# Calculating a rectifier's resistance in a PSU

Discussion in 'General Electronics Chat' started by richard3194, Apr 25, 2015.

1. ### richard3194 Thread Starter Active Member

Oct 18, 2011
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2
One can calculate or measure a rectifier's static resistance or dynamic resistance.

Let's imagine though, we have a simple PSU - consisting of a secondary outputing 150V RMS sinusoidal, feeding a 1N4006 rectifier/diode in series with a load consisting of a 1800R resistor, which is shunted with a 8uF capacitor.

The rectifier/diode will offer a resistance to current flow. But, it won't be the static resistance, nor will it be the dynamic resistance. So, what resistance will it be? What is it called?

I guess if you measure the mean voltage across the diode, and divide it by the mean current flowing through the diode, you can calculate the resistance the rectifier is offering.

Can you calulate this resistance purely by circuit analysis? That is from circuit component parameters. Thank you. Rich

2. ### Papabravo Expert

Feb 24, 2006
11,150
2,177
One of the problems with averaging a periodic waveform over a complete cycle is that the answer is often zero. As you are no doubt aware an average value of zero is not very useful when it comes to calculations. I'm curious to explore your concept of static and dynamic resistance as it relates to rectifiers, and in particular why you think either or both are unsuitable for your purposes.

Last edited: Apr 25, 2015
3. ### #12 Expert

Nov 30, 2010
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I would go at it from the graph of the diode characteristics.
I would call it part of the limiting impedance to the charging pulses.

4. ### Papabravo Expert

Feb 24, 2006
11,150
2,177
I think that is the definition of "dynamic resistance", the slope (or it's reciprocal) of the V-I characteristic curve is how I thought dynamic resistance was defined.

$R_d = {\Delta V} / {\Delta I}$

5. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
Static resistance: A circuit arrangement would be such, that the voltage across the diode, thus the current, would be stable. So, the forward resistance would be fixed.

Dynamic resistance #1: If you explain the operation of a diode, you can show, by picking out spots on a graph, the dynamic nature of diode forward resistance.

Dynamic resistance #2: A circuit arrangement would be such, that the voltage across the diode would be varying. So, the forward resistance would vary, i.e. be dynamic. That's what we have in the circuit I described. Maybe then a certain understanding of the term "dynamic resistance" is good for the problem of determing the diode forward resistance in the circuit.

I wonder if what I'm calling "Dynamic resistance #2", is often called Effective Supply Resistance or Impedance. I might be mistaken here.

I've been looking at valve/tube type U52. There is a graph with "Effective Valve Resistance". I think it's the same as dynamic resistance #1, I thought it was "Dynamic resistance #2".

Last edited: Apr 25, 2015
6. ### KLillie Member

May 31, 2014
126
14
Sounds like a calculus problem.

7. ### #12 Expert

Nov 30, 2010
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The impedance of the supply can be a lot more than the impedance of the diode, especially in a low current circuit that a 1N4006 would be in. The transformer impedance can be significant.

In fact, the entire impedance goes all the way back to the generation station, but measuring after the circuit breakers should include all significant sources and some insignificant sources.

8. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
I think that equation is the dynamic resistance of the diode.

What I'm after is the dynamic resistence, or even just the resistance of the rectifier, with a varying voltage. Not an actual characterisic of the diode.

9. ### Papabravo Expert

Feb 24, 2006
11,150
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The terms diode and rectifier are often used interchangeably to mean the single component. I think your use of the word rectifier implies the configuration of one or more diodes in a half-wave, full-wave, or bridge topology. That was not clear to me from your original post. In any case the job of conduction is passed back and forth among the diodes. When they conduct the resistance is low and when they block the resistance is high. I guess I have no clue what your looking for and/or why so I'll defer to the smarter people around here.

10. ### #12 Expert

Nov 30, 2010
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I think you need to reconsider your question. The rectifier is made of diodes. Their apparent resistance does not depend on the voltage of the power supply, only on the voltage across the diode during current conduction. The resistance of the rectifier can not be divorced from the characteristics of the diodes. If there are 2 diodes conducting at a given time, the answer is 1+1 = 2
One apparent diode resistance plus one apparent diode resistance equals 2 apparent diode resistances (in series).

11. ### Roderick Young Member

Feb 22, 2015
409
169
The ideal diode equation is

$I = I_0[e^{{qV}/{kT}} - 1]$

where I is the current and V is the voltage. T is the temperature, and kT/q is about 26 mV at room temperature.
differentiating,

${dI}/{dV} = ({I_0q}/{kT})e^{{qV}/{kT}}$

or

${dV}/{dI} = R_{dynamic} = ({kT}/{qI_0})e^{-{qV}/{kT}}$

As you can see, the dynamic resistance is not a constant, nor is any derivative of it a constant. However, it becomes vanishingly small at higher voltages. Also remember that in the real world, the diode burns out once there is too much voltage across it (= too much power through it).

If you are applying a signal V(t), you could integrate with respect to time, then divide by t to get an average.

Last edited: Apr 26, 2015
cmartinez and KLillie like this.
12. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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Here is your circuit: I show two cycles after giving the capacitor time to charge up. Note the voltage at V(x) and V(y). I also plot the difference between V(x) and V(y). The diode conducts only when this difference is positive. I also plot the current through the diode I(D1), which also shows when the diode conducts.

Since you're interested in the dynamic impedance of the diode, it only makes sense to talk about that while the diode is forward-biased. I use the cursors to zoom in on the region when the diode is conducting between 21.125ms and 25.812ms. I repeat the sim over that interval, and plot the expression (V(x)-V(y))/I(D1), which you will recognize. Note that this expression blows up when I(D1) ~= 0, so that is why we need to focus only when the diode is conducting (forward-biased)...

So during the time that the diode is forward-biased, its effective impedance is between ~2Ω and 45Ω.

13. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
Yes, very good. As I say, the resistance value I'm after isn't actually a characteristic of the rectifier/diode. It's simply the resistance or impedance the rectifier is presenting to the passage of current. The value will of course be a function of the diode parameters and the voltage waveform across the diode. This is why I stated it won't be your textbook static nor dynamic resistance that I'm looking for. I believe these terms are parameters of the diode.

You have stated two seperate things, dynamic impedance and effective impedance. What is it that I'm after? I think effective impedance, but I'm not 100% sure.

Whatever it is I want, I think the program you are using probably gives the mean voltage across the diode and the mean current through it. So, I guess, Vdmean / Idmean = Rdmean. This is what I'm seeking. Whatever it is called. Maybe just mean diode impedance.

Last edited: Apr 26, 2015
14. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
To me, same thing.
No. I plotted the instantaneous Vf/Id (varies over time from 2Ω to over 45Ω). The average of these values are pretty meaningless...

Look at this comparison:

I replaced the diode with a magic switch that has a resistance of 1Ω when V(x)>V(z). Note that the voltage across the two filter capacitors are virtually identical. What does this say about the effective impedance of the diode?

Last edited: Apr 26, 2015
15. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
I have an AVO VCM MKI valve (tube) tester. It tests tube tectifiers by loading the rectifier with a specific load, which is dependent on how much current a good rectifier should be capable of delivering. You test calibration of the MKI, by placing in series a DD058 diode and a 150R resistor (standard rectifier). Then the instructions tell you that if you put in circuit a U52 rectifier you should get a certain meter reading (actually a meter reading range).

The VCM meter shows average current.

In an attempt to get a more accurate reading when I test my 5R4 tubes, I'm trying to work out how I might achieve these more accurate results. And it seemed to me that I might venture if I could work out the average effective diode impedance of the standard rectifier, and the 5R4. I mean if the standard rectifier gave a figure of 75 on the meter scale, my hope was that I could calculate the expected meter reading for a brand new 5R4, having had reference to it's effective resistance.

But anyway, if you were to place a 5R4, in Pspice, or whatever, loaded with 1800R shunted with 8uF, Pspice ought to be able to indicate the average voltage across the rectifier, and the average current. This would, I think, enable one to calculate the tube's effective resistance. Which I think needs to be a mean value. Is this correct?

16. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
I don't have Pspice installed, but something called PSU Designer II. This shows average voltage across D1 and average current thru D1. A circuit consisting of a 1N4006 (D1) in series with a 270R, loaded as per the circuit I earlier described. Average D1 current is given as 58.662 mA, and average D1 voltage as -122.95. V/I gives 2095R. That does not seem right.

17. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
I showed you that the forward resistance of a Si diode during the part of the AC cycle where it conducts is a couple of Ω. AFAICR, the effective plate resistance of a 5U4, 5R4 is about 200Ω

18. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
Are you averaging the forward voltage divided by the forward current for the small part of the AC cycle where the diode conducts, or is the average being computed for the entire period of one AC cycle?

19. ### richard3194 Thread Starter Active Member

Oct 18, 2011
68
2
Let me cycle thru this: If AC is applied to a resistor, it doesn't matter that the current reverses. Usually you are interested in the RMS current and voltage, because then you can work out power. But, if your task was to calculate the average voltages or currents, you would multiply the peak voltage or current by 0.637. Average voltage divided by average current, equals the resistance. It's the same value as the RMS voltage divided by RMS current.

With diodes it's different. The current only flows in the forward direction. If a diode was a linear device, like a resistor, the effective average voltage across the diode would remain the same, but the effective average current would be half. What this means is, effective resistance of a linear diode marked with a forward resistance of 100R, would appear to be 200R to the external circuit.

Proof: Connect a fictitious linear diode with a forward resistance of 100R across a sinewave voltage of 6.37V average. The current would be 31.85mA. It's 31.85mA, because actually across the 100R is 3.185V, not 6.37V, because there is nothing for half of a cycle across the 100R. So, the 100R diode looks like 200R to the external supply. If you take voltage and divide by current, you would get the effective resistance of the diode - 200R.

Of course, there are no linear diodes. Real diodes don't have a set forward resistance. The effective resistance of a diode is dependent upon the current waveform. The resistance varies as current varies, but there will be an effective resistance, which is peculiar to the circuit in question.

The question is, if you measure the average voltage across the diode, and the average current, does that give the effective resistance of the diode. It does in my fictitious linear diode scenario. In that scenario I did not divide the voltage by 2. The current reflected the fact that there was only effectively 3.185V average across the 100R forward resistance of the diode.

Last edited: Apr 28, 2015
20. ### #12 Expert

Nov 30, 2010
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In any real circuit, the peak current is diminished by the peak voltage across the diode. That lost voltage is what does not show up in the remaining circuits. It is not the average voltage loss across the diode that does not get through, it is the peak voltage loss that does not get through the diode.

If you post a circuit wherein it matters whether the average lost voltage is relevant, somebody might be able to relate to your quest.