Cap value for timing circuit

John Fields <jfields@austininstruments.com> wrote in
news:u9osc5h0oodogdtbfdgo9us2hg152hq0j4@4ax.com:

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid@invalid.net> wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.
Click to expand...
So what IS the story, exactly? Right now you've focussed on the results of a
load (the resistor), so you've said things that might be taken to discredit
the rest of us who specified that either peak has an absolute value of
voltage across a capacitor that when unloaded, must be considered for its
safety. While a half-wave rectified form is a more complex wave whose RMS
value needs a different calculation, in this case it's the separation between
peaks that matters.

Many PSU capacitor failures seems to be a result of people underspecifying
the working voltage while assuming they'll be 'safely' loaded to prevent the
peak from dictating terms more than the RMS. It's not a safe assumption.
 
On Thu, 08 Oct 2009 22:27:38 -0500, Lostgallifreyan <no-one@nowhere.net>
wrote:

John Fields <jfields@austininstruments.com> wrote in
news:u9osc5h0oodogdtbfdgo9us2hg152hq0j4@4ax.com:

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid@invalid.net> wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.



So what IS the story, exactly? Right now you've focussed on the results of a
load (the resistor), so you've said things that might be taken to discredit
the rest of us who specified that either peak has an absolute value of
voltage across a capacitor that when unloaded, must be considered for its
safety.

While a half-wave rectified form is a more complex wave whose RMS
value needs a different calculation, in this case it's the separation between
peaks that matters.

Many PSU capacitor failures seems to be a result of people underspecifying
the working voltage while assuming they'll be 'safely' loaded to prevent the
peak from dictating terms more than the RMS. It's not a safe assumption.
Click to expand...
---
Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.
 
Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.
Click to expand...
No worries, mate! I'm learned(er).

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.
Click to expand...
Wow. That's probably why the original 150v cap is no good after 40 years. It
should have been spec'd higher. Is 160 good enough for this application?

Thanks.
 
DaveC <invalid@invalid.net> wrote in
news:0001HW.C6F54A01045E00D4B08A39AF@news.eternal-september.org:

Is 160 good enough for this application?
Click to expand...
Doubtful, note the post from Hal Murray, about how much line voltage can
vary. (Even at exactly 110V which it likely never is for long, you're pushing
to to within 5V of a dangerous limit and trusting capacitor voltage tolerance
is a Very Bad Idea, unless you're going to stress test samples in multiple
copies of a design).

You might get by for a while if the cap is run as cool as possible, but bear
in mind that the last one failed, and you now have a very good reason to
suspect it failed from being run too close to maximum levels. So leave a
generous margin this time.
 
On Fri, 9 Oct 2009 20:13:53 -0700, DaveC <invalid@invalid.net> wrote:

Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.

No worries, mate! I'm learned(er).

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.

Wow. That's probably why the original 150v cap is no good after 40 years. It
should have been spec'd higher. Is 160 good enough for this application?
Click to expand...
---
Dunno...

Can you post a schematic?
 
Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing
Frank

"Lostgallifreyan" <no-one@nowhere.net> wrote in message
news:Xns9C9E2D5A92AE2zoodlewurdle@216.196.109.145...
DaveC <invalid@invalid.net> wrote in
news:0001HW.C6F2193A039E92E2B08A39AF@news.eternal-september.org:

What was the existing cap?

DaveC <invalid@invalid.net> wrote in
news:0001HW.C6F178710378E3BEB08A39AF@news.eternal-september.org:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. :) Heat could well be right. Also accelerated
loss
of electrolyte. Did the cap look like it had burst from inside earlier
than
any final appearance of internal gubbins? Another possibility is the
voltage
it saw. When you say 110 volts, you mean the mains, right? If so it will
be
rectified and that cap will be seeing the peak value, not the RMS, so
around
125V when under no load, so you really do need it to have a higher voltage
rating than that.
Click to expand...
 
"Frank S" <items4sale@bellsouth.net> wrote in
news:hbgaft$14h4$1@adenine.netfront.net:

Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing
Click to expand...
I agree. Was just saying that length of time is like the length of a piece of
string. Like someone else here said, some electrolytic caps work fine in gear
tens of years old. Valve/tube gear even, where you have enough heat to
accerate drying. It sort of suggests that looking at lengths of time is
missing plenty, there's a lot of difference between electrolytics, they're
not at all consistent.
 
"Lostgallifreyan" <no-one@nowhere.net> wrote in message
news:Xns9CA92AB5B28E8zoodlewurdle@216.196.109.145...
"Frank S" <items4sale@bellsouth.net> wrote in
news:hbgaft$14h4$1@adenine.netfront.net:

Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing


I agree. Was just saying that length of time is like the length of a piece
of
string. Like someone else here said, some electrolytic caps work fine in
gear
tens of years old. Valve/tube gear even, where you have enough heat to
accerate drying. It sort of suggests that looking at lengths of time is
missing plenty, there's a lot of difference between electrolytics, they're
not at all consistent.
Click to expand...
And there's how hard the user pushes them as regards the voltage and
temperature
ratings.
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top