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I would like to have the envelope from a signal using a simple circuit composed of a diode and a low pass filter.

I know :

*Tam > t >> Tcarrier

*The diode must be a germanium diode because of the Vthreshold

I use:

*A germanium diode

*R = 1k

*C = 4.7nF

The wave is generate from a GBF :

*fcarrier = 1MHz *fam = 20 kHz

My circuit:

enter image description here

I took a capture from the oscloscope measuring the signal accross the resistance. My signal is shown bellow :

enter image description here

My questions are:

  • How can I have a clean modulating signal?
  • Some books talk about a High pass filter, why it could be necessary? (I tried and it's not better)
  • The circuit is currently designed for a specific carrier but if my carrier has a defined range (in my case I will have a range between 1MHz to 7MHz), how could I adapt the circuit (or is there another circuit)?
  • Is the type of the capacitor change something?

Thank you,

Farad

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  • \$\begingroup\$ No schematic and no idea where you are measuring these traces. \$\endgroup\$ – Brian Drummond Jan 5 '16 at 13:23
  • \$\begingroup\$ You are right, I updated the question \$\endgroup\$ – uFarad Jan 5 '16 at 13:40
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The frequency ratio between your carrier and your modulation is only 50:1, which is less than 6 octaves (less than 2 decades). You'll need a much better filter than a single-pole RC network to effectively separate the two.

The output of the diode alone is a half-wave rectified version of the original signal. While the average value of the original signal was 0, the average value of the rectified signal is proportional to the envelope of the signal. It is the job of the output filter to separate this average value from all of the other higher-frequency components.

To eliminate obvious distortion on the oscilloscope, you'll need to attenuate the carrier voltage by a factor of 100 or more relative to the baseband signal, which is a power ratio of 40 dB. To accomplish this within a 50:1 frequency ratio, you'd need a filter with a minimum slope of

$$\frac{40 dB}{\log_{10}(50)} = 24 dB/decade$$

This assumes that the cutoff frequency of the filter is exactly equal to the highest frequency of the baseband signal, which means that that frequency will also be attenuated by 3 dB. The point is, you'll need at least a two-pole filter, and you'll need something more sophisticated than a resistor and capacitor connected directly to the diode, which creates asymmetries in the time-domain response because of its varying source impedance.

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  • How can I have a clean modulating signal?

The main problem is that your RC filter is too "heavy" and you are getting slew rate limiting on the lower half of the demodulated signal. Try lowering R or limiting your maximum modulation frequency to less than 20 kHz. Then try adding a further op-amp low pass filter to remove more carrier leak-through.

It's also possible that the diode you are using is not really that good - the waveform seems to be suffering from extended diode reverse recovery time so, what diode are you using?

  • Some books talk about a High pass filter, why it could be necessary? (I tried and it's not better)

A high pass filter sounds like a band-aid to a poorly designed demodulator filter.

  • The circuit is currently designed for a specific carrier but if my carrier has a defined range (in my case I will have a range between 1MHz to 7MHz), how could I adapt the circuit (or is there another circuit)?

Demodulation is normally done after an intermediate frequency stage (try researching I.F. or ask a new question about this because it is a large subject that I have just introduced). Because demodulation is done on an I.F. signal you don't need to adjust the demodulator at all BUT I note that your target band is 1MHz to 7MHz therefore you cannot realistically use standard I.F. techniques and you might have to develop a synchronous demodulation method (something else to look-up because it is also too big to cover here along with the basic stuff you are asking about).

  • Is the type of the capacitor change something?

I have no idea what you mean.

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  • \$\begingroup\$ Might the "high pass filter" be on the output to remove DC bias (i.e. a coupling cap) ? \$\endgroup\$ – Brog Jan 5 '16 at 14:04
  • \$\begingroup\$ @Brog it all depends where the high pass filter is (unspecified by OP). \$\endgroup\$ – Andy aka Jan 5 '16 at 14:46
  • \$\begingroup\$ Thank you very much for your quick answer and sorry for my late comment, I tried to lower the resistance which doesn't change the problem... I use a TFK BYT 52 diode (from a circuit I found in my house) Do you have any advice to command a diode? (caracteristics, material etc) What does I.F mean please (English is not my language, I don't understand this acronym) ? I will post a new question but I would like to done some research about it before! About the capacitor I was wondering if ceramics or chimical capacitor will change something? Thanks again for your help! \$\endgroup\$ – uFarad Jan 6 '16 at 17:49
  • \$\begingroup\$ You need to use a diode a lot faster than the BYT52 if your carrier is 1MHz - the reverse recovery time is 200ns and that is what is causeing all the spiky lumps on top of the waveform - try a BAS16 - it has a Trr of 6ns. 1N4148 is about the same so that is also a choice. Try any low size schottky diode also. This solves half the problem. When you have got this show a new picture of the waveform. \$\endgroup\$ – Andy aka Jan 6 '16 at 21:18
  • \$\begingroup\$ Thank you very much for you answer, I'll try and then show you the new waveform! \$\endgroup\$ – uFarad Jan 21 '16 at 9:24

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