# failure of the low-ESR electrolytic capacitor in second winding

E

#### Eeyore

Jan 1, 1970
0
Winfield said:
But, taking for example a boost converter (or an
equivalent flyback), operating in continuous mode
(which is much more efficient than discontinuous),
RMS current = 1.0 * load current if the duty cycle
is 50%.

This is only true if the current flow is perfectly rectangular.

I would certainly NOT expect that to be true for a flyback converter.

Graham

W

#### Winfield Hill

Jan 1, 1970
0
Eeyore said:
This is only true if the current flow is perfectly rectangular.
I would certainly NOT expect that to be true for a flyback converter.

It's a simplification, but not far off the mark if the
converter is operating well into the continuous mode.
Low-power offline flyback converters usually operate in
discontinuous mode, I believe, and an accurate formula
is therefore more complicated, but I think this gets
us into the ballpark, which should be good enough for
usenet. If the transformer turns ratio was optimally
selected, can't we expect that the switcher will be
operating near 50% duty at its nominal full current.

J

#### Jamie

Jan 1, 1970
0
Eeyore said:
Winfield wrote:

Win !

Graham
now that's a subject you're acquainted with..

J

#### Jamie

Jan 1, 1970
0
Robert said:
Put a current sensor in series with the capacitor and then you will
actually, I've always used low valued R's in series with the + line
before the caps if there is a feed back for regulation involved. Still
keeping with the low ESR caps nets good results.

T

#### The Phantom

Jan 1, 1970
0
We don't know what kind of smps Simon has, do we?

But, taking for example a boost converter (or an
equivalent flyback), operating in continuous mode
(which is much more efficient than discontinuous),
RMS current = 1.0 * load current if the duty cycle
is 50%. It's lower for duty cycles higher than 50%,
and higher below 50%, e.g., rising to 1.5 Iout for
D = 25%, according to a quick notepad calculation.

But one thing, since power loss goes as I^2 R, a
low duty cycle, with very high charging currents
(e.g. 3x load current for D = 25%), will actually
cause more heating than the simple RMS calculation
implies, and means a greater safety margin should
be required.

I assume you are considering the simplest (not realistic) case of a purely
rectangular current pulse delivered to the diode. In that case, wouldn't a
D of 25% imply that the peak of the current pulse would be 4x the load
current?

When you say "more heating", are you only talking about the heating of the
capacitor due to the ripple current in its ESR? Then shouldn't a simple
RMS calculation give the correct heating? Isn't that the point of using
the RMS value of current?

You're quite right about the relationship of capacitor ripple current to
load current. I've posted some graphs over on ABSE. Will somebody else
undertake the calculations to verify my results?

J

#### Jamie

Jan 1, 1970
0
Eeyore said:
Jamie wrote:

Yes !

I haven't seen anything from you that shows even the tiniest sign of knowledge
or experience of electronics to date.

Shorted electrolytic capacitor my arse !

Do please explain the shorting mechasnism won't you ?

Graham
I would but I don't have the time to get through your thick skull and
the fact that you have a social disorder because a few here know more
than you! Actually, a high percentage of them.

http://www.cde.com/news/acq/semco/
The company they acquired I worked at as a consultant (secondary
adventure) and was offered to relocate to, where else (India). Of
course, I'm still here because unlike you, I know better.

I was the chief EE, and tech that took care of the equipment and
R&D.
You want a scary thought to put in your little mind? The majority of
the components we made were for military/goverment applications!. things
like the patriot missiles? get the picture? not too much got sold to
other markets.

I hope Cornell Dubilier enjoys their purchase.

P

#### Phil Allison

Jan 1, 1970
0
"Eyesore"
With a load current of 1.5 - 1.7A, your 2 x 1000uF capacitors connected in
parallel will not have an adequate ripple current rating.

** That very much depends on the actual temperature of the caps in
operation.

The quoted 1040mA figure relates to an EXTRAORDINARY cap temp of 105C !!

At the far more likely operating temp of say 50C, the rating goes up to 2
amps

http://www.hitano.com.tw/PDF/EXR_070706.pdf

Note also, the quoted 0.076 ohm ESR is a "max" or worst case and is quoted
at 20C and 100kHz.

Now, the ESR of an electro cap *falls* with rising temperature, so at 50C
the real ESR value of typical examples ought to be under 0.03 ohms. With 2
amps rms of ripple, the dissipation is then a puny 120mW - raising the caps
temperature by only 10 degrees C or so.

IMO the reported cap failures are down to abuse ( very high ambients,

The "fix" is probably adding a small 12 volt fan to the PSU.

....... Phil

E

#### Eeyore

Jan 1, 1970
0
Jamie said:
actually, I've always used low valued R's in series with the + line
before the caps if there is a feed back for regulation involved.

More gibberish from Jamie (the perennially clueless) as ever.

E

#### Eeyore

Jan 1, 1970
0
Phil said:
"Eyesore"

** That very much depends on the actual temperature of the caps in
operation.

For sure that should be factored in as well.

I did ask the OP for info about operating temp but he doesn't appear to have the
info.

Graham

P

#### Phil Allison

Jan 1, 1970
0
"Eeysore"
For sure that should be factored in as well.

** Big FUCKNING SHAME how YOU failed to take ANY any account of the
matter.

Graham Stevenson = Shit for brians SNIPPER !!!!!!!!!!

----------------

** That very much depends on the actual temperature of the caps in
operation.

The quoted 1040mA figure relates to an EXTRAORDINARY cap temp of 105C !!

At the far more likely operating temp of say 50C, the rating goes up to 2
amps

http://www.hitano.com.tw/PDF/EXR_070706.pdf

Note also, the quoted 0.076 ohm ESR is a "max" or worst case and is quoted
at 20C and 100kHz.

Now, the ESR of an electro cap *falls* with rising temperature, so at 50C
the real ESR value of typical examples ought to be under 0.03 ohms. With 2
amps rms of ripple, the dissipation is then a puny 120mW - raising the caps
temperature by only 10 degrees C or so.

IMO the reported cap failures are down to abuse ( very high ambients,

The "fix" is probably adding a small 12 volt fan to the PSU.

E

#### Eeyore

Jan 1, 1970
0
Phil said:
"Eeysore"

** Big FUCKNING SHAME how YOU failed to take ANY any account of the
matter.

It was one of the first questions I asked.

Graham

P

#### Phil Allison

Jan 1, 1970
0
"Eeysore OVER SNIPPING POMMY CUNTHEAD "

It was one of the first questions I asked.

** Not one bit relevant to my point.

Graham Stevenson = ASD fucked MORON !!!!!!!!!!!!!!!

His crap ANALYSIS was FALSE as it did not take account of the matter.

At all.

So, go get fucking fucked you -

LYING CRIMINAL ARSEHOLE

----------------

** That very much depends on the actual temperature of the caps in
operation.

The quoted 1040mA figure relates to an EXTRAORDINARY cap temp of 105C !!

At the far more likely operating temp of say 50C, the rating goes up to 2
amps

http://www.hitano.com.tw/PDF/EXR_070706.pdf

Note also, the quoted 0.076 ohm ESR is a "max" or worst case and is quoted
at 20C and 100kHz.

Now, the ESR of an electro cap *falls* with rising temperature, so at 50C
the real ESR value of typical examples ought to be under 0.03 ohms. With 2
amps rms of ripple, the dissipation is then a puny 120mW - raising the caps
temperature by only 10 degrees C or so.

IMO the reported cap failures are down to abuse ( very high ambients,

The "fix" is probably adding a small 12 volt fan to the PSU.

--------------------

....... Phil

....... Phil

W

#### Winfield

Jan 1, 1970
0
The said:
I assume you are considering the simplest (not realistic) case
of a purely rectangular current pulse delivered to the diode.
In that case, wouldn't a D of 25% imply that the peak of the
current pulse would be 4x the load current?

The inductor and diode have 4x the output current,
but 1x of this goes into the load, so the capacitor
only has 3x while its being charged. However, the
correct multiplier is 1.73, not 1.5, and for the

W

#### Winfield

Jan 1, 1970
0
The inductor and diode have 4x the output current,
but 1x of this goes into the load, so the capacitor
only has 3x while its being charged. However, the
correct multiplier is 1.73, not 1.5, and for the
[OOPS - off it went before I was finished!]

I take back my casual dismissal of the higher capacitor
dissipation for the actual case of discontinuous current.
The multiplier for 50% rises from 1 to about 1.25, and
for 25% to over 2x, which is certainly very significant.

turns ratio, etc. One certainly doesn't want to fall
below 50% at full output, if it can be avoided.

Yes, of course, that's right. I stand corrected.

BTW, D is usually defined as the PWM switch-on time,
rather than the PWM power-delivery fraction as I've
used it here, so readers should reverse the D values.

What were your formulas? The plots vs duty-cycle are
indeed useful, but I do have one comment: showing the
current rising from 0 to full for 0 to 100% duty cycle
is misleading. In reality the power supply might reach
full output at a 50% power-delivery fraction, etc., the
other 50% being used to charge the inductor/transformer.

Moreover, as we've used it here, not knowing the supply
design details, a more serious case occurs if we assume
that full power is reached for a lesser fraction than 50%,
such as 25%. That's when things get tough for the poor
capacitor, and surely would represent poor power-supply
for a simple inductor boost regulator, e.g., 25% duty
for a 1:3 step-up ratio at maximum power.

T

#### The Phantom

Jan 1, 1970
0
The ripple current for a 1A7 output flyback will depend on duty cycle
and secondary inductance. If you're quoting measured secondary rms
current, however, then the capacitor rms ripple current should equal:

Icaprms = ( Isecrms^2 ­ Ioutdc^2 )^0.5

This expression does not imply that "...the ripple current in the capacitor
is not greater than the output current...". It merely implies that both
the ripple current in the cap and the output current are less than the
secondary RMS current.

For example, suppose Isecrms = 5. Then either (Icaprms = 3 AND Ioutdc = 4)
or (Icaprms = 4 AND Ioutdc = 3) will satisfy the expression.

T

#### The Phantom

Jan 1, 1970
0
Ok, I see. I overlooked your intended meaning for the phrase "charging
current". You meant specifically current into the capacitor and not total
diode current. That was unclear to me.
However, the correct multiplier is 1.73

Yes, this is what my first graph would show if I extended the vertical
scale a little.
, not 1.5, and for the
[OOPS - off it went before I was finished!]

I take back my casual dismissal of the higher capacitor
dissipation for the actual case of discontinuous current.
The multiplier for 50% rises from 1 to about 1.25, and
for 25% to over 2x, which is certainly very significant.

turns ratio, etc. One certainly doesn't want to fall
below 50% at full output, if it can be avoided.

Yes, of course, that's right. I stand corrected.

BTW, D is usually defined as the PWM switch-on time,
rather than the PWM power-delivery fraction as I've
used it here, so readers should reverse the D values.

I agree that's what is usually meant by D, but in this discussion we've
been concentrating on the output part of the supply, and I think it's
reasonable to use D for the on time of the diode as long as everybody
understands that's what we mean. I used it that way in my plots on ABSE,
and I understood that's what you meant in your posts here. Until now,
anyway. I'm going to stick with that meaning, because I think it will just
add confusion to change it now.

I explained what I was calculating in words. I used the usual
calculations. For example, the RMS current in the capacitor is the square
root of the integral (over 1 period) of the capacitor current of 1 amp for
D % of the period, squared, and so forth for the other calculations. I
have made mistakes doing this sort of thing in the past, and that's why I
invited others to verify the calculations. I think it would be better if
anybody who did so, came up with their own formulas, the better to provide
verification.
The plots vs duty-cycle are
indeed useful, but I do have one comment: showing the
current rising from 0 to full for 0 to 100% duty cycle
is misleading. In reality the power supply might reach
full output at a 50% power-delivery fraction, etc., the
other 50% being used to charge the inductor/transformer.

I should have explained even more precisely, something I'm criticizing
you for. This lack of precision in description is the cause for a lot of
the disagreement in this group! :-(

In the plots, I am considering a supply without feedback, and I made the
peak current into the diode at the beginning of a pulse always 1 amp. In
other words, during the on time of the primary side switch, the inductance
is always charged up to the same current, regardless of duty cycle. I
wanted to normalize things. The main purpose of the plots is to show that
the ripple current in the cap is not always greater or always less than the
output current, a point you were making and others were disputing.
Moreover, as we've used it here, not knowing the supply
design details, a more serious case occurs if we assume
that full power is reached for a lesser fraction than 50%,
such as 25%. That's when things get tough for the poor
capacitor, and surely would represent poor power-supply
for a simple inductor boost regulator, e.g., 25% duty
for a 1:3 step-up ratio at maximum power.

I'll do some more plots, normalized to constant power delivered to the

W

#### Winfield Hill

Jan 1, 1970
0
The said:
I'll do some more plots, normalized to constant power

I look forward to that.

T

#### The Phantom

Jan 1, 1970
0
I look forward to that.

The OP really needs to make some measurements. He should get a scope photo
of the secondary winding current, and he should get some of the temperature
indicating labels another poster mentioned and stick them on the caps.

Otherwise, we're all just speculating.

L

#### legg

Jan 1, 1970
0
Utter and complete nonsense.

You first sunshine.

What I stated is 100% correct.

I'm staggered quite frankly that so few people here know how to design something
as simple as a PSU competently.

I mean SERIOUSLY ! Not to know how to determine the required ripple current
rating of a capacitor ? Bizarre.

Find refresher penance attached on a.b.s.e.

Backwards waveforms, but the rms effects are the same. Other errors?

RL

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