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I'm designing an AM detector that uses a 2N3904 transistor in common collector configuration. Pardon me if my design's silly because I'm still quite a novice with using BJT transistors as amplifiers, but after a bit of calculations and a lot more trial and error, this is what I've been able to come up with:

enter image description here

In my simulation, I made the input signal (V1) a 1 MHz sine wave with an amplitude of 15 mV modulated at 1 kHz (imported from a CSV file), as shown in the plot below:

enter image description here

I measured the output signal and while it does seem to have the right frequency of 1 kHz and be pretty free of the carrier frequency of 1 MHz, the signal only has an amplitude of about 0.8 mV. That's less than a tenth of the magnitude of the input signal!

enter image description here

Would anyone be able to recommend any modifications that I should make to my circuit so that the output signal has a larger magnitude? (While still being "smooth" and having the right frequency, of course.)

kene02
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  • Are you only concerned with 1kHz? or perhaps audio range up to 20kHz? R2C2 product is chosen to reproduce the highest frequency expected, while suppressing the 1MHz carrier. Your R2*C2 product is too high, even for 1 kHz. However, it does suppress the 1MHz carrier very well. – glen_geek May 13 '23 at 15:15
  • @glen_geek Yes all the audio frequencies (around 20 Hz - 20 kHz) should be reproduced – kene02 May 13 '23 at 15:20
  • You have the transistor biased for a collector current of about \$2 \mu \mathrm A \$ -- the 2N3904 isn't really designed for collector currents below 1mA. It'll go down that far, but the gain and bandwidth will suffer. Why the low current? Are you trying for a super-high impedance input? – TimWescott May 13 '23 at 15:23
  • "1 kHz sine wave with an amplitude of 15 mV modulated at 1 MHz". It's the other way around: 1 MHz sine wave with an amplitude of 15 mV modulated at 1 kHz. – Davide Andrea May 13 '23 at 16:37
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    @Davide Andrea Thanks. I stand corrected. – kene02 May 14 '23 at 00:48
  • @TimWescott Thanks for letting me know. I found through trial and error that the higher the emitter resistance was, the better the circuit suppressed the 1MHz carrier signal and I sort of used trial and error again to find biasing resistor values that wouldn't cause the output signal to clip. – kene02 May 14 '23 at 00:57
  • @glen_geek By the way, correct me if I'm wrong, but from your comment, I'm just going to assume that you're telling us to use the low pass filter cut-off frequency equation $$f_c=\frac{1}{2{\pi}RC}$$ to evaluate the value of R2. In that case, assuming we want fc = 20 kHz and C2 = .01 uF, R2 would have to be 796 ohms = 820 ohms (nearest E12 value). When I made R2 = 820 ohms in my circuit, it increased the amplitude of the output signal to about 9 mV but now it doesn't filter out the 1 MHz carrier wave. – kene02 May 14 '23 at 06:40
  • You can always add an additional low-pass filter after the detector: choose a compromise between detector efficiency & carrier attenuation. Perhaps your expectations are too high - with 15mV signal, silicon junctions (600mV) are operating in square-law territory: at these low amplitudes, demodulated output is tiny. Amplify 1MHz signal first, then demodulate. Detector efficiency will improve greatly. – glen_geek May 14 '23 at 12:55

2 Answers2

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Would anyone be able to recommend any modifications that I should make to my circuit so that the output signal has a larger magnitude? (While still being "smooth" and having the right frequency, of course.)

Expectations are high: for such tiny AM source signal (15mV), a silicon transistor junction is not an efficient detector no matter how you bias it. It is operating in its square-law region for such small signals...if signal amplitude is cut by half, the detector's output reduces to one-quarter.

For more efficient AM detection that is also more linear, first boost the amplitude of the signal presented to the detector. Shown below is a demo AM detector simulation as input signal levels are varied. Perhaps not optimized, but it demonstrates how square-law hurts you... schematic one transistor AM detector


 AM 1MHz in    detector_out
     
     7.5mV        2mV
     15mV         7.8mV
     30mV         26.6mV
     60mV         75.6mV
     120mV        185mV

Doubling input from 7.5mV to 15mV causes output to nearly quadruple from 2mV to 7.8mV. This is square-law in action.
At the higher end, doubling input from 60mV to 120mV causes output to increase by a factor of 2.45. This is somewhere between a square-law detector and a linear detector...a linear detector would increase by a factor of two.

TLDR: A proper AM detector requires large input amplitude. One also must consider what load impedance must be driven at detector output: ideally, the resistance that the detector drives is high.

glen_geek
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  • Thanks. I'll try putting a common emitter amplifier before the detector to bump up the input signal to an amplitude of about 100 mV. – kene02 May 18 '23 at 13:11
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Try biasing Q1 near to cutoff .An easy way on the sim is to greatly reduce R3 like starting with 68K and optimise this .In real life I use a diode connected BJT to give better thermal stability.R2 could also come down or C2 could come down if slope distortion is an issue.

Autistic
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