RF transmitter explanation

Discussion in 'Wireless & RF Design' started by tmarc, Jul 23, 2017.

  1. tmarc

    Thread Starter New Member

    Jul 23, 2017
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    0
    Hi all,

    I'm trying to figure out the AM radio-transmitter diagram attached below, but can't seem to wrap my head around it completely.
    My current understanding is as follows:

    1) When power is connected, the CB junctions of (Q1,Q2,Q3) are reverse biased. Q2-BE becomes forward biased with a base-current that depends on the resistance of R2. The value of R2 is chosen such to make Q2 (and therefore Q1) non-conducting when Q3 is off.

    2) The DC-surge induces oscillations in the L2-C3(C4) tank circuit, which filter out to leave only the resonant frequency (~200kHz). This gives a damped sine-wave, unless compensated for the losses.

    3) When the keying switch is closed, the 555 sends out a ~1kHz (50%) pulse-train (0-Vin) into the base of Q3. The LC-Q3 combination then acts as an oscillator (due to the gain of Q3). The waveform generated by the LC-oscillator is then "presented" to the base of Q2, and through BE(Q2) also to the base of Q3. Q3 is then triggered to conduct according to the signal set by the LC circuit.

    stx.png
    However, I don't yet have the complete picture. Could somebody please help me clarify the following:

    1) What is the voltage at the base of Q2 relative to ground, when the keying-switch is open? How does this change when the keying-switch is closed?
    Is it so that, when the keying-switch is open, the voltage at the base of Q2 is 0 (because almost all of the 12V drops accross R2 since the remaining path to ground is essentially through 2 BE diodes). And then that, upon closing the switch, Q3 starts conducting, causing R2 and R8 to form a resistive divider, which changes the voltage at the base of Q2?

    2) How EXACTLY do the LC oscillations activate Q2 to conduct? As I understand, transistors are current-controlled devices which start to conduct when CB is reverse-biased and BE is forward-biased (~0.6V<). Then, upon closing the keying-switch, the voltage at the base should be increased (resistive divider through R8?), thereby increasing the base-current into Q2. This should happen "in accordance with the waveshape'' at the LC-tank, but I don't understand exactly how the LC wave influences the base of Q2.

    ps: my EE knowledge is mostly self-taught, so please forgive me if I make some (obvious) mistakes or use the wrong terminology, thank you
     
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,184
    1,119
    Hello,

    Did you actually build this circuit or just simulate it? I have having trouble believing it actually works right.
    What is that "Translator" there and how is it positioned relative to the inductor L2?

    Assuming it does work, that would mean that L1,L2, C3 cause a higher frequency oscillation, and as the modulating signal turns Q3 on and off, that causes the oscillation amplitude to be higher or lower, and that changes the drive level getting to Q2.

    That's a simple analysis, and it is hard to get very specific with this circuit because we'd have to know the self inductance of all inductors and the mutual inductance between L1 and L2 to understand it better.
     
  3. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
    6,385
    1,479
    Following the dots on L1/L2 it looks like the feedback is negative o_O
    Also, the RFC seems odd. I would expect the left end of L1 to be low impedance to ground but the RFC makes it high impedance o_O (again).

    But then again I have no idea what a 'translator' is in this context.
     
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