What role does the capacitor play in an AM demodulating circuit? [closed]

Question

The goal of this question. Demodulate the AM signal.

(Please assume a diode demodulating AM radio of the most imaginable simplicity.)

I am assuming an alternating current sine wave is coming into a wire from the antenna? In order for an envelope to coexist with the sine wave the amplitude is varying I assume?

If the antenna is the correct length then resonance of some fundamental frequency of the EM wave allows the signal to increase in strength in the antenna before it enters the diode? An assumption on my part.

The signal is then converted to DC but since the strength of the signal determines the fluctuations in the "envelope" the voice information is in "duplicate" so the bottom half is cut out of the rectifier , no? From this point I am lost.

The sketch shows a capacitor in parallel. I know what a capacitor does in principle. Is it attempting to maintain the voltage as the current changes since capacitors don't like to have their voltage changed and resist it, no? Please complete my picture to demodulate the AM signal.

Answer 1

We need to start at the beginning.

MODULATION

At the source the music signal (represented by a sine wave) is combined (used to modulate) a high frequency carrier signal at the transmitter.

• The music signal contains low frequencies (as humans hear from about 20hz to 20Khz)
• The carrier signal is much higher frequency (say 100 KHz).

note: the carrier signal is a fixed frequency as each radio station operates at a different frequency.

TRANSMISSION

This modulated signal is now transmitted a long way over the air.

RECEPTION

Somewhere far away, say 20 km, we now need a way of recovering that original music signal from the 'received signal' we receive at the antenna.

To do this we are going to use a circuit called an envelope detector.

The envelope detector is first going to convert the AC signal into a DC signal.

RECTIFICATION

Imagine the received signal (the modulated signal) is a 'messy' sine wave that varies 10V peak to peak (it goes from -5V to +5V).

We are going to ignore the negative half of this 'messy' sine wave.

A diode only allows current to flow one way, and hence has the effect of producing a constant voltage

• the -5V to 5V signal has an average voltage of 0V
• the 0V to 5V signal has an average voltage of 2.5V

DEMODULATION

Now imagine that this 'messy' sine wave is not a 'true sine wave' but is actually made up by a more complicated waveform that contains the original message signal (at low frequencies) and the 'messy' bits at high frequencies.

The envelope detector is now going to remove the high frequencies from the low frequencies. This is done with a lowpass filter (resistor and capacitor). A lowpass filter lets low frequencies pass but stops high frequencies from passing.

Now we have recovered the original signal!

note:

Sometimes a picture can really help understanding:

I have tried to give you the bigger picture and have skipped some details here.

Answer 2

The capacitor is not required for the demodulator to work. However it does improve its operation is two ways:-

1. By charging up to the peak signal amplitude and holding most of that charge between peaks, it increases the audio output voltage.

Without any capacitance after the diode, the average output voltage of a half wave rectifier is Vpeak*0.318. With a suitably sized capacitor the audio output voltage increases closer to the peak value. The result is a stronger audio signal, potentially as much as 10*log(3.18²) = 10dB louder.

In practice audio circuits have some capacitance of their own, which will affect the output to varying degrees depending on the rf frequency and complex impedance of the load. For example a magnetic headphone (commonly used in 'crystal' sets) has high inductance which resonates with stray capacitance at some unspecified frequency. An audio amplifier may be connected through a signal cable whose capacitance depends on length, and the amp's input capacitance may vary depending on the volume control setting. Having a parallel capacitor in the detector circuit reduces the effect of varying load capacitance and inductance, making the audio signal amplitude more stable and predictable.

However if the capacitance is too high then it will fill in the gaps between peaks enough to distort the troughs in the lower part of the audio waveform and reduce the amplitude at higher frequencies. To faithfully reproduce the original modulation there needs to be a balance between filter capacitance and load impedance.

2. The capacitor filters out the rf frequency.

You can't hear rf frequencies but your amplifier might. Without low pass filtering the rf component of the signal is higher than the audio component. If the amplifier doesn't filter this out internally then its power consumption will be increased and it could produce distortion and spurious frequencies (which could be audible). Effective power output will also be reduced because the amp will overload at a lower audio signal level.

Even if the amp can handle this rf leakage without distortion, you now have high power rf signals in the audio circuit, which could feed back into sensitive rf stages to cause distortion or even oscillation. Since only a small amount of capacitance couples rf signals quite well, the amount of feedback could change depending on the position of speaker or headphone wires or the operator's body.

Answer 3

This circuit is a peak detector, with decay rate fast enough to allow the audio/music to appear on output.

^{simulate this circuit – Schematic created using CircuitLab}

Here are some useful waveforms, showing a charge-fast, discharge-slow behavior