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### Network # Voltage Measurement on a Capcitor

J

#### James Howe

Jan 1, 1970
0
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.

B

#### Brian

Jan 1, 1970
0
James Howe said:
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.

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

B

#### Bob

Jan 1, 1970
0
James Howe said:
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.

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

R

#### Richard The Troll

Jan 1, 1970
0
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.

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!'

Well, you'd better! ;-)

Cheers!
Rich

J

#### John Larkin

Jan 1, 1970
0
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

J

#### John Larkin

Jan 1, 1970
0
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.

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

R

#### R. Steve Walz

Jan 1, 1970
0
James said:
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

B

#### Brian

Jan 1, 1970
0
John Larkin said:
-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

J

#### John Larkin

Jan 1, 1970
0
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

J

#### James Howe

Jan 1, 1970
0
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

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.

B

#### Brian

Jan 1, 1970
0
John Larkin said:
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

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