Ro, I get that Ro=83M Ohms as displayed below:

enter image description here

Can somebody please tell me why am getting such a big difference between my calculated and simulated values for Ro?? Please help me figure out if there is something wrong with my analytical equations or the way I set up my circuit in LTSpice. Thank you for your help in advance.

  • 2
    \$\begingroup\$ I think that the small signal circuit is incorrect. The gm's should not be shorted (it means shorting C and E) so I can't recognize the small signal transistors in them. \$\endgroup\$
    – Sven B
    Commented Jun 19, 2018 at 6:18
  • \$\begingroup\$ Indeed it is incorrect, Q1 and Q2 have their base and collector shorted (they're diodes), their s.s. circuit is a gm between B,C and E, that behaves as a 1/gm resistor. Any Ro is simply in parallel with that. \$\endgroup\$ Commented Jun 19, 2018 at 6:25
  • \$\begingroup\$ @Bimpelrekkie, thank you for you comment. So, if am understanding you correctly, we pretty much don't need the left half of the circuit. We can treat it as a resistance of 1/gm and attach that to the base of Q1. If that is the case, then I can treat the right half of the circuit as Common Base amplifier with a resistance of 1/gm attached to the base of Q1. Am I correct? \$\endgroup\$
    – JordenSH
    Commented Jun 19, 2018 at 15:11

2 Answers 2


First, notice that in the text we can clearly read that for \$I_{C1} = 1\textrm{mA}\$ we have \$V_{BE} = 0.7\textrm{V}\$ But now we have \$I_{C2} = 10 \mu \textrm{A}\$ So the transistor \$V_{BE}\$ will be equal to:

$$\Delta V_{BE} = V_T \cdot \textrm{ln}\left(\frac{I_{C1}}{I_{C2}}\right) = 25\textrm{mV}\cdot \textrm{ln}\left(\frac{1\textrm{mA}}{10\mu\textrm{A}}\right) \approx 115.13\textrm{mV} $$

So we have \$V_{BE} = 0.7\textrm{V} - 115.13\textrm{mV}\approx 585\textrm{mV} \$

And the resistor will be in the range of \$ R = \frac{585\textrm{mV}}{10\mu\textrm{A}} \approx 58.5\textrm{k}\Omega \$

Now the small signal equivalent circuit. By now it should be obvious for you that \$Q_1\$ and \$Q_2\$ are diode-connected BJT's. And their small-signal equivalent circuit is

\$r_d = \frac{1}{gm}||r_\pi||r_o \approx \frac{1}{gm} = \frac{10\mu A}{25mV}= 2.5\textrm{k}\Omega\$

\$r_o = \frac{V_A}{I_C} = \frac{40V}{10\mu A} = 4\textrm{M}\Omega \$

\$r_\pi = \frac{\beta}{g_m} = \frac{100}{0.4mS} = 250\textrm{k}\Omega\$

Hence the small signal circuit will look like this:


simulate this circuit – Schematic created using CircuitLab

I omitted \$r_{o1}\$ and \$r_{o2}\$ on purpose (\$r_{o}>>\frac{1}{g_m}\$).

For this circuit we can write this KVL

$$V_X = (I_X - g_m \cdot v_{be})r_o + I_x\left(R||(r_\pi+r_{d1}+r_{d2})\right) $$

If we use

\$Rz = R||(r_\pi+r_{d1}+r_{d2})\$ and \$R_B = r_\pi+r_{d1}+r_{d2}\$

will have

$$V_X = (I_X - g_m \cdot v_{be})r_o + I_x Rz$$

And vbe voltage is equal to:

\$\large v_{be} = - I_X \cdot Rz \cdot \frac{r_\pi}{R_B}\$

So our equation becomes:

$$V_X = \left(I_X - g_m \cdot(- I_X \cdot Rz \cdot \frac{r_\pi}{R_B})\right)r_o + I_x Rz$$

$$V_X = \left(I_X -(- I_X \cdot Rz \cdot \frac{r_\pi}{R_B} \cdot g_m)\right)r_o + I_x Rz$$

$$V_X = \left(I_X + I_X \cdot Rz \cdot \frac{r_\pi}{R_B} \cdot g_m\right)r_o + I_x Rz$$

$$V_X = I_X \left(r_o + Rz \cdot \frac{r_\pi}{R_B} \cdot g_m \cdot r_o + Rz \right)$$

$$R_{out} = \frac{V_X}{I_X} = r_o + Rz \cdot \frac{r_\pi}{R_B} \cdot g_m \cdot r_o + Rz$$

$$ = \frac{V_X}{I_X} = r_o \left(1 + Rz \cdot \frac{r_\pi}{R_B} \cdot g_m \right) + Rz$$

$$R_{out} = r_o\left(1 + R||(r_\pi+r_{d1}+r_{d2}) \cdot \frac{r_\pi}{r_\pi+r_{d1}+r_{d2}} \right)+R||(r_\pi+r_{d1}+r_{d2}) $$

And we are done here.

$$R_{OUT} = 4\textrm{M}\Omega\left( 1 + \frac{1}{\frac{1}{58.5\textrm{k}\Omega} + \frac{1}{255\textrm{k}\Omega}}\cdot \frac{250\textrm{k}\Omega}{255\textrm{k}\Omega} \cdot 0.4\textrm{mS}\right)+\frac{1}{\frac{1}{58.5\textrm{k}\Omega} + \frac{1}{255\textrm{k}\Omega}}=78.688\textrm{M}\Omega$$

  • \$\begingroup\$ Thank you so much for your answer. This makes things so much more clear. I was super confused about too many things. So based on your answer I can conclude that my LTSpice set up is correct and my mistake is in the hand calculations. Now based on your answer, (1) can you please tell me where does the term (1+gm3*ro3) in R_out come from? (2) how are you allowed to combine ( r_pi+ 2/gm) in parallel with R? Since the first is in the base and the latter is in the emitter? Somehow my book says in the chapter notes we have to use the resistance reflection rule. Thank you for your clarification \$\endgroup\$
    – JordenSH
    Commented Jun 19, 2018 at 16:04
  • \$\begingroup\$ Take a look at MY small signal diagram, as you can see rpi3 is in series with (1/gm1+1/gm2) and all of this three resistors are in parallel with R resistor. Do you see it? And don't you recognize this circuit? \$\endgroup\$
    – G36
    Commented Jun 19, 2018 at 17:19
  • \$\begingroup\$ But was wondering on the fact that the series resistances (rpi3+2/gm) are in the base and R is in the emitter, even though they are connected to the same node, they are at two different terminals. My book suggests the use of the resistance reflection rule, if I understood it correctly. So could they be combined together in parallel if they are at two different terminals?? Thank you for clarifying, am here to learn \$\endgroup\$
    – JordenSH
    Commented Jun 19, 2018 at 17:25
  • \$\begingroup\$ @Raykh I update my answer. Can you show me an example with resistance reflection rule? \$\endgroup\$
    – G36
    Commented Jun 19, 2018 at 18:55
  • \$\begingroup\$ I don't know if I can post a picture here in the comments, but the resistance reflection rule pretty much says that if you want to reflect the resistance Ze from the emitter to the base Zb, then Zb=(beta+1)*Ze \$\endgroup\$
    – JordenSH
    Commented Jun 19, 2018 at 21:50

I have been out of college for a while. But what I remember is that resistance looking into the collector, ro, is normally taken to be Va/Ic where Va is the Early voltage and Ic is the DC collector current.

This is why they gave you Va.

So ro = 40V / 10uA = 4MOhm.


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