0
\$\begingroup\$

I'm trying to solve this problem where I find Vout/Vin if rπ if infinite. I also have to conside early effect. When I try to solve it, I get Vπ = 0 and that makes gm zero, a bit confused on that.

enter image description here

\$\endgroup\$
4
  • 3
    \$\begingroup\$ I think they only mean to suggest that \$\beta \to \infty\$ so that you don't have to worry about base recombination current. I don't think they mean to suggest anything else by it. \$\endgroup\$ – jonk Mar 24 at 4:08
  • \$\begingroup\$ Useless comment by me, I frankly hate common base (or common gate, same thing). IIRC however gm is not relevant in this case \$\endgroup\$ – Lorenzo Marcantonio Mar 24 at 7:06
  • \$\begingroup\$ Just a stylistic comment: it is "Early effect", with a capital "E", and not early effect. An early effect was present at the beginning, but it does not appear any more. The Early effect is a phenomenon named after James M. Early. \$\endgroup\$ – Horror Vacui Apr 14 at 21:32
  • \$\begingroup\$ What is the problem you are trying to solve? There is only a picture, but it is not clear what you are trying to solve. I am not sure how you got to the conclusion that VBE=0. It can not be. Vb is fixed, and Vin varies. This variation will generate a gm*vin current through the transistor. \$\endgroup\$ – Horror Vacui Apr 14 at 21:35
1
\$\begingroup\$

I don't care who is explaining gm in transistors circuits because it is incorrect. That is why you are stumped.

Common base circuit, is voltage amplifier that has the current gain (approx) as 1 but is a calculation of Alpha Trans-resistance Factor (instead of Beta) is explained:

formula

Of course you looking for voltage gain, it has no trans conductance because its a factor of a voltage drop across the emitter and collector resistors in parallel when you look at the AC analysis equivalent.

enter image description here

enter image description here

Of course, the gain formula's expression:

enter image description here

Images from: Electronics Tutorials - Common Base Amplifier

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Only for DC voltage, the CB stage voltage gain is equal to RC/RE. But this is not true for an AC signal. So before you give a bad answer next time try to analyze the AC equivalent circuit yourself. Don't rely on the "knowledge" you find on the web. \$\endgroup\$ – G36 Mar 24 at 7:48
  • \$\begingroup\$ @David M., Your formula is a very rough approximation only. It applies for RE>>ré only. The correct expression is Av=RC/(RE+ré). \$\endgroup\$ – LvW Mar 24 at 8:23
  • \$\begingroup\$ @LvW But this is not the CE amplifier. In this case, when driven from the ideal voltage source the gain is \$\Large A_V = \frac{R_C}{r_e} \times \frac{\beta}{\beta +1} \$ \$\endgroup\$ – G36 Mar 24 at 8:41
  • 1
    \$\begingroup\$ G36, thank you for correction. In my comment, I have assumed that the resistor RE is between signal input and emitter node (as indicated by the approximation Av=RC/RE in the above answer). You are correct - for the shown circuit, the gain is practically Av=gm*Rc=Rc/ré. \$\endgroup\$ – LvW Mar 24 at 9:01
  • \$\begingroup\$ G36, Just because I find and use some one else's picture don't mean anything. This calculation with trans conductance in transistors is actually new to me, as the formulas I was taught and use are the ones above over 30 years ago. Technically, yes its similar, because trans-conductance is an inverse function or trans-resistance.That is why that formula kind of works. But the transistor formula is the correct one to use because of the trans-resistance properties that wouldn't be included into the modified trans conductance formula. \$\endgroup\$ – David Mikeska Mar 26 at 17:27

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.