I am trying to solve the following question related to cascode circuit. Where T5 has Vbe=0.7V

enter image description here

I have answered the questions as follows.

- Explain how the circuit works? The circuit has a DC operating point which gives the AC parameters of the transistors such as the transconductance. A AC current will flow through the transistor and the output voltage will be given by the voltage dropped at R4. The input impedance of the circuit is high but it's not depending on the bias current due to the gate current is really small, thus the rd can be regard as infinite.

- Explain the FET T3 is biased at the red point in figure 3 characteristics. how do I explain this, does this has something to do with calculating the Q point in the load line?

- Calculate the value of R3 using the data from figure 3.

Vgs = Id * R3 there fore,

R3 = Vgs/Id = 0.4V/10mA = 40 ohm

**- **Determine in a graphical way from the characteristics from figure 3 the transconductance of T3?****

gm = delta Id/delta Vgs (keeping Vds constant)

gm = (13.8-6.3)/(0.2-0.6)

gm = 7.5mA/0.4 = 18.75 mA/V

- Calculate the voltage gain of the circuit?

Vout = R4 * id

We know, id = gm * Vgs = gm*(vin - id*R3) = (gm *vin)/(1+R3*gm)

Replacing the value of id we get,

Vout = (gm*R4*vin)/(1+R3*gm)

Vout/vin = (gm * R4) /(1+R3*gm)

Av = (18.75mA/V * 3.9K ) / (1 + 40*18.75mA/V)

 = 41.78

- Judge the power supply rejection of circuit 3? This is a small signal analysis so i think,

PSRR = 20log ((delta Supply / delta Vout)*AV)

but i don't have the supply voltage and Vout so how can I calculate PSRR?


1 Answer 1


The BJT (T5)is acting as an emitter follower. With 6V8 at its base (Zener D1, R5) the emitter will be 0.7V lower giving 6.1V. This is the Drain-Source voltage of T3 and is shown as the red square on the graph giving Ids as 10mA.

  • \$\begingroup\$ Thanks JIm for the explanation and what about the other's are my explanation and calculations right? \$\endgroup\$ Sep 28, 2014 at 23:11
  • 1
    \$\begingroup\$ @Jim, T5 is acting as an emitter follower in setting up the dc bias, but it's probably worth noting that as far as the signal path is concerned, T5 is in common base configuration. \$\endgroup\$
    – The Photon
    Sep 29, 2014 at 16:54
  • 1
    \$\begingroup\$ T5 is "acting as an emitter follower" and has a signal output at the collector node? I ratherr think, it is in common-base configuration. \$\endgroup\$
    – LvW
    Oct 28, 2014 at 17:45
  • \$\begingroup\$ ... and that is why I say acting rather than is because it is an easy way to explain why you get the bias (DC) operating point at the drain as 6V. From an AC (signal) point of view it is a common base configuration. \$\endgroup\$ Feb 4, 2015 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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