Electro ripple current hint

Discussion in 'Electronic Design' started by Phil Allison, Jul 5, 2013.

1. Phil AllisonGuest

** 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

2. Tim WilliamsGuest

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

3. Tim WilliamsGuest

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

4. P E SchoenGuest

"John Larkin" wrote in message
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

5. Phil AllisonGuest

"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

6. Guest

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).

7. Phil AllisonGuest

"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

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

8. P E SchoenGuest

"Phil Allison" wrote in message
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

9. John SGuest

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.

10. Phil AllisonGuest

"John S"
** Not for long.

As usual, reality has gone missing.

Can I remind you:

1. Electro ESR has a large negative tempco.

2. 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