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.
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
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.
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