Film capacitor as power-supply filter

On Wed, 23 Oct 2019 04:48:25 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

On Tue, 22 Oct 2019 18:00:15 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

jlarkin@highlandsniptechnology.com wrote:

On Tue, 22 Oct 2019 07:15:45 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

jlarkin@highlandsniptechnology.com wrote:

On Mon, 21 Oct 2019 07:57:11 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

jlarkin@highlandsniptechnology.com wrote:

The issue was series strings of electrolytic caps, where the total
supply voltage is more than the rated voltage of the caps. Film
caps can blow up in that situation. What do lytics do?

Answer: their IV curves make them safely self-equalize.

How does self-equalization work, and why does the I-V curve have to
be exponential?

The cap with more voltage across it would conduct way more current,
which would integrate to lower voltage on itself and more on others.
The end result is obviously equal currents, but *less* current than
there would be if the voltages were evenly distributed in the
string.

When you first apply voltage across the cap, the current decays
exponentially.

Not exponentially, because the cap leakage is nonlinear.

So both caps may have a high current, but they both decay quickly.

If voltage is applied quickly, the voltage across each cap is inverse
on C.

As the leakage time constants lick in, the voltage redistributes into
the minimum-current situation. It might take hours to mostly settle
down, or maybe days.

The best test would be to find two caps with different leakage
current.

Put them in series, apply voltage, and plot the voltage at the
junction.

This could be done with a digital scope.

Betcha they don't equalize.

Of course they don't equalize to equal voltage. They do magically find
the voltage distribution that minimizes the leakage current.

Does that voltage distribution mean that one capacitor may have a
voltage that exceeds its rating?

Possibly, but since the current is low, it does no harm.

Current damages electrolytic caps. As noted, the string current is
lower than it would be if the cap voltages were forced equal.

Allowing the capacitor to exceed the rated working voltage is not a
suitable engineering goal.

Now you're getting prissy. When people can't explain things, they fall
back on "good engineering practice" as their reason for doing, or
usually not doing, things.

You can use your method. I'll stick with the industry standard bleeder
resistors where the bleed current is 10 times the maximum capacitor
leakage spec

Fine. Add fans as needed.

A typical spec is

I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

The leakage current is measured at 20C by applying the rated voltage to
the capacitor through a series resistor of 1000 Ohms. The leakage current
is the value 2 minutes after the capacitor has reached the rated voltage.

This test requires the capacitor to be already properly formed.

Example for a 1000uf, 450V capacitor

I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA

So the bleeder current should be 45mA, or

R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is

P = 45e-3 * 450 = 20.25 Watts for each resistor.

Make that LOTS of fans!

In addition to supplying a known voltage across the capacitor, this also
provides a means of discharging the capacitors, which your method ignores.

The bleeder time constant is

RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe level one minute
after power off.

Your method would allow a dangerous or lethal voltage to remain on the
capacitor for an unknown amount of time. That is hardly an example of good
engineering.

Presumably something would discharge the entire string; that's a
separate issue. Caps leak like sieves in the reverse direction.



--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

On 23 Oct 2019 06:17:05 -0700, Winfield Hill <winfieldhill@yahoo.com>
wrote:

Steve Wilson wrote...

A typical spec is
I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

That's a common, but horrible formula, especially if
the units aren't stated. Should be farads and amps.

Example for a 1000uf, 450V capacitor
I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA
So the bleeder current should be 45mA, or
R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is
P = 45e-3 * 450 = 20.25 Watts for each resistor.

Wasting 40W in a pair of bleeder resistors is not an
attractive situation, leading one to change the "rule".

I routinely make HV supplies, with big caps for high
pulse output currents, but with a maximum dissipation
of under 10W. Wouldn't want to waste more than 4W.

The bleeder time constant is
RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe
level one minute after power off.

I designed a fairly simple HV discharge circuit, that
works automatically when the operating voltages are
removed, fully discharges the HV in under 5 seconds.

I like depletion fets to discharge caps linearly. May as well throw in
an LED in series.

On my big (as in BIG) gradient amps, under the cover we had a bright
LED and a warning sign and a discharge pushbutton. The depletion fet
insures that the LED is bright almost all the way to zero volts.



--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[Steve Wilson]

jlarkin@highlandsniptechnology.com wrote:

Allowing the capacitor to exceed the rated working voltage is not a
suitable engineering goal.

Now you're getting prissy. When people can't explain things, they fall
back on "good engineering practice" as their reason for doing, or
usually not doing, things.

Your goal of trying to minimize the capacitor leakage current is absurd.

You are leaving dangerous voltages on the caps.

You can use your method. I'll stick with the industry standard bleeder
resistors where the bleed current is 10 times the maximum capacitor
leakage spec

Fine. Add fans as needed.

Or a discharge switch, or some other discharge method.

A typical spec is

I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

The leakage current is measured at 20C by applying the rated voltage to
the capacitor through a series resistor of 1000 Ohms. The leakage
current is the value 2 minutes after the capacitor has reached the rated
voltage.

This test requires the capacitor to be already properly formed.

Example for a 1000uf, 450V capacitor

I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA

So the bleeder current should be 45mA, or

R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is

P = 45e-3 * 450 = 20.25 Watts for each resistor.

Make that LOTS of fans!

In addition to supplying a known voltage across the capacitor, this also
provides a means of discharging the capacitors, which your method
ignores.

The bleeder time constant is

RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe level one minute
after power off.

Your method would allow a dangerous or lethal voltage to remain on the
capacitor for an unknown amount of time. That is hardly an example of
good engineering.

Presumably something would discharge the entire string; that's a
separate issue. Caps leak like sieves in the reverse direction.

No, it's the same issue. You are leaving dangerous voltages on the caps.

Assuming something would discharge the caps is an error.

With no bleeder resistors, you still leave -330.42V on the caps, assuming a
supplied voltage of 800V. There is no way to determine the discharge time.

Version 4
SHEET 1 1648 680
WIRE 720 -256 656 -256
WIRE 768 -256 720 -256
WIRE 864 -256 832 -256
WIRE 928 -256 864 -256
WIRE 1104 -256 928 -256
WIRE 656 -240 656 -256
WIRE 928 -240 928 -256
WIRE 1104 -208 1104 -256
WIRE 656 -144 656 -160
WIRE 928 -144 928 -176
WIRE 928 -112 928 -144
WIRE 1104 -112 1104 -128
WIRE 928 -32 928 -48
FLAG 656 -144 0
FLAG 928 -32 0
FLAG 1104 -112 0
FLAG 864 -256 D1C1
FLAG 928 -144 C1C2
FLAG 720 -256 V2
SYMBOL voltage 656 -256 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
WINDOW 3 -61 161 Left 2
SYMATTR InstName V2
SYMATTR Value PULSE(800 0 1 10m 10m 20 100 1)
SYMBOL cap 912 -240 R0
SYMATTR InstName C1
SYMATTR Value 1000uf
SYMBOL cap 912 -112 R0
SYMATTR InstName C2
SYMATTR Value 100uf
SYMBOL res 1088 -224 R0
SYMATTR InstName R1
SYMATTR Value 2k
SYMBOL diode 768 -240 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value 1N4007
TEXT 720 -352 Left 2 !.tran 0 10 0 1m
TEXT 720 -384 Left 2 ;'Capacitor Discharge
TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

 

On Wednesday, 23 October 2019 17:07:35 UTC+1, Steve Wilson wrote:

jlarkin@highlandsniptechnology.com wrote:

Allowing the capacitor to exceed the rated working voltage is not a
suitable engineering goal.

Now you're getting prissy. When people can't explain things, they fall
back on "good engineering practice" as their reason for doing, or
usually not doing, things.

Your goal of trying to minimize the capacitor leakage current is absurd.

You are leaving dangerous voltages on the caps.

You can use your method. I'll stick with the industry standard bleeder
resistors where the bleed current is 10 times the maximum capacitor
leakage spec

Fine. Add fans as needed.

Or a discharge switch, or some other discharge method.

A typical spec is

I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

The leakage current is measured at 20C by applying the rated voltage to
the capacitor through a series resistor of 1000 Ohms. The leakage
current is the value 2 minutes after the capacitor has reached the rated
voltage.

This test requires the capacitor to be already properly formed.

Example for a 1000uf, 450V capacitor

I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA

So the bleeder current should be 45mA, or

R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is

P = 45e-3 * 450 = 20.25 Watts for each resistor.

Make that LOTS of fans!

In addition to supplying a known voltage across the capacitor, this also
provides a means of discharging the capacitors, which your method
ignores.

The bleeder time constant is

RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe level one minute
after power off.

Your method would allow a dangerous or lethal voltage to remain on the
capacitor for an unknown amount of time. That is hardly an example of
good engineering.

Presumably something would discharge the entire string; that's a
separate issue. Caps leak like sieves in the reverse direction.

No, it's the same issue. You are leaving dangerous voltages on the caps.

Assuming something would discharge the caps is an error.

With no bleeder resistors, you still leave -330.42V on the caps, assuming a
supplied voltage of 800V. There is no way to determine the discharge time..

snip

40w of bleeder is excessive. Your approach may work but it's a prodigious energy waste. There are assorted other ways to do things. Eg smaller resistors, fet constant current leaks, checking there's always some load on the circuit etc. Maybe permit a longer discharge time & justify it based on access time if that's permitted where you are.


NT

 

[John Larkin]

On Wed, 23 Oct 2019 16:07:29 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

jlarkin@highlandsniptechnology.com wrote:

Allowing the capacitor to exceed the rated working voltage is not a
suitable engineering goal.

Now you're getting prissy. When people can't explain things, they fall
back on "good engineering practice" as their reason for doing, or
usually not doing, things.

Your goal of trying to minimize the capacitor leakage current is absurd.

You are leaving dangerous voltages on the caps.

You can use your method. I'll stick with the industry standard bleeder
resistors where the bleed current is 10 times the maximum capacitor
leakage spec

Fine. Add fans as needed.

Or a discharge switch, or some other discharge method.

A typical spec is

I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

The leakage current is measured at 20C by applying the rated voltage to
the capacitor through a series resistor of 1000 Ohms. The leakage
current is the value 2 minutes after the capacitor has reached the rated
voltage.

This test requires the capacitor to be already properly formed.

Example for a 1000uf, 450V capacitor

I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA

So the bleeder current should be 45mA, or

R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is

P = 45e-3 * 450 = 20.25 Watts for each resistor.

Make that LOTS of fans!

In addition to supplying a known voltage across the capacitor, this also
provides a means of discharging the capacitors, which your method
ignores.

The bleeder time constant is

RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe level one minute
after power off.

Your method would allow a dangerous or lethal voltage to remain on the
capacitor for an unknown amount of time. That is hardly an example of
good engineering.

Presumably something would discharge the entire string; that's a
separate issue. Caps leak like sieves in the reverse direction.

No, it's the same issue. You are leaving dangerous voltages on the caps.

Assuming something would discharge the caps is an error.

With no bleeder resistors, you still leave -330.42V on the caps, assuming a
supplied voltage of 800V. There is no way to determine the discharge time.

That's a wonderful simulation. Negative 330.42 volts across an
electrolytic cap.

The group thanks you for the laugh.

--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[Steve Wilson]

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

That's a wonderful simulation. Negative 330.42 volts across an
electrolytic cap.

The group thanks you for the laugh.

You might have looked at the sym more closely. The lower capacitor was
100uF instead of 1000uf. Typo - missing a zero.

This shows a problem if the capacitors are not equal. You need to add a
dump diode across the lower capacitor to prevent reverse bias on the
electrolytic.

The sym assumes a constant load. You still have no protection if the load
is intermittent or shuts down when power is lost.

Here's the result when the capacitors are not matched.

Version 4
SHEET 1 1648 680
WIRE 720 -256 656 -256
WIRE 768 -256 720 -256
WIRE 864 -256 832 -256
WIRE 928 -256 864 -256
WIRE 1168 -256 928 -256
WIRE 656 -240 656 -256
WIRE 928 -240 928 -256
WIRE 1168 -208 1168 -256
WIRE 656 -144 656 -160
WIRE 928 -144 928 -176
WIRE 1008 -144 928 -144
WIRE 1072 -144 1008 -144
WIRE 928 -112 928 -144
WIRE 1072 -112 1072 -144
WIRE 1168 -112 1168 -128
WIRE 928 -32 928 -48
WIRE 1072 -32 1072 -48
FLAG 656 -144 0
FLAG 928 -32 0
FLAG 1168 -112 0
FLAG 864 -256 D1C1
FLAG 1008 -144 C1C2
FLAG 720 -256 V2
FLAG 1072 -32 0
SYMBOL voltage 656 -256 R0
WINDOW 123 0 0 Left 2
WINDOW 3 -61 161 Invisible 2
SYMATTR Value PULSE(800 0 1 10m 10m 20 100 1)
SYMATTR InstName V2
SYMBOL cap 912 -240 R0
SYMATTR InstName C1
SYMATTR Value 1000uf
SYMBOL cap 912 -112 R0
SYMATTR InstName C2
SYMATTR Value 900uf
SYMBOL res 1152 -224 R0
SYMATTR InstName R1
SYMATTR Value 2k
SYMBOL diode 768 -240 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value 1N4007
SYMBOL diode 1056 -48 M180
WINDOW 0 24 64 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D2
SYMATTR Value 1N4007
TEXT 720 -352 Left 2 !.tran 0 10 0 1m
TEXT 720 -384 Left 2 ;'Capacitor Discharge
TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

 

[John Larkin]

On Wed, 23 Oct 2019 18:12:10 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

That's a wonderful simulation. Negative 330.42 volts across an
electrolytic cap.

The group thanks you for the laugh.

You might have looked at the sym more closely. The lower capacitor was
100uF instead of 1000uf. Typo - missing a zero.

This shows a problem if the capacitors are not equal. You need to add a
dump diode across the lower capacitor to prevent reverse bias on the
electrolytic.

Why? If the caps are about equal, capacitive division works for the
transient cases. You are proposing adding a lot of parts that nobody
actually uses, to solve a problem that doesn't exist.

In a series string of similar electrolytic caps, they automatically
voltage-equalize in both directions.

This is not a problem. Sure, discharge the entire series string for
safety, as you would if there were one big cap. But no electrolytic
cap in a series string is ever going to have much negative voltage
across it.

I did a leakage test on a real cap, beyond the positive rated voltage.
You might volunteer to do a similar leakage test, but for negative
voltage. I should have done that when I had it set up.

Other people could measure some alum caps too, to get a better
sampling.

--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[Steve Wilson]

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

On Wed, 23 Oct 2019 18:12:10 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

You might have looked at the sym more closely. The lower capacitor was
100uF instead of 1000uf. Typo - missing a zero.

This shows a problem if the capacitors are not equal. You need to add a
dump diode across the lower capacitor to prevent reverse bias on the
electrolytic.

Why? If the caps are about equal, capacitive division works for the
transient cases. You are proposing adding a lot of parts that nobody
actually uses, to solve a problem that doesn't exist.

One diode is a lot of parts?

The typical electrolytic capacitance tolerance is -50% to +100%. You are
not going to get matched caps often.

The problem is near then end of the discharge. With unbalanced caps, the
lower cap would be reverse biased. The diode prevents that from happening.

In a series string of similar electrolytic caps, they automatically
voltage-equalize in both directions.

We have been through this. You previously said the capacitor leakage
current minimizes itself automatically after a few hours or days.

How do they equalize voltages? You need bleed resistors to do that.

This is not a problem. Sure, discharge the entire series string for
safety, as you would if there were one big cap. But no electrolytic
cap in a series string is ever going to have much negative voltage
across it.

The reverse bias occurs near the end of the discharge. Run the sym and
disconnect the dump diode. Then change C1 to 2000uF. Look at the voltage at
the junction.

> I did a leakage test on a real cap, beyond the positive rated voltage.

Yes, we all remember your test. You also stated the reading was very noisy,
which meant the measurement was bad.

You might volunteer to do a similar leakage test, but for negative
voltage. I should have done that when I had it set up.

There is no point. You can damage electrolytics by applying reverse bias.
That's what the dump diode is for.

Other people could measure some alum caps too, to get a better
sampling.

What's the point?

 

[John Larkin]

On Wed, 23 Oct 2019 20:27:32 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

On Wed, 23 Oct 2019 18:12:10 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

You might have looked at the sym more closely. The lower capacitor was
100uF instead of 1000uf. Typo - missing a zero.

This shows a problem if the capacitors are not equal. You need to add a
dump diode across the lower capacitor to prevent reverse bias on the
electrolytic.

Why? If the caps are about equal, capacitive division works for the
transient cases. You are proposing adding a lot of parts that nobody
actually uses, to solve a problem that doesn't exist.

One diode is a lot of parts?

The typical electrolytic capacitance tolerance is -50% to +100%. You are
not going to get matched caps often.

The problem is near then end of the discharge. With unbalanced caps, the
lower cap would be reverse biased. The diode prevents that from happening.

In a series string of similar electrolytic caps, they automatically
voltage-equalize in both directions.

We have been through this. You previously said the capacitor leakage
current minimizes itself automatically after a few hours or days.

How do they equalize voltages? You need bleed resistors to do that.

Hopeless.

This is not a problem. Sure, discharge the entire series string for
safety, as you would if there were one big cap. But no electrolytic
cap in a series string is ever going to have much negative voltage
across it.

The reverse bias occurs near the end of the discharge. Run the sym and
disconnect the dump diode. Then change C1 to 2000uF. Look at the voltage at
the junction.

I did a leakage test on a real cap, beyond the positive rated voltage.

Yes, we all remember your test. You also stated the reading was very noisy,
which meant the measurement was bad.

My HP power supply and Fluke DVM are bad?

No, the current leakage of a wet aluminum cap is noisy. There's
complex chemistry going on inside.

You might volunteer to do a similar leakage test, but for negative
voltage. I should have done that when I had it set up.

There is no point. You can damage electrolytics by applying reverse bias.
That's what the dump diode is for.

I didn't think you would actually measure anything.

Where will you get an ideal diode to prevent reverse bias on the cap?
At -0.6, it will explode and kill everyone in the building.

I need to run some polymer aluminum caps from +20 to -6 volts, so I
tested some for a couple of months. Now I know.

--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[gray_wolf]

On 10/20/2019 3:53 PM, Phil Hobbs wrote:

On 2019-10-20 03:44, Steve Wilson wrote:
jlarkin@highlandsniptechnology.com wrote:

Unless a part is very expensive, may as well test it to destruction.

Be careful doing that with electrolytics. The process of generating Al2O3
gives off hydrogen gas:

2Al + 3H2O --> Al2O3 + 3H2 (Gas) + 3e

At high rates this can exceed the capacity of the capacitor to reabsorb and
the hydrogen is released outside the capacitor. If the cap is in an enclosed
space, a dangerous buildup of hydrogen gas can provide the basis for an
explosion.

A second problem is high current can cause heating which can boil the water
in the electrolyte. The buildup of steam can rupture the case and result in
an explosion.

I had the unfortunate experience of an electrolytic explosion once. The cap
was failing but I didn't know about it since the unit was in its case. When
the cap blew, it left a dent in the 1/16 inch lid that clearly outlined the
top of the capacitor. The inside of the unit was a mess of electrolyte and
shredded aluminum foil. This was in the old days when the electrolyte was
liquid and would slosh around when shaken. But don't underestimate the power
of a steam explosion, and beware of caps that start to get hot.


Old time electros were known as "confetti generators" for that reason.

Cheers

Phil Hobbs

When I was a kid and in the Boy Scouts, at summer camp, I and a few others
were known to have put an un-opened can of beans in the 'enemy's' campfire.
It didn't take too long for a big bang to occur and the hot coals scattered about.
Thankfully no one was ever hurt but a few tents developed small holes in the top.

Another trick I learned in the air force was taking a small camera flash bulb,
the kind that had the small wires bent up the side of the flattened base and
used for contacts could be reformed to connect the leads to the filament pins of a
7 or 9 pin miniature tube and placed in the target's radio.

When turned on, there was a nice flash an a small stream of smoke coming from
the melted plastic film on the bulb. All those present would shout "WTF was that."
Some of the guys who did this also worked on the B-52s weapons systems.
Apparently they never thought to take this to another level. -0)

 

[Steve Wilson]

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

On Wed, 23 Oct 2019 20:27:32 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

In a series string of similar electrolytic caps, they automatically
voltage-equalize in both directions.

We have been through this. You previously said the capacitor leakage
current minimizes itself automatically after a few hours or days.

How do they equalize voltages? You need bleed resistors to do that.

Hopeless.

You stated:

--------------------------------------------------------------------
From: jlarkin@highlandsniptechnology.com
Newsgroups: sci.electronics.design
Subject: Re: Film capacitor as power-supply filter
Date: Mon, 07 Oct 2019 19:45:57 -0700

That is precisely the charm of wet aluminum caps. The series string
optimizes itself for minimum possible leakage current.
--------------------------------------------------------------------

How does it equalize voltages?

I did a leakage test on a real cap, beyond the positive rated voltage.

Yes, we all remember your test. You also stated the reading was very
noisy, which meant the measurement was bad.

My HP power supply and Fluke DVM are bad?

You stated:

--------------------------------------------------------------------
From: John Larkin <jlarkin@highland_atwork_technology.com>
Newsgroups: sci.electronics.design
Subject: Re: Film capacitor as power-supply filter
Date: Wed, 16 Oct 2019 16:48:02 -0700

Here's possibly the only curve like this ever posted online:

https://www.dropbox.com/s/i4wwttdgqycz9rv/Alum_Leakage_63u.JPG?raw=1

The data is crude, because there is reforming and dielectric
absorption going on, and I don't have a month to play with this. But
the curve is clearly radically upward. Not a zener, more like an MOV.

The current is very noisy above maybe 70 volts. I see what looks like
brief high current spikes.
--------------------------------------------------------------------

No, the current leakage of a wet aluminum cap is noisy. There's
complex chemistry going on inside.

Not much current leakage below the rated voltage. The cap would filter it.

You might volunteer to do a similar leakage test, but for negative
voltage. I should have done that when I had it set up.

There is no point. You can damage electrolytics by applying reverse
bias. That's what the dump diode is for.

I didn't think you would actually measure anything.

No point. you can damage the cap.

Where will you get an ideal diode to prevent reverse bias on the cap?
At -0.6, it will explode and kill everyone in the building.

Diode D2 is a 1N4007. It won't explode.

I noticed a problem with the sym. The line that reads

TEXT 1016 -352 Left 2 !.ic V(c1c2) @0

should read

TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

If you try to run the sym, you will get an error:

"Trouble parsing initial condition value for V(c1c2)"

The fact you never reported an error says you never ran the sym.

Hardly any point in discussing the results of the sym if you never run it.

 

[Steve Wilson]

Winfield Hill <winfieldhill@yahoo.com> wrote:

Steve Wilson wrote...

A typical spec is
I = 0.01CV or 3uA, whichever is greater

Where
I : Max. leakage current (uA) at 20C after 2 minutes
C : Nominal capacitance (uF)
V : Rated voltage (V)

That's a common, but horrible formula, especially if
the units aren't stated. Should be farads and amps.

Example for a 1000uf, 450V capacitor
I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA
So the bleeder current should be 45mA, or
R = 450 / 45e-3 = 10,000 Ohms

The bleeder dissipation is
P = 45e-3 * 450 = 20.25 Watts for each resistor.

Wasting 40W in a pair of bleeder resistors is not an
attractive situation, leading one to change the "rule".

I routinely make HV supplies, with big caps for high
pulse output currents, but with a maximum dissipation
of under 10W. Wouldn't want to waste more than 4W.

The bleeder time constant is
RC = 10,000 * 1000e-6 = 10 seconds.

So the capacitor voltage should be reduced to a safe level one minute
after power off.

I designed a fairly simple HV discharge circuit, that
works automatically when the operating voltages are
removed, fully discharges the HV in under 5 seconds.

can you post the LTspice file?

 

On Thu, 24 Oct 2019 06:01:55 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

John Larkin <jlarkin@highland_atwork_technology.com> wrote:

On Wed, 23 Oct 2019 20:27:32 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

In a series string of similar electrolytic caps, they automatically
voltage-equalize in both directions.

We have been through this. You previously said the capacitor leakage
current minimizes itself automatically after a few hours or days.

How do they equalize voltages? You need bleed resistors to do that.

Hopeless.

You stated:

--------------------------------------------------------------------
From: jlarkin@highlandsniptechnology.com
Newsgroups: sci.electronics.design
Subject: Re: Film capacitor as power-supply filter
Date: Mon, 07 Oct 2019 19:45:57 -0700

That is precisely the charm of wet aluminum caps. The series string
optimizes itself for minimum possible leakage current.
--------------------------------------------------------------------

How does it equalize voltages?

I did a leakage test on a real cap, beyond the positive rated voltage.

Yes, we all remember your test. You also stated the reading was very
noisy, which meant the measurement was bad.

My HP power supply and Fluke DVM are bad?

You stated:

--------------------------------------------------------------------
From: John Larkin <jlarkin@highland_atwork_technology.com
Newsgroups: sci.electronics.design
Subject: Re: Film capacitor as power-supply filter
Date: Wed, 16 Oct 2019 16:48:02 -0700

Here's possibly the only curve like this ever posted online:

https://www.dropbox.com/s/i4wwttdgqycz9rv/Alum_Leakage_63u.JPG?raw=1

The data is crude, because there is reforming and dielectric
absorption going on, and I don't have a month to play with this. But
the curve is clearly radically upward. Not a zener, more like an MOV.

The current is very noisy above maybe 70 volts. I see what looks like
brief high current spikes.
--------------------------------------------------------------------

No, the current leakage of a wet aluminum cap is noisy. There's
complex chemistry going on inside.

Not much current leakage below the rated voltage. The cap would filter it.

You might volunteer to do a similar leakage test, but for negative
voltage. I should have done that when I had it set up.

There is no point. You can damage electrolytics by applying reverse
bias. That's what the dump diode is for.

I didn't think you would actually measure anything.

No point. you can damage the cap.

Where will you get an ideal diode to prevent reverse bias on the cap?
At -0.6, it will explode and kill everyone in the building.

Diode D2 is a 1N4007. It won't explode.

I noticed a problem with the sym. The line that reads

TEXT 1016 -352 Left 2 !.ic V(c1c2) @0

should read

TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

If you try to run the sym, you will get an error:

"Trouble parsing initial condition value for V(c1c2)"

The fact you never reported an error says you never ran the sym.

If you ran your own sim, why did you not fix the error?

Hardly any point in discussing the results of the sym if you never run it.

I ran your first one and decided that it was silly. Which it certainly
was. -330 volts across an electrolytic cap!

OK, you buy a bunch of giant wirewound resistors, mounting brackets
for same, diodes, fans, blast shields, warning signs, covers with
interlock switches, safety goggles, rubber glove dispensers, fire
extinguishers, and tranquilizers so you can sleep.

I'll put electrolytic caps in series and not worry.



--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

On Thu, 24 Oct 2019 16:06:54 -0000 (UTC), Steve Wilson <no@spam.com>
wrote:

jlarkin@highlandsniptechnology.com wrote:

On Thu, 24 Oct 2019 06:01:55 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

I noticed a problem with the sym. The line that reads

TEXT 1016 -352 Left 2 !.ic V(c1c2) @0

should read

TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

If you try to run the sym, you will get an error:

"Trouble parsing initial condition value for V(c1c2)"

The fact you never reported an error says you never ran the sym.

If you ran your own sim, why did you not fix the error?

It didn't occur on my copy. Only on the posted version.

Hardly any point in discussing the results of the sym if you never run
it.

I ran your first one and decided that it was silly. Which it certainly
was. -330 volts across an electrolytic cap!

As I have already pointed out, the bottom cap had a typo.

Instead of 1,000uF, it was 100uF. Obviously something was wrong with the
sym. It took a bit of sleuthing to find the problem.

You should have noticed this when you looked at it.

OK, you buy a bunch of giant wirewound resistors, mounting brackets
for same, diodes, fans, blast shields, warning signs, covers with
interlock switches, safety goggles, rubber glove dispensers, fire
extinguishers, and tranquilizers so you can sleep.

I have since developed a method that requires no bleeder resistors, no
drain while power is applied, and will drain 800V off 2,000 uF of caps in 2
seconds when power is turned off.

I'll put electrolytic caps in series and not worry.

1. You have no discharge path when power is turned off. Plenty of loads do
not have constant power drain. This can leave dangerous voltages on the
caps.

I have already noted the obvious fact that it's sometimes reasonable
to discharge the entire series string, as if it were a single cap.

2. You can have reverse current on the lower cap when there is a constant
load and power is turned off. Reverse current is bad for electrolytics.

Reverse current? Caps always have "reverse current" during discharge.
The current is equal everywhere in a series circuit. All series caps
have the same current.

Does ripple damage caps?



--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[Steve Wilson]

jlarkin@highlandsniptechnology.com wrote:

On Thu, 24 Oct 2019 06:01:55 -0000 (UTC), Steve Wilson <no@spam.com
wrote:

I noticed a problem with the sym. The line that reads

TEXT 1016 -352 Left 2 !.ic V(c1c2) @0

should read

TEXT 1016 -352 Left 2 !.ic V(c1c2) =400

If you try to run the sym, you will get an error:

"Trouble parsing initial condition value for V(c1c2)"

The fact you never reported an error says you never ran the sym.

If you ran your own sim, why did you not fix the error?

It didn't occur on my copy. Only on the posted version.

Hardly any point in discussing the results of the sym if you never run
it.

I ran your first one and decided that it was silly. Which it certainly
was. -330 volts across an electrolytic cap!

As I have already pointed out, the bottom cap had a typo.

Instead of 1,000uF, it was 100uF. Obviously something was wrong with the
sym. It took a bit of sleuthing to find the problem.

You should have noticed this when you looked at it.

OK, you buy a bunch of giant wirewound resistors, mounting brackets
for same, diodes, fans, blast shields, warning signs, covers with
interlock switches, safety goggles, rubber glove dispensers, fire
extinguishers, and tranquilizers so you can sleep.

I have since developed a method that requires no bleeder resistors, no
drain while power is applied, and will drain 800V off 2,000 uF of caps in 2
seconds when power is turned off.

> I'll put electrolytic caps in series and not worry.

1. You have no discharge path when power is turned off. Plenty of loads do
not have constant power drain. This can leave dangerous voltages on the
caps.

2. You can have reverse current on the lower cap when there is a constant
load and power is turned off. Reverse current is bad for electrolytics.

 

[Steve Wilson]

Steve Wilson <no@spam.com> wrote:

jlarkin@highlandsniptechnology.com wrote:

On Thu, 24 Oct 2019 16:06:54 -0000 (UTC), Steve Wilson <no@spam.com
wrote:
I'll put electrolytic caps in series and not worry.

1. You have no discharge path when power is turned off. Plenty of loads
do not have constant power drain. This can leave dangerous voltages on
the caps.

I have already noted the obvious fact that it's sometimes reasonable
to discharge the entire series string, as if it were a single cap.

Then you need a dump diode.

2. You can have reverse current on the lower cap when there is a
constant load and power is turned off. Reverse current is bad for
electrolytics.

Reverse current? Caps always have "reverse current" during discharge.
The current is equal everywhere in a series circuit. All series caps
have the same current.

When you have mismatched caps with the higher value on top, the top cap
will try to push the voltage on the bottom cap negative. A dump diode
will prevent the bottom cap from going negative.

This doesn't help when the top cap has a lower value than the bottom
one.

As I pointed out, the capacitor tolerance is -50% to +100%, so your
chances of getting matched caps is very low.

Does ripple damage caps?

Of course it can if it too high, or causes reverse current through the
cap at or near zero volts.

A bit more explanation. When the cap voltage goes negative, the current
through the cap reverses.

This causes the formation of a thin aluminum oxide layer on the cathode.

This causes another capacitor to form on the cathode. This capacitor is in
series with the capacitor on the anode, so the total capacitance is
reduced.

So passing current through the cap when the voltage is near zero is a bad
idea.

The dump diode will prevent this from happening. It bypasses the current
from going through the cap.

 

[Steve Wilson]

jlarkin@highlandsniptechnology.com wrote:

On Thu, 24 Oct 2019 16:06:54 -0000 (UTC), Steve Wilson <no@spam.com
wrote:
I'll put electrolytic caps in series and not worry.

1. You have no discharge path when power is turned off. Plenty of loads
do not have constant power drain. This can leave dangerous voltages on
the caps.

I have already noted the obvious fact that it's sometimes reasonable
to discharge the entire series string, as if it were a single cap.

Then you need a dump diode.

2. You can have reverse current on the lower cap when there is a
constant load and power is turned off. Reverse current is bad for
electrolytics.

Reverse current? Caps always have "reverse current" during discharge.
The current is equal everywhere in a series circuit. All series caps
have the same current.

When you have mismatched caps with the higher value on top, the top cap
will try to push the voltage on the bottom cap negative. A dump diode will
prevent the bottom cap from going negative.

This doesn't help when the top cap has a lower value than the bottom one.

As I pointed out, the capacitor tolerance is -50% to +100%, so your chances
of getting matched caps is very low.

> Does ripple damage caps?

Of course it can if it too high, or causes reverse current through the cap
at or near zero volts.

 

[Steve Wilson]

Steve Wilson <no@spam.com> wrote:

Does ripple damage caps?

Of course it can if it too high, or causes reverse current through the
cap at or near zero volts.

A bit more explanation. When the cap voltage goes negative, the current
through the cap reverses.

This causes the formation of a thin aluminum oxide layer on the cathode.

This causes another capacitor to form on the cathode. This capacitor is
in series with the capacitor on the anode, so the total capacitance is
reduced.

So passing current through the cap when the voltage is near zero is a
bad idea.

The dump diode will prevent this from happening. It bypasses the current
from going through the cap.

Best Supporting Reference:

https://www.mouser.com/pdfDocs/UCC_ElectrolyticCapacitorTechnicalNotes.pdf

Page 10

Applying a reverse voltage will cause chemical reactions (formation of
dielectric) to occur on the cathode foil, and, as is the case with
overvoltage, the leakage current will rapidly increase with heat and gases
generating and thus the internal pressure increases. The reactions are
accelerated by the voltage, current density and ambient temperature. It
may also accompany a reduction in capacitance and an increase in tan h as
well as an increase in the leakage current. An example of capacitor
reverse-voltage characteristics is shown in Fig. 27. A reverse voltage of
as small as 1V can cause the capacitance to decrease. A reverse voltage of
2 to 3V can shorten lifetime due to a reduction in capacitance, increase
in tan h, and/or increase in leakage current. A reverse voltage of even
higher value can open the pressure relief vent or lead to destructive
failures (Fig. 27).

NOTES:

There are many references on the web to reversed voltage on electrolytic
capacitors. Many of them are just plain silly, such as the production of
oxygen on the cathode. This is impossible due to the electrolysis equation:

2Al + 3H2O --> Al2O3 + 3H2 (Gas) + 3e

This shows that hydrogen is produced, not oxygen.

However, the web links almost universally indicate that reversing the
polarity of an aluminum electrolytic cap will damage it.

Also, the original PDF would not allow copying the above paragraph. I
simply passed it to "Remove restrictions in PDF files", at

https://online2pdf.com/remove-pdf-restrictions

Now I can copy information freely from the PDF.

 

[Phil Allison]

Steve Wilson wrote:

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

The end of the series capacitor saga.

** Wish it was the end of PITAs like you.

If the capacitor values can be held to within 25% of each other, then the
bleed resistors needed to set the capacitor voltages to 50% of VCC can be
reduced to 1.6W each.

** There is no general need for bleed resistors at all.

When the AC supply is cut off, the existing load bleeds them down very nicely.

Only in a few cases can the load disappear completely in normal use.



...... Phil

 

[Steve Wilson]

The end of the series capacitor saga.

If the capacitor values can be held to within 25% of each other, then the
bleed resistors needed to set the capacitor voltages to 50% of VCC can be
reduced to 1.6W each.

The rational is the typical leakage spec is a maximum value. For example,

I = 0.01 * C * V (in microamps)
= 0.01 * 1000 * 450
= 4.5mA

To allow worst-case conditions, the industry standard is to use 10 times
this current for the bleed resistors. This results in high power dissipated
in the bleed resistors.

However, manufacturers try to minimize the leakage current, and often
produce capacitors with leakage currents one hundred times lower.

This means the bleed resistor values can be much higher and the power
dissipated can be much lower.

This also allows a simple way to discharge the capacitor string within 2
seconds after power off. Simply use a relay to switch a dump resistor
across the supply voltage when power is turned off. When power is on, the
dump resistor is switched off and does not load the supply. As soon as
power is turned off, the relay turns off and the normally closed contacts
switch the dump resistor across the supply.

Cheap dump diodes (1N4007) are still needed to prevent reverse voltage on
the caps, which can damage them.

Here is the ASC file that shows how this works.

Version 4
SHEET 1 1532 0
WIRE 752 -256 688 -256
WIRE 768 -256 752 -256
WIRE 864 -256 832 -256
WIRE 928 -256 864 -256
WIRE 1088 -256 928 -256
WIRE 1248 -256 1088 -256
WIRE 1472 -256 1248 -256
WIRE 688 -240 688 -256
WIRE 928 -240 928 -256
WIRE 1088 -240 1088 -256
WIRE 1248 -240 1248 -256
WIRE 1472 -176 1472 -256
WIRE 688 -144 688 -160
WIRE 928 -144 928 -176
WIRE 1008 -144 928 -144
WIRE 1088 -144 1088 -176
WIRE 1088 -144 1008 -144
WIRE 1248 -144 1248 -160
WIRE 1248 -144 1088 -144
WIRE 1248 -128 1248 -144
WIRE 928 -112 928 -144
WIRE 1088 -112 1088 -144
WIRE 1472 -48 1472 -96
WIRE 928 -32 928 -48
WIRE 1088 -32 1088 -48
WIRE 1248 -32 1248 -48
FLAG 688 -144 0
FLAG 928 -32 0
FLAG 1248 -32 0
FLAG 864 -256 D1C1
FLAG 1008 -144 C1C2
FLAG 752 -256 V1
FLAG 1088 -32 0
FLAG 1472 -48 0
SYMBOL voltage 688 -256 R0
WINDOW 123 0 0 Left 2
WINDOW 3 -61 161 Invisible 2
WINDOW 39 0 0 Left 2
SYMATTR Value PULSE(0 800 1 1 10m 50 100 1)
SYMATTR InstName V1
SYMBOL cap 912 -240 R0
SYMATTR InstName C1
SYMATTR Value 1250uf
SYMBOL cap 912 -112 R0
SYMATTR InstName C2
SYMATTR Value 1000uf
SYMBOL res 1232 -144 R0
SYMATTR InstName R2
SYMATTR Value 100k
SYMBOL diode 768 -240 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value 1N4007
SYMBOL diode 1072 -48 M180
WINDOW 0 24 64 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D3
SYMATTR Value 1N4007
SYMBOL diode 1072 -176 M180
WINDOW 0 24 64 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D2
SYMATTR Value 1N4007
SYMBOL res 1232 -256 R0
SYMATTR InstName R1
SYMATTR Value 100k
SYMBOL res 1456 -192 R0
SYMATTR InstName Rd
SYMATTR Value 1k
TEXT 720 -352 Left 2 !.tran 0 100 0 1m
TEXT 720 -384 Left 2 ;'Capacitor Discharge
TEXT 1016 -352 Left 2 !.ic V(c1c2) = 0