# Electro ripple current hint

• posted
** Hi,

electro cap makers specify the maximum, safe "ripple current" for each category and size. Normally, two figures are given - one at AC supply frequencies and another at a much higher frequency - reflecting the differing ESR values at the two frequencies. In order to measure this current, a low value resistor needs to be installed in series with the particular electro and the resulting voltage measured with a true RMS volt meter or maybe a DSO that can do the same task.

Finding a suitable resistor and fitting it in series could prove very tedious (or even unsafe) in practice. However, there is a simple way around this that works for low and AC supply frequency measurements.

The classic formula " I = C dv/dt " shows the current flowing in an electro is proportional to the dv/dt of the voltage across the terminals any point in time - so if you derive a current proportional to dv/dt, then its RMS value is a measure of the ripple current.

A series RC network does the job, long as its time constant is short in relation to the high frequency components in the current wave. For AC supply frequencies a value of 100uS proves to be OK.

All you need then is a 100nF film cap and a 1kohm resistor, connected in series across the electro and simply measure the RMS voltage appearing across the resistor (Vrms). The voltage wave replicates the electro current wave almost exactly.

The necessary scale factor is simply the electro's value in uF multiplied by

10,000 ( as a 1mA current in the RC network requires a dv/dt of 10,000 V/S.)

Electro ripple amps = C x Vrms x 10,000

BTW:

While ripple current in electros operating at AC supply frequencies is not regularly an issue - it can become one when the supply impedance is unusually low, as with direct mains operation or with large transformers and small value caps.

In these cases, the peak charging current can exceed the average load current by a factor of more than 10:1 , exaggerating the RMS current value considerably.

Larger than usual ripple voltage percentages also increase peak and RMS ripple currents, particularly if combined with the above.

.... Phil

• posted

What about the voltage drop in the ESR?

• posted

If you guess the time constant correctly, you already know (that was the point).

Works for inductive shunt resistors at high frequency, too:

Iin o | R

+---RRR--+---o out | | R Rs | R | | === C L Ls | L | | | +--------+---o GND | o Iout

R*C = Ls/Rs

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://www.seventransistorlabs.com/```
• posted

The ESR bump may well dominate the integrated-current waveform, and it wrecks the math.

The problem with ripple current is that it cooks the caps. The capacitive component dissipates no power. So ESR really matters. It's circular to guess the time constant to guess the ESR.

What's more sensble is to just look at the waveform across the cap, with a scope, and estimate the current waveform.

Better yet, scope the voltage drop in a trace between the rectifier and the cap and use that to estimate the current waveform.

That's lowpass; I did that in by boost switcher, posted recently. But it requires you to know the Rs and Ls values, or to put in a current step and tweak the RC.

• posted

Hold that thought.....

Unfortunately, copper traces have an excellent Q and make better inductors than resistors at, oh, just a few kHz really. Copper is hopelessly inductive at SMPS frequencies. You'd have to know Rs and Ls very well indeed to compensate that trace well enough to know anything. Circular reasoning. :)

Anyways, you'd have to be a pretty awful PCB designer to leave any length between rectifier and cap. :)

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://www.seventransistorlabs.com/```
• posted

Phil was talking about low frequency stuff, like AC-line power. I think.

But we're talking numbers in the ballpark of 1 nh per inch and maybe a few milliohms per inch for a power trace. Tau in the hundreds of nanoseconds.

Copper is hopelessly

Well, he is in the audio business.

• posted

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS, with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and it seems to work well. However, with an ESR of 1.8 ohms, it has an error of

60%. But that is an extreme case at which there is 8W dissipation in the

capacitor. Also, if you use a 1 uF capacitor instead of 100 nF, the result then becomes much closer. But then there is a large error at lower ESR values.

Paul

Version 4 SHEET 1 1368 680 WIRE 304 128 112 128 WIRE 320 128 304 128 WIRE 432 128 384 128 WIRE 496 128 432 128 WIRE 528 128 496 128 WIRE 576 128 528 128 WIRE 672 128 576 128 WIRE 672 144 672 128 WIRE 112 160 112 128 WIRE 320 208 272 208 WIRE 432 208 432 128 WIRE 432 208 384 208 WIRE 576 208 576 128 WIRE 496 224 496 128 WIRE 672 256 672 208 WIRE 112 272 112 240 WIRE 272 272 272 208 WIRE 272 272 112 272 WIRE 320 272 272 272 WIRE 416 272 384 272 WIRE 672 288 672 256 WIRE 304 368 304 128 WIRE 320 368 304 368 WIRE 416 368 416 272 WIRE 416 368 384 368 WIRE 496 368 496 288 WIRE 496 368 416 368 WIRE 576 368 576 288 WIRE 576 368 496 368 WIRE 592 368 576 368 WIRE 672 368 592 368 WIRE 592 432 592 368 FLAG 592 432 0 FLAG 528 128 out FLAG 672 256 ripple SYMBOL voltage 112 144 R0 WINDOW 123 0 0 Left 2 WINDOW 39 -30 182 Left 2 WINDOW 0 -15 57 Left 2 WINDOW 3 -85 151 Left 2 SYMATTR SpiceLine Rser=50m SYMATTR InstName V1 SYMATTR Value SINE(0 175 60 0 0 0 100) SYMBOL diode 384 288 M270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D2 SYMATTR Value MUR460 SYMBOL diode 320 224 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D3 SYMATTR Value MUR460 SYMBOL polcap 480 224 R0 WINDOW 3 24 64 Left 2 SYMATTR Value 560u SYMATTR InstName C1 SYMATTR Description Capacitor SYMATTR Type cap SYMATTR SpiceLine V=600 Irms=2.9 Rser=1.8 Lser=0 SYMBOL res 560 192 R0 SYMATTR InstName R1 SYMATTR Value 50 SYMBOL diode 320 144 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D1 SYMATTR Value MUR460 SYMBOL diode 384 384 M270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D4 SYMATTR Value MUR460 SYMBOL res 656 272 R0 SYMATTR InstName R2 SYMATTR Value 1k SYMBOL cap 656 144 R0 SYMATTR InstName C2 SYMATTR Value 100n TEXT 64 360 Left 2 !.tran 200m startup TEXT 112 424 Left 2 ;Electro ripple amps = C x Vrms x 10,000 TEXT 752 112 Left 2 ;560u * 1.075 * 10,000 = 6.02 TEXT 752 144 Left 2 ;for R1 = 50, I(C1) = 6.19 A RMS TEXT 440 312 Left 2 ;0.018 ohms TEXT 752 360 Left 2 ;for R1 = 500, I(C1) = 1.21 A RMS TEXT 752 328 Left 2 ;560u * 0.199 * 10,000 = 1.114 TEXT 752 432 Left 2 ;for R1 = 500, ESR(C1)=0.18, I(C1) = 1.102 A RMS TEXT 752 400 Left 2 ;560u * 0.196 * 10,000 = 1.098 TEXT 752 176 Left 2 ;560u * 1.070 * 10,000 = 5.99 TEXT 752 208 Left 2 ;for R1 = 50, ESR(C1)=0.18, I(C1) = 5.99 A RMS TEXT 752 272 Left 2 ;for R1 = 50, ESR(C1)=1.8, I(C1) = 5.33 A RMS TEXT 752 240 Left 2 ;560u * 1.394 * 10,000 = 7.81

• posted

"P E Schoen" "Phil Allison"

I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS, with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and it seems to work well. However, with an ESR of 1.8 ohms, it has an error of

60%.

** Simulations need to use realistic data or they are just plain stupid.

Obviously, you did not look up the ESR or ripple current figures of any

*real* 560 uF, 250V electro.

A figure of 0.25 ohms (100/120Hz ) is about right at room temp for standard grade, 560uF, 250V electros - dropping sharply with increasing temp. But a figure of 1.8 ohms indicates a faulty cap.

Also, your 50ohm load draws over 3 amps resulting in circa 35V p-p ripple - which is rather high.

Was your source resistance for 120VAC zero ?

If so, that was crazy too.

But that is an extreme case at which there is 8W dissipation in the capacitor.

** That cap would have exploded in minutes - so is utterly irrelevant.

You need to do some REAL tests with REAL electros.

Cos simulations are mostly bullshit.

... Phil

• posted

Those inductance figures seems to be off by one order of magnitude.

For narrow PCB tracks 1 nH/mm would be typical, so 0.4 nH/mm for a wide trace would be believable (10 nH/inch).

• posted

I wouldn't use narrow tracks between a rectifier and filter caps.

But I did say "ballpark". The copper trace is a good current shunt in an AC rectifier-capacitor supply.

• posted

"P E Schoen"

I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS, with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and it seems to work well. However, with an ESR of 1.8 ohms, it has an error of

60%. But that is an extreme case at which there is 8W dissipation in the capacitor.

** That 8W figure cannot be right - cap ripple current always exceeds DC load current.

With a 3.1 amp load, the power must exceed 17 watts.

To get the real answer I used the following:

240/120V at 800VA step-down tranny

470uF, 350V electro ( tested as 530 uF )

1.2kW, 240V heater for the 50 ohms load.

1.8 ohm, 50 watt metal clad resistor.

Results:

I ripple rms = 4.3 amps, power loss = 33watts

BTW:

The ripple voltage wave showed a big bump at the end of the charging period - the usual sign of an electro that is completely worn out.

... Phil

• posted

exceeds DC

Yup, I ran it again and I got 32.6 watts, with 4.35 A for I(C) and 3.02 A I(R).

I thought it seemed a bit "off". Even 0.18 ohms gives 13 watts.

But that's not quite right either. I(C1) is 6 amps, and thus the power should be 36*0.18 = 6.48W.

The problem seems to be due to the way LTSpice computes power in a capacitor and you must set the sample window to an exact multiple of half cycles, because otherwise the reactive VA dominates the equation. Hmm. Even that

does not seem to work...

Paul

• posted

No, I don't think so. Try this, Paul:

.tran 0 .5 .4 10u

This will give you a whole number of cycles to work with and will drop the initial surge.

Then you can use the nice power and average functions available to do what you want.

It is.

• posted

"John S"

** Not for long.

As usual, reality has gone missing.

Can I remind you:

1. Electro ESR has a large negative tempco.

1. A 560uF, 250V electro with a 6A ripple rating ( if they exist) does not have a 0.18 ohm ESR, but something a lot less.

Recall what I said about simulations ?

It's a mug's game.

.... Phil

• posted

In LT Spice, I pull out inductor and capacitor ESR as separate components, so I can measure their power dissipation un-tangled from the reactive part.

I was recently simulating a flyback switcher and probed the transformer primary, to see the losses there, expecting milliwatts. I got watts. Of course, all the load power is flowing into the primary with no visible exit!

I pulled out the copper ohms as a separate resistor and got the expected dissipation.

In the case at hand, I'd just scope the waveform across the cap and tease out the components. Maybe even Spice it and fiddle C and ESR to match what I see. The shape of the waveform immediately after the rectified peak tells a lot about C and ESR.

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