Capacitor and resistor in parallel as a shunt

Discussion in 'General Electronics Chat' started by atldave, Aug 24, 2010.

  1. atldave

    Thread Starter New Member

    Jul 26, 2010
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    I need help understanding this part of a circuit.

    In the diagram I uploaded, you'll see an AM demodulator. After the diode is a resistor and a capacitor as a shunt for getting rid of the carrier, the cap will discharge through the resistor and the resistor affects the time value of the cap. I know that much.

    I also calculated the impedance of the capacitor between 3 - 30MHz: 5.3Ω - 0.53Ω respectively. I also understand that the resistance of the resistor will decrease as the frequency increases.

    What I don't understand:
    1. Do you calculate the time value of the cap & resistor combination as if they're in series?

    2. How would I calculate those values to a particular frequency range?

    3. Could someone give me a detailed explanation of what's happening?

    4. Suggested references: books or web to explain this.


    Thank you for any help or comments (such as if I'm completely off on what's going on)
     
  2. sceadwian

    New Member

    Jun 1, 2009
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    You're definitely mistaken, the resistor has nothing to do with the capacitor directly. The capacitor will shunt all higher frequencies to ground and let lower frequencies go. The only reason for the resistor being there is to prevent the capacitor from charging to a high DC voltage when there is no load attached to it, keep in mind this will also universally shunt some power to ground.
     
    Last edited: Aug 24, 2010
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  3. beenthere

    Retired Moderator

    Apr 20, 2004
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  4. atldave

    Thread Starter New Member

    Jul 26, 2010
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    Thanks!

    Finally some calculations!!

    I'm very interested in any books that cover that material - with the calculations.
     
    Last edited: Aug 24, 2010
  5. sceadwian

    New Member

    Jun 1, 2009
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    There's not much to calculate... The only thing of any real importance is the capacitance value which acts as the most basic form of low pass filter that can exist, it's impedance at the pass frequency, and at the unwanted signal are the only calculations you really need. The resistor is a simple a parasitic added as I said simply to prevent the capacitor from charging to a high DC voltage and damaging the load when the device is first connected, depending on the usage it may not even be required. There's not much more to learn from the circuit other than that.
     
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  6. marshallf3

    Well-Known Member

    Jul 26, 2010
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    The capacitive reactance (fancy term for the resistance of a capacitor in an AC circuit) will decrease as the frequency increases but a pure resistor won't change at all with frequency.

    Unfortunately most resistors nowadays are metal or carbon film; they're built with a long "trace" of resistive material wound spirally around a ceramic core. This causes them to exhibit inductance thus inductive reactance, which increases with frequency. The higher the value of the resistor the higher the internal inductance thus the more pronounced the effect.

    I once had to redesign a circuit a couple of older engineers couldn't get to work correctly in the upper end of the audio region and it was due to them not taking this effect into their calculations. Seeing that replacing them all with carbon resistors was pretty much out of the question (solid carbon resistors are non-inductive) I pretty much had to redesign a lot of the circuit, adding and changing capacitors and resistors until it worked properly. Luckily I had an audio sweep generator and real time audio spectrum analyzer at my disposal.

    This is probably absolutely of no concern at all in your circuit unless by rare chance the resistor and capacitor just happen to resonate somewhere near the carrier frequency.
     
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  7. gootee

    Senior Member

    Apr 24, 2007
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    The simple explanation is that without the resistor, it would only be a "peak detector", at least for some range of frequencies. i.e. If the capacitor were ideal, its voltage would be a record of only the highest voltage that had made it past the diode.

    That's fine, UNLESS you want it to also be able to track DOWNWARD-trending amplitudes, which you DO want for an envelope detector.

    So the resistor is added in parallel, to bleed current from the cap, which allows the voltage to fall after a peak.

    The trick is picking the C and R so that it tracks the envelope amplitude up and down reasonably well over the whole frequency range of interest (and doesn't track the details of the underlying signal or carrier).
     
    Last edited: Aug 25, 2010
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  8. atldave

    Thread Starter New Member

    Jul 26, 2010
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    Thank you! That's what I've looking for and I'm also looking for calculations on that very thing. How would one know what capacitor and resistor combination to put there without copying someone else or guessing. Or, for that matter, to see what audio frequencies you're getting.

    It also clears up something that I've been having a horrible time understanding. Every radio electronics book or site I've seen has glossed over this - they just explain that it's demodulating and grabbing the audio and then moving on. I've seen some equations that describe some of the relationship between the carrier and audio (Hayward) but nothing that connects math with devices.
     
    Last edited: Aug 26, 2010
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    That is very misleading. The RC time constant (and therefore the resistor value) is very important in an envelope detector. You apparently need to read up on envelope detectors. See the link posted by beenthere, as a start.
    You also seem to imply that the cap will charge up to a dangerously high voltage if the resistor is absent. In fact, it will only charge to the highest peak value of the input (minus the diode drop). This is true whether the resistor is present or not.
     
  10. gootee

    Senior Member

    Apr 24, 2007
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    Well, the time constant of the parallel RC part is just R * C. But I think that if you include a source resistance it becomes Tc = (RsR/Rs+R)C, and Vo = [Vi*R÷(Rs+R)](1-e^-t/Tc).

    The only ones I have ever designed were to produce the envelope of an audio-frequency signal that could arbitrarily be sawtooth, triangle, or sine, with variable desired amplitude, and variable frequency from 30 Hz to 22 kHz, which was surprisingly difficult, in that case, because it had to interact in a feedback control loop that was supposed to be able to change the amplitude from anywhere between 1V and 30V p-p, to anywhere else in that range, within less than 100-200 ms. That response time turned out to be impossible if using a fixed R (in parallel with the C), for those frequencies and waveforms. So I ended up using a voltage-controlled R that was automatically lowered when chasing the amplitude downward and then raised very high when trying to accurately track a stable target amplitude for a while. I think I could actually do it within 50 ms, but went back to 200 ms to get no overshoot with the simple feedback system I wanted to use.

    Anyway, at the time, I basically just "winged it" by using iterative LTspice simulations, starting with a few different available values for the small red WIMA polypropylene box capacitors that I wanted to use. (I know, I know, simulation should not be used as a design tool, they say. But it is often the fastest way, for me. And that experience is ALSO the only reason I had "the simple explanation" that I originally posted.) [It might not matter for your application, but, if you want to be able to handle small-amplitude signals, i.e. down to 0 Volts, you can use an "ideal diode" opamp rectifier circuit, in place of the diode. And you might want a buffer amplifier for the output.]

    But, anyway, if you know what frequency ranges of modulation and carrier you want to be able to handle, you can get a good idea of (i.e. calculate) how small the time-constant would have to be by knowing how fast the voltage would have to be able to fall in order to follow the downward-going parts of the envelope well-enough. (It's pretty easy to derive the maximum slew-rate of a sine wave given the frequency and amplitude.) And you should also be able to calculate how much carrier "ripple" any time constant would allow, for any carrier frequency.

    Cheers,

    Tom
     
    Last edited: Aug 26, 2010
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