# How does this AM Modulator work?

I have been given a question and tasked with identifying the various components. However, this circuit doesn't really seem to make sense to me.

Error: RC RE inverted

simulate this circuit – Schematic created using CircuitLab

It is given that the input signal is a sin wave that goes from 12v to 8v (4v PP). I understand that for modulation, you essentially have to generate a carrier wave that is mixed in with the input. Op amps won't work for such a high frequency so we have BJT's instead.

1. For a sin carrier wave to be generated, we need a band pass filter at the output of the amplifier fed back to the inpit? But from what I see, that seems to be a butterworth low pass filter

2. I assume the LC network is going to cause some drop in gain. So assuming I calculate it, must I set the gain for the BJT amplifier to match the drop of the LC network?

I forget the name of that one, but that circuit is an oscillator. (Is this a commonly accepted variant of the 3R, 3C phase-shift oscillator maybe?)

It becomes an AM modulator because its power supply is your input signal - so obviously its output Amplitude will be Modulated by the input signal.

The point of that LC network is to provide a 180 degree phase shift at the frequency you want to oscillator to work at.

The transistor in this circuit is acting as the amplifier in the circuit so yes you are correct that need to calculate its Rc & Re resistor values to produce a gain which compensates for the attenuation in the LC stage at your frequency of interest. (You've labeled Rc & Re backwards btw).

• I still don't understand though, that's a low pass filter either way right. Inverted the frequency response of a low pass graph does not give a sin wave. Only the inverted frequency response of a band pass filter would give a sin wave – Raaj Nov 1 '14 at 16:13
• For one thing, I see two capacitors in series with the feedback loop, so they will tend to high-pass the signal. More importantly, it's the phase shift across the loop that's important. The circuit will oscillate at the frequency where the phase shift is 180 degrees. If the amplitude is reduced due to filter effects, the gain of the BJT makes up for the loss. – gbarry Nov 1 '14 at 18:16
• Can you explain to me about why the 180 degree phase shift in the signal is important? – Raaj Nov 2 '14 at 5:08

The LC network along with those by-pass capacitors constitute a 180o phase shift at some certain frequency. This frequency is the oscillator's oscillating frequency. The magnitude of the output signal is directly related to the supply voltage level, which is given as input.

Therefore, the oscillator oscillates at a fixed frequency and generates a sinusoidal "carrier" waveform. The input signal affects the envelop of the output waveform.

Yes, the LC network itself will behave a low-pass filter. However, if you pay attention to those by-pass capacitors as well, all together they form a band-pass scheme.

By the way, this is not an "AM transmitter"; it won't transmit anything without an antenna. This is only an "AM waveform generator circuit".

Why don't you redraw your circuit in CircuitLab, and also add a simulation of it?

• can't seem to figure out how to run the simulation though. it says bode plot not found – Raaj Nov 2 '14 at 8:06