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I found following schematic of a half wave rectifier (from here) and I was wondering how to choose the resistor, and I have following questions:

  • Is it correct that this resistor is basically functioning as a pulldown for when the AC is in the reverse polarity (top negative)?
  • When connecting the output of this rectifier to the input of another device I think we would lose current across the other device when R1 is too small, so my guess is that we have to choose R1 at least as large as the input impedance of the connected device (or even a lot larger), is this correct? And if so, is there a more precise way to compute a value?

schematic

simulate this circuit – Schematic created using CircuitLab

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4 Answers 4

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Is it correct that this resistor is basically functioning as a pulldown for when the AC is in the reverse polarity (top negative)?

Correct. And it is only there to discharge any capacitance in your measuring circuit which might otherwise give a false reading. (The input capacitance on your measuring circuit - an oscilloscope, for example - would tend to hold charge between DC pulses.)

When connecting the output of this rectifier to the input of another device I think we would lose current across the other device when R1 is too small, so my guess is that we have to choose R1 at least as large as the input impedance of the connected device (or even a lot larger), is this correct?

If you have a load then R1 is not required. The load will ensure that the voltage falls to zero.

And if so, is there a more precise way to compute a value?

If you're charging a capacitor in your circuit then look at the discharge time constant.

enter image description here

Figure 1. Capacitor discharge curve. The capacitor discharges by 63% per RC time period. [Image by @Transistor.]

A handy rule of thumb is that the "time constant" of an RC circuit is given by multiplying R and C. τ=RC. After τ the voltage will have decayed by 63%. After 3τ it will have decayed by 95% and 5τ, 99%.

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  • \$\begingroup\$ In words it’s e^-N so for N=5 the remainder is 0.67% or 1% \$\endgroup\$ Commented Feb 25, 2020 at 15:03
  • \$\begingroup\$ And may the parallel resistor be especially useful if my only load is an led (with current limiting resistor in series) and my AC voltage is high? So that no reverse potential is across the led? In practice, I have a capacitor there, so if its size is enough for a full cycle, there shouldn’t be really any reverse voltage. But if it wasn’t there/wasn’t enough? \$\endgroup\$
    – Adam
    Commented Jun 6, 2020 at 9:00
  • \$\begingroup\$ Hi, Adam. This seems to be a new question. Post it as a new question with a schematic (there's a schematic button on the editor toolbar) and reference this one if it's relevant. \$\endgroup\$
    – Transistor
    Commented Jun 6, 2020 at 10:18
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  1. Yes, pull-down resistor (level 0, to ground);
  2. The value of the resistor should be as large as possible if linearity of the rectifier is intended;
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  • \$\begingroup\$ Thanks for answering, can you clarify: What are we trading in if we choose a very high value for the resistor? Is there a value that is too high? (I mean if we have an "infinite" resistor it doesn't do its job as a pull down anymore.) \$\endgroup\$
    – flawr
    Commented Feb 25, 2020 at 9:19
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I guess you want some nasty fuzz-type distortion in a synth or sound amplifying circuit. You are not worrying about wasted energy or other things which are a problem in high power AC rectifiers which handle watts or kilowatts, you are going to handle micro- or milliwatts.

Pull-down action of the resistor = OK

You must have so big resistance that the signal source isn't pulled on its knees but the resistance must be low enough to sink the leakage of the diode and what's stored in the non-ideal capacitances in the circuit. One of the capacitances is in the diode. This makes easily high current 50...60Hz rectifier diodes useless in high speed pulse circuits. I guess 1N4148 works well in whole audio band with say 3Vpp signals when R=500...500000 Ohms.

There's a trap. If the signal source has an output series capacitor, it will get charged and it stops the whole signal. To prevent it you should have also a resistor to GND at the left end of the diode.

Learn to use circuit simulators. This site has quite good one for free.

Audio signals are often only 1Vpp or less. The voltage drop of the diode (about 0,7V) can cause serious volume loss. You should consider to use a rectifier circuit which has an operational amplifier to compensate the voltage drop in the diode. An example: OpAmp Precision Half Wave Rectifier

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  • \$\begingroup\$ Thanks for pointing out the problem of capacitors. And I'll definitely have to look into circuit simulators, I didn't know this was a thing! Right now I was mainly looking into passive circuits (that don't require external power), so I'd limit myself to circuits without OpAmps, but it is definitely nice to know for a future project:) \$\endgroup\$
    – flawr
    Commented Feb 25, 2020 at 11:18
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As a large signal AM detector the voltage drop depends on the diode type and current. As shown with 0.6V silicon signal diode drop means the draw was around 1mA peak.

If 10uA is ok for stray current noise immunity, this drop reduces in half with logarithmic proportions.

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