Voltage Measurement on a Capcitor

J

James Howe

Guest
I'm analyzing a circuit which contains a capacitor. The capacitor is fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe
 
On 12 Dec 2004 18:13:47 EST, "James Howe"
<concentricnews@wingspread.imap-mail.com> wrote:

I'm analyzing a circuit which contains a capacitor. The capacitor is fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.
Click to expand...

A DC voltmeter reads the average voltage. The average voltage of an 8
volts p-p sawtooth, riding on a base of -12, should be

-12 + (0.5 * 8) = -8,

close to what you're seeing.

John
 
"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The capacitor is
fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures
around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook
up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe
Click to expand...
Your volt meter is probably reading the RMS voltage, which is the peak
voltage (of the sawtooth), divided by the square root of 3 (with a little
bit if DC voltage added in too).
Brian
 
"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The capacitor is fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.

--
James Howe
Click to expand...
If you're using a voltmeter when you get the -7.5V reading then you're
probably seeing either the average or rms value (depending on the type of
voltmeter you have). The meter's response is too slow for you to see the
variations in the voltage -- whereas the scope is fast enough.

Also, be aware that the meter and/or scope can affect the reading if the
current source is small (in value). With low-amperage current sources the
current drawn by the measuring equipment can have an effect on the actual
voltage in the circuit. For example, I recently was measuring the voltage on
a capacitor that was tied to a non-varying current source. When I hooked up
a voltmeter, I saw the capacitor's voltage changing -- even though I knew
that its voltage had stopped changing (before I connect the meter). The
voltmeter was drawing current from the capacitor and thus caused its voltage
to change.

Bob
 
On Sun, 12 Dec 2004 18:13:47 -0500, James Howe wrote:

I'm analyzing a circuit which contains a capacitor. The capacitor is fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.
Click to expand...
The average voltage at that point, relative to whatever point the other
meter lead is probing.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'
Click to expand...
Well, you'd better! ;-)

Cheers!
Rich
 
On Sun, 12 Dec 2004 18:11:41 -0600, "Brian" <bellis350@comcast.net>
wrote:

"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The capacitor is
fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures
around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook
up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe

Your volt meter is probably reading the RMS voltage, which is the peak
voltage (of the sawtooth), divided by the square root of 3 (with a little
bit if DC voltage added in too).
Brian
Click to expand...

-7.5 is an interesting "RMS" voltage!

John
 
James Howe wrote:
I'm analyzing a circuit which contains a capacitor. The capacitor is fed
a constant current and is discharged periodically. The capacitor is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a flat
line which I presume would be the -11.94 v source voltage. When I hook up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.
James Howe
------------------------------
Click to expand...
Do you have the AC switch on on your scope inputs??

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public
 
"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:vctpr0lpe777mglabf0r6hkvgceflghtpt@4ax.com...
On Sun, 12 Dec 2004 18:11:41 -0600, "Brian" <bellis350@comcast.net
wrote:


"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in
message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The capacitor
is
fed
a constant current and is discharged periodically. The capacitor is
a
non-polarized .01uf capacitor connected on one side to a -12v supply
and
the other side to the constant current source. If I measure voltage
on
one pin, I get approximately -7.5v, the other pin measures
around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a
flat
line which I presume would be the -11.94 v source voltage. When I
hook
up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure
out,
being new to both electronics and using oscilloscopes, is what
the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer
I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe

Your volt meter is probably reading the RMS voltage, which is the peak
voltage (of the sawtooth), divided by the square root of 3 (with a
little
bit if DC voltage added in too).
Brian



-7.5 is an interesting "RMS" voltage!

John
Click to expand...
If you subtract the effective RMS voltage of a sawtooth wave that has a
peak voltage of about 8 volts, from the - 12 volts DC, you come out with an
effective RMS voltage of 7.38 volts. While this is not an AC voltage, it
would take an AC voltage of this value to get the same heating effect. Since
his meter is set to read DC voltage, he will get a negative reading.
12V - 4.62V = 7.38V
Brian
 
On Sun, 12 Dec 2004 22:21:57 -0600, "Brian" <bellis350@comcast.net>
wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:vctpr0lpe777mglabf0r6hkvgceflghtpt@4ax.com...
On Sun, 12 Dec 2004 18:11:41 -0600, "Brian" <bellis350@comcast.net
wrote:


"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in
message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The capacitor
is
fed
a constant current and is discharged periodically. The capacitor is
a
non-polarized .01uf capacitor connected on one side to a -12v supply
and
the other side to the constant current source. If I measure voltage
on
one pin, I get approximately -7.5v, the other pin measures
around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see a
flat
line which I presume would be the -11.94 v source voltage. When I
hook
up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high. I
expected to see both of these outcomes. What I'm trying to figure
out,
being new to both electronics and using oscilloscopes, is what
the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer
I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe

Your volt meter is probably reading the RMS voltage, which is the peak
voltage (of the sawtooth), divided by the square root of 3 (with a
little
bit if DC voltage added in too).
Brian



-7.5 is an interesting "RMS" voltage!

John


If you subtract the effective RMS voltage of a sawtooth wave that has a
peak voltage of about 8 volts, from the - 12 volts DC, you come out with an
effective RMS voltage of 7.38 volts. While this is not an AC voltage, it
would take an AC voltage of this value to get the same heating effect. Since
his meter is set to read DC voltage, he will get a negative reading.
12V - 4.62V = 7.38V
Brian
Click to expand...
But DC voltmeters (usually) read average, not RMS.

The RMS value of this waveform (an 8v p-p sawtooth whose max negative
excursion is -12) must be greater than 8, because the average is 8.
The RMS is actually 8.33.

John
 
On Sun, 12 Dec 2004 17:56:23 -0800, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On 12 Dec 2004 18:13:47 EST, "James Howe"
concentricnews@wingspread.imap-mail.com> wrote:

I'm analyzing a circuit which contains a capacitor. The capacitor is
fed a constant current and is discharged periodically. The capacitor
is a
non-polarized .01uf capacitor connected on one side to a -12v supply and
the other side to the constant current source. If I measure voltage on
one pin, I get approximately -7.5v, the other pin measures around
-11.94. [...] What I'm trying to figure out,
being new to both electronics and using oscilloscopes, is what the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the answer I'll
probably go 'Doh!'

Thanks.


A DC voltmeter reads the average voltage. The average voltage of an 8
volts p-p sawtooth, riding on a base of -12, should be

-12 + (0.5 * 8) = -8,

close to what you're seeing.

John
Click to expand...

Ok, that's what I suspected. There was some doubt about whether the cap
was really discharging completely back to -12 before ramping again, but
given the wave is 8v p-p and the -7.x v is close to the average if the
swing was from -12 to -4 I feel pretty comfortable that the cap is
draining to -12 before recharging.

Thanks.


--
James Howe

Contact: http://public.xdi.org/=James.Howe
 
"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:a17qr05coqb4vqpa1ckkpf80aujohp4t28@4ax.com...
On Sun, 12 Dec 2004 22:21:57 -0600, "Brian" <bellis350@comcast.net
wrote:


"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:vctpr0lpe777mglabf0r6hkvgceflghtpt@4ax.com...
On Sun, 12 Dec 2004 18:11:41 -0600, "Brian" <bellis350@comcast.net
wrote:


"James Howe" <concentricnews@wingspread.imap-mail.com> wrote in
message
news:eek:psiw366cti1k3p1@eaglefeather.riskmetrics-aa.riskmetrics.com...
I'm analyzing a circuit which contains a capacitor. The
capacitor
is
fed
a constant current and is discharged periodically. The
capacitor is
a
non-polarized .01uf capacitor connected on one side to a -12v
supply
and
the other side to the constant current source. If I measure
voltage
on
one pin, I get approximately -7.5v, the other pin measures
around -11.94.
When I hook an oscilloscope up to the pin with the -11.94, I see
a
flat
line which I presume would be the -11.94 v source voltage. When
I
hook
up
to the -7.5v side, I see a sawtooth ramp which is 8 volts high.
I
expected to see both of these outcomes. What I'm trying to
figure
out,
being new to both electronics and using oscilloscopes, is what
the -7.5v
volt meter value represents.

I'm sure this is a really dumb question and when I see the
answer
I'll
probably go 'Doh!'

Thanks.

--
James Howe

Contact: http://public.xdi.org/=James.Howe

Your volt meter is probably reading the RMS voltage, which is the
peak
voltage (of the sawtooth), divided by the square root of 3 (with a
little
bit if DC voltage added in too).
Brian



-7.5 is an interesting "RMS" voltage!

John


If you subtract the effective RMS voltage of a sawtooth wave that has
a
peak voltage of about 8 volts, from the - 12 volts DC, you come out with
an
effective RMS voltage of 7.38 volts. While this is not an AC voltage, it
would take an AC voltage of this value to get the same heating effect.
Since
his meter is set to read DC voltage, he will get a negative reading.
12V - 4.62V = 7.38V
Brian


But DC voltmeters (usually) read average, not RMS.

The RMS value of this waveform (an 8v p-p sawtooth whose max negative
excursion is -12) must be greater than 8, because the average is 8.
The RMS is actually 8.33.

John
Click to expand...
Your Right.
 
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