Cascode Amp questions

  1. Hello all. I have been looking into RF amps. I have come across the following cascode amp (from ARRL handbook 2014, p 12.10, FIG. 12.11. Thanks to Baluncore for the great reference). Here is a simplified version made with Qucs: [​IMG]

    My questions are regarding the capacitors, C1 and C2. Why are they there?

    I know that if we were to have only a common source (CS) amplifier the bypass capacitor C2 will provide a larger gain for the small signal required. But this is not the case here, if I am not mistaken, in a cascode configuration, the CS phase does not provide amplification, it provides high input impedance, right? If yes, then why do we care about the small signal gain at that point. Is it not the common gate stage that provides amplification?

    I have no idea why C1 is there! I know R1 and R2 set up the Voltage bias at T2, I also know that for small signal, the gate will look like a ground, what is the point of C1?

    Thank you.
     
  2. jcsd
  3. NascentOxygen

    Staff: Mentor

    It is not "common source" if the source is not "common" or at ground to both input and output, so the source is connected to ground via that capacitor. Similarly, with the biasing at gate of T2 it can't be common gate unless the gate is grounded at signal frequencies. The gate won't 'look like a ground' if there are resistors in the way.

    Edit: you are probably going to say because the gate draws no current, what does it matter if there is series resistance. I'd think you don't want any extraneous signal to be induced on the wiring to the gate, and any internal leakage signal making its way to the gate should be shunted to ground via very low impedance so it can't create its own gate voltage.
     
    Last edited: Aug 8, 2014
    1 person likes this.
  4. Amazing, that makes a lot of sense. Thank you so much.

    Edit: Thanks for the "edit." I was not wondering about it but it helps that you mentioned it.
     
  5. Hey again. Sorry for resurrecting this thread but it isn't that old and I have a question pertaining to the same circuit. So after some simulations, I got the output to look like it should but I have two questions. Here is the circuit I used in LTSpice:
    [​IMG]

    Q1] Why is it that J1 doesn't have to be biased? For a common source amplifier one would have to bias the gate appropriately to not have signal distortion. At first I tried having the signal ride on DC but had weird results (it would alter my input frequency!). Why is it that J1 does NOT have to be biased?

    Q2] What is the job of L1 and R2? I know at DC L1 will seem to be a short, and it will not see R2. I also know at certain frequencies L1 will seem to be a very high impedance and only see R2. What is the point of this?

    Thank you.
     
    Last edited: Aug 11, 2014
  6. Baluncore

    Baluncore 3,022
    Science Advisor

    You have two distinct paths. One is the DC bias, the other is the RF signal.
    Those paths are separated by the L and C components.

    J1 is a depletion mode FET. The gate is DC biassed to ground by R1.
    The current bias is set by the negative gate voltage divided by the value of R5.
    C1 gives the V3 input AC signal on the gate something solid for the source to work against by hiding the DC value of R1 behind the low impedance of C1 at RF.

    I presume your RF output is appearing across R6.
    L1 is an RF choke that isolates the RF output signal from the low impedance power supply bypass cap C5.
    L1 also carries the DC bias current around R2 and so reduces the need for a higher supply voltage.
    R2 sets, or limits, the output stage RF gain.
    R2 also prevents self resonance of L1, RF chokes were once often wound on the resistors.
     
    1 person likes this.
  7. Great information. Thank you for answering my question and more.

    The only thing that I have a hard time understanding is when you say: "The current bias is set by the negative gate voltage divided by the value of R5."

    Maybe if I work out the math of the circuit I will understand better what you meant... Thanks again for all the help.
     
  8. Baluncore

    Baluncore 3,022
    Science Advisor

    The Gate-Source voltage of an MPF102 is specified at VDS = 15 Vdc, ID = 0.2 mAdc as being between −0.5 and −7.5 Vdc.

    The gate of the depletion mode FET is therefore more negative than the source, so a source resistor can make a very simple DC current regulator that sets the cascode chain bias current.

    The gate-source voltage appears across R5 and sets the bias. Gain is then approximately R5 / R2.
     
  9. Interesting, so if J2 is biased correctly it is seen as a short (a wire) looking into its source? Hence the gain of approximately R5/R2?
     
  10. Baluncore

    Baluncore 3,022
    Science Advisor

    Sorry, I got that wrong, twice. Without C1 the RF gain would be controlled by the R2 / R5 resistor ratio.
    But with C1 present the RF gain is a function of the Drain current to Gate-Source voltage relationship, (Gm), and the value of R2.

    J2 is a source follower, the gate is fixed so there is little voltage variation on the source.
    That prevents voltage changes on the drain of J1.
    So the “miller” capacitance between the gate and drain of J1 is not such a frequency limiting factor.

    Cascode isolates the J1 gate input from the J2 drain output.
    That is why the cascode arrangement is used. It works well at VHF and UHF.
     
  11. Thank you so much!
     
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