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| Lostgallifreyan... |
 Posted: Thu Oct 08, 2009 10:30 am |
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DaveC <invalid at (no spam) invalid.net> wrote in
news:0001HW.C6F35F0303EAED94B08A39AF at (no spam) news.eternal-september.org:
[quote:2d72e5cced]peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V
JF
This is for half-wave rect.?
Thanks.
[/quote:2d72e5cced]
Yes, under no load either peak of a sine wave defines the maximum (absolute)
value for voltage across a capacitor. Full wave rectification just means more
energy and less ripple present for a given load. |
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| Mike Paff... |
| Posted: Thu Oct 08, 2009 10:31 am |
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On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid at (no spam) invalid.net> wrote:
[quote:76039dd1fa]peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V
JF
This is for half-wave rect.?
Thanks.
[/quote:76039dd1fa]
Half-wave and full-wave rectification result in the same peak voltage.
It's just that for full-wave, the peaks occur twice as often. |
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| Hal Murray... |
| Posted: Thu Oct 08, 2009 11:41 am |
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[quote:6e6dcc96a8]peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V
[/quote:6e6dcc96a8]
My line voltage runs well above 110.
PG&E says they target for a min of 114 and a max of 126.
That's at the handoff point. I think the idea is that
you get a min of 110 if you allocate a few volts for
drop on the wiring within your house.
Here is PG&E's blurb:
http://www.pge.com/tariffs/tm2/pdf/ELEC_RULES_2.pdf
--
These are my opinions, not necessarily my employer's. I hate spam. |
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| Wild_Bill... |
| Posted: Thu Oct 08, 2009 3:24 pm |
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Yep, yep.. half-wave is a single rectifier, full-wave is 2 rectifiers often
with a center-tapped AC source, and full-wave-bridge rectification is 4
rectifiers.
--
Cheers,
WB
..............
"Mike Paff" <paffm at (no spam) yahoo.com> wrote in message
news:lt4sc59ujefvrdpkjaldeobqupgspd813n at (no spam) 4ax.com...
[quote:f9306c5e52]On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid at (no spam) invalid.net> wrote:
peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V
JF
This is for half-wave rect.?
Thanks.
Half-wave and full-wave rectification result in the same peak voltage.
It's just that for full-wave, the peaks occur twice as often.[/quote:f9306c5e52] |
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| John Fields... |
| Posted: Thu Oct 08, 2009 4:34 pm |
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On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid at (no spam) invalid.net> wrote:
[quote:f6bde68663]peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V
JF
This is for half-wave rect.?
[/quote:f6bde68663]
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. |
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| Lostgallifreyan... |
| Posted: Thu Oct 08, 2009 9:27 pm |
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John Fields <jfields at (no spam) austininstruments.com> wrote in
news:u9osc5h0oodogdtbfdgo9us2hg152hq0j4 at (no spam) 4ax.com:
[quote:7e439e87fe]On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid at (no spam) 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.
[/quote:7e439e87fe]
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. |
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| John Fields... |
| Posted: Fri Oct 09, 2009 5:16 am |
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On Thu, 08 Oct 2009 22:27:38 -0500, Lostgallifreyan <no-one at (no spam) nowhere.net>
wrote:
[quote:ece6ae7cdd]John Fields <jfields at (no spam) austininstruments.com> wrote in
news:u9osc5h0oodogdtbfdgo9us2hg152hq0j4 at (no spam) 4ax.com:
On Thu, 8 Oct 2009 09:18:59 -0700, DaveC <invalid at (no spam) 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.
[/quote:ece6ae7cdd]
---
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. |
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| DaveC... |
| Posted: Fri Oct 09, 2009 9:13 pm |
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[quote:272fb6ebe5]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.
[/quote:272fb6ebe5]
No worries, mate! I'm learned(er).
[quote:272fb6ebe5]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.
[/quote:272fb6ebe5]
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. |
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| Lostgallifreyan... |
| Posted: Fri Oct 09, 2009 9:47 pm |
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DaveC <invalid at (no spam) invalid.net> wrote in
news:0001HW.C6F54A01045E00D4B08A39AF at (no spam) news.eternal-september.org:
[quote:aedb8e506f]Is 160 good enough for this application?
[/quote:aedb8e506f]
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. |
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| John Fields... |
| Posted: Sat Oct 10, 2009 5:43 am |
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On Fri, 9 Oct 2009 20:13:53 -0700, DaveC <invalid at (no spam) invalid.net> wrote:
[quote:e9a63f5ce2]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?
[/quote:e9a63f5ce2]
---
Dunno...
Can you post a schematic? |
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| Frank S... |
| Posted: Sun Oct 18, 2009 6:04 pm |
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Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing
Frank
"Lostgallifreyan" <no-one at (no spam) nowhere.net> wrote in message
news:Xns9C9E2D5A92AE2zoodlewurdle at (no spam) 216.196.109.145...
[quote:eac024475d]DaveC <invalid at (no spam) invalid.net> wrote in
news:0001HW.C6F2193A039E92E2B08A39AF at (no spam) news.eternal-september.org:
What was the existing cap?
DaveC <invalid at (no spam) invalid.net> wrote in
news:0001HW.C6F178710378E3BEB08A39AF at (no spam) 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.[/quote:eac024475d] |
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| Lostgallifreyan... |
| Posted: Sun Oct 18, 2009 9:11 pm |
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"Frank S" <items4sale at (no spam) bellsouth.net> wrote in
news:hbgaft$14h4$1 at (no spam) adenine.netfront.net:
[quote:a0e726e085]Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing
[/quote:a0e726e085]
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. |
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| whisky-dave... |
| Posted: Wed Oct 21, 2009 8:11 am |
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"Lostgallifreyan" <no-one at (no spam) nowhere.net> wrote in message
news:Xns9CA92AB5B28E8zoodlewurdle at (no spam) 216.196.109.145...
[quote]"Frank S" <items4sale at (no spam) bellsouth.net> wrote in
news:hbgaft$14h4$1 at (no spam) 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.
[/quote]
And there's how hard the user pushes them as regards the voltage and
temperature
ratings. |
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