1
\$\begingroup\$

I am trying to find the voltage gain of the circuit below. I drew the small-signal model for it (shown below). I know the formula for finding the voltage gain of a common-emitter BJT with and without degeneration (also shown below). The problem is, the formula applies when Vout and Vin are connected to the same transistor, but the circuit I'm dealing with has the Vout on the transistor Q2 and Vin on the transistor Q1. According to the small-signal model, I've found the voltage gain as gm1*(RC//Rπ2). I've used the formula for the common-emitter BJT without degeneration, but Q2 has degeneration and the resistor for Vout is RD. I'm not sure if the voltage gain I found is correct. I would really appreciate it if I can get an answer on what I should do to find the voltage gain for all BJT circuits in general. Thank you in advance.

Circuit

enter image description here

Small-Signal Model

enter image description here

Formulas for Finding the Voltage Gain of Common-Emitter BJTs

enter image description here

\$\endgroup\$
12
  • \$\begingroup\$ Hint: The top transistor isn't in a common-emitter configuration. \$\endgroup\$
    – Hearth
    Jun 12 at 18:33
  • \$\begingroup\$ Is it in common-base configuration, since there is no voltage applied to its base? \$\endgroup\$
    – Kalamakra
    Jun 12 at 18:46
  • 1
    \$\begingroup\$ Yes, and together these make up a common circuit called the cascode, which may be a useful search term. Notice what the configuration means for the collector voltage of the input transistor, and what that means for the configuration's bandwidth. Q2's base really should be tied somewhere other than Vcc, though. (of course, Q1 should be biased to linearize the gain and protect it from overvoltage, or Vin should be a current, so this is clearly meant to be a didactic circuit more than a practical one) \$\endgroup\$
    – Hearth
    Jun 12 at 19:14
  • \$\begingroup\$ I've looked at our course book and found nothing similar to this circuit (Vout and Vin on different transistors). I have no idea what to do to solve a question like this. I also thought the voltage gain as gm1*(RC//Rπ2//RD). Could you tell me what I should do to find the voltage gain (if I am wrong that is) because our course book has different solutions for cascade circuits. \$\endgroup\$
    – Kalamakra
    Jun 12 at 19:22
  • \$\begingroup\$ Cascode, not cascade. It's not too common nowadays outside of RF circuitry and other things that need the high bandwidth it provides, but it used to be much more common back when analog circuit design was more prevalent. \$\endgroup\$
    – Hearth
    Jun 12 at 19:26
0
\$\begingroup\$

Your theoretical circuit can be seen as a two-stage amplifier.

The first stage is nothing more than a common-emitter amplifier. And the second stage is a common-base amplifier.

The first stage gain can be found using this equivalent circuit:

enter image description here

As you can see the input impedance of a common-base amplifier is \$r_{\pi 2}|| \frac{1}{g_{m2}}\$

output resistance of BJT

Therefore, we can solve for a voltage gain:

$$V_{C1} = - I_{C1}(r_{\pi 2}|| \frac{1}{g_{m2}}||R_C) = - V_{in}*g_{m1}*(r_{\pi 2}|| \frac{1}{g_{m2}}||R_C)$$

Thus,

$$\frac{V_{C1}}{V_{in}} = -g_{m1}*(r_{\pi 2}|| \frac{1}{g_{m2}}||R_C)$$

Now the common-base stage gain can be found using this equivalent circuit:

enter image description here

As you can see \$V_{C1} = V_{in}\$ thus:

$$V_{OUT} = -I_{C2}*R_D = - (-V_{in})*g_{m2}*R_D = V_{in}*g_{m2}*R_D$$

And finally we have:

$$\frac{V_{OUT}}{V_{IN}} = g_{m2}R_D$$

And this is the end of this simple analysis.

\$\endgroup\$
15
  • \$\begingroup\$ I am sorry, I know I am taking your time with this. On our midterm exam we got this question and I don't think I got the right answer. So, I am trying to understand the concept well so if we get a similar question on our final I won't have any problems solving it. Now, what I don't understand is you said "... common-base amplifier is rπ2||(1/gm2)" What about RC, shouldn't that be in parallel with the output resistance as well making the output resistance RC||rπ2||(1/gm2)? Furthermore, how did we find (1/gm2)? Also, thank you for your time, I really do appreciate it. \$\endgroup\$
    – Kalamakra
    Jun 19 at 11:31
  • \$\begingroup\$ Also another question, why was RC removed in the second figure (common-base equivalent circuit)? Shouldn't it be common-base with degeneration? \$\endgroup\$
    – Kalamakra
    Jun 19 at 11:49
  • \$\begingroup\$ @Kalamakra What is the input impedance of a common-base amplifer? Do you know how to find it? \$\endgroup\$
    – G36
    Jun 19 at 15:00
  • \$\begingroup\$ It is 1/gm but in our case, shouldn't the input impedance for the common-base be RC||1/gm? \$\endgroup\$
    – Kalamakra
    Jun 19 at 15:13
  • \$\begingroup\$ Well, the exact solution shows that \$ \Large r_{in} = \frac{1}{g_m} || r_\pi = \frac{1}{ g_m + \frac{1}{r_{\pi}}} = \frac{r_{\pi}}{g_mr_{\pi} +1} = \frac{r_{\pi}}{\beta +1} = r_e\$ \$\endgroup\$
    – G36
    Jun 19 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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