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Cant figure out transistor operation in this cct

Discussion in 'General Electronics Discussion' started by davenn, Aug 25, 2012.

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  1. davenn

    davenn Moderator

    Sep 5, 2009
    hi guys,
    will be the first to admit that my transistor theory is a bit weak at times....

    have been sorting out a piece of circuitry for another guy and I cant quite explain the circuit operation of 2 of the transistors.....


    The circuit for for power control of an RF amplifier, a detected and rectified voltage comes down to the base of Q211 from a reflected power sensor. and via Q211,212 and 213 the corrent to the base of Q1 is controlled. This controls the voltage/current from the emitter of Q1 up to the final output power transistor (not shown).
    Under normal conditions, where there is little or no reflected power, Q211 isnt conducting Q212 is conducting via the current flowing from the 8V TX rail via R230 and 231 to its base. Q213 is fully conducting allowing Q1 to fully conduct and supply full voltage to the final transistor.
    What I cant fully explain is the interaction of Q212 and 213.
    when Q 212 is conducting so is Q213 and visa versa .... but how is Q212 turning Q213 on and off ?


    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Q212/213 form a Sziklai pair and turn on and off a bit like a darlington. Ic for Q212 is Ib for Q213.

    Q212 is biased on by R23, with Q211 robbing it of base current when it is turned on.
  3. duke37


    Jan 9, 2011
    The gain of the pair will be R233/R232 +1 = 47
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Does that qualify as a Sziklai pair? It certainly works in a similar way, but with the presence of R233 and R232 I wouldn't have thought it would fit that definition...?

    I would describe the circuit as a DC amplifier, with its input at Q212 base, and output at Q213 collector (which goes straight to Q1 base). It is non-inverting, and has a gain of 1 + (R233 / R232), as duke37 pointed out, which is 47.8. It also has an input offset of about 0.6V, from Q212's Vbe drop, which I will assume is exactly 0.6V.

    Q212's collector current biases Q213 ON, and Q213's collector voltage is applied to Q212's emitter via a voltage divider (R233 and R232) and tends to bias Q212 OFF. This is local negative feedback, and it gives the circuit voltage gain.

    Start with powering up in the normal state, where Q211 is off and Q212's base will be pulled up towards the +8VTX rail. This makes Q212 conduct heavily, which causes Q213 to conduct heavily, which makes Q213's collector voltage rise, biasing Q1 on as well.

    Even when Q213 collector has reached its maximum of +13.8V (or nearly, because of the small Vce drop in Q213), Q212 emitter will only sit at 0.23V (because of the voltage divider action of R233 and R232), and Q212 will be heavily biased ON by current through R231. Q212's base voltage will rise to about 0.83V (the emitter voltage plus the base-emitter drop), but there will still be plenty of current through R231 to keep both Q212 and Q213 saturated. Q1 base will be at say 13.7V, and the transmitter will run at maximum power.

    Now assume the reflected power increases, and biases Q211 ON. Q211's collector voltage drops. Let's assume it drops down to 0.7V.

    Now Q212 will not have enough bias to stay turned ON, because its emitter was 0.23V and its base is now only 0.7V. So Q212 starts to turn off, and so does Q213. At a certain point, an equilibrium will be reached where Q212's emitter is exactly one Vbe drop below its base. In this case, that will be at 0.1V. At that voltage, Q213's collector voltage will be 4.78V and the transmitter will run at reduced power.

    Those are just two scenarios to show how the circuit works. The circuit continuously regulates its output (to Q1 base) according to the inverted reflected power signal at Q211 collector, to ensure that the transmitter power is reduced (fairly aggressively) when the reflected power starts to get too high.

    Because reducing the transmitter power will also reduce the reflected power, an equilibrium will be reached, and so the reflected power is limited.

    The Q212/213 gain stage has a fairly high voltage gain, so it is sensitive to a small voltage range at Q211 collector. If there is no antenna mismatch, reflected power will be low, Q211's collector will be high, and the Q212/213 circuit saturates and runs the transmitter output stage at maximum power. In fact as long as Q211's collector voltage is more than about 1V, the Q212/213 circuit will be saturated and Q1's base will be almost 13.8V.
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Probably not, but close enough that you can infer the general operation.

    At least, that's what I think...
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