In the circuit below, base of Q2 is grounded.

Even though base of Q2 is grounded in the circuit, the ac equivalent circuit(T model) seems to suggest that one end of the current source Q2 is also grounded(Dotted region)!

Does this mean \$r_e+R_E\$ are shorted by a wire?
And does the output \$v_{out}\$ equal the voltage across the current source?

I know both above statements are wrong, but I don't have a good feeling why. Any help?

In short: Where is base in T model? Looks base,emitter,collector are all meeting at a point. How to interpret T model correctly? Can't we use the T model "as-it-is" in circuit analysis?

enter image description here

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  • \$\begingroup\$ Why shorted? I do not see it. \$\endgroup\$
    – G36
    Commented Feb 20, 2022 at 17:05
  • 1
    \$\begingroup\$ Now in your case, the current will follow via in the opposite direction \$r_e2\$ ( From Vin via \$r_{e1}\$ and into \$r_e2\$ ) the base current will be the difference between IE and IC current. But if IC = is 0A then IE = IB. This will Turn_ON the CCCS so the collector current will start to flow in the opposite to the marked direction. Do you see it? \$\endgroup\$
    – G36
    Commented Feb 20, 2022 at 17:31
  • 1
    \$\begingroup\$ Yes, but the IE current is the current that controls IC current. So as long as IE is flowing IC also must flow too. \$\endgroup\$
    – G36
    Commented Feb 20, 2022 at 17:41
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    \$\begingroup\$ i think so... \$\frac{i_cR_c}{i_e(r_e+r_e)}\approx \frac{R_c}{2r_e}\$ Thanks again it is clear now xD \$\endgroup\$
    – across
    Commented Feb 20, 2022 at 17:57
  • 1
    \$\begingroup\$ @across You can see what I develop as the large-scale answer here. The factor, \$\tanh x = x-\frac13 x^3+\frac{2}{15}x^5-\frac{17}{315}x^7+...\$, can be simplified to \$\tanh x = x\$ in the small scale. Knowing that \$r_e^{\:'}=\frac{I_{_\text{E}}}{V_T}\$ and that \$I_{_\text{E}}=\frac12 I_{_{R_{_\text{E}}}}\$, your result follows. \$\endgroup\$
    – jonk
    Commented Feb 20, 2022 at 18:59

1 Answer 1


Let me show you how we can find the voltage gain for a CB stage (common-base amplifier).

The small-signal model will look like this:


simulate this circuit – Schematic created using CircuitLab

We can see that:

$$V_{IN} = I_{IN}* r_e $$

And that: $$I_B + I_{IN} + I_C = 0$$ $$I_B = - I_C - I_{IN} =- \beta I_B - I_{IN}$$

So, the base current is equal to:

$$I_B = -\frac{I_{IN}}{\beta + 1}$$

And the output voltage is

$$V_{OUT} = - I_C*R_C = -(-\frac{I_{IN}}{\beta + 1})* \beta * R_C = I_{IN}R_C * \frac{\beta}{\beta +1} $$

So, the votlage gian is:

$$\frac{V_{OUT}}{V_{IN}} = \frac{I_{IN}R_C * \frac{\beta}{\beta +1}}{I_{IN}* r_e} = \frac{R_C}{re} \frac{\beta}{\beta +1}$$

No back to your circuit.


simulate this circuit

This time I used a slightly modified T-model. So that you can see that the emitter current controls the collector current \$I_C = I_E \times \alpha = I_E \times \frac{\beta}{\beta + 1}\$

Also notice that \$Q_1\$ is working here as an emitter follower and \$Q_2\$ is a common-base amplifier. Therefore we can find the voltage gain by inspection.

$$\frac{V_{O}}{V_{IN}} = \frac{R_E||r_{e2}}{r_{e1} + R_E||r_{e2}} \times\frac{R_C}{r_{e2}} \frac{\beta_2}{\beta_2 +1} \approx \frac{R_C}{2r_e} $$


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