Capacitor on the output

[Ed]

I've noticed on some circuits that op-amps sometimes have a capacitor on the
output (in series). For example, the circuit I'm looking at has an LM741
op-amp with a 0.02uF capacitor on the output. What is the purpose of this
capacitor? Is it just to filter out noise? How do you calculate the value
of the required capacitor (does some magic formula exist)?

 

[John Popelish]

Ed wrote:

I've noticed on some circuits that op-amps sometimes have a capacitor on the
output (in series). For example, the circuit I'm looking at has an LM741
op-amp with a 0.02uF capacitor on the output. What is the purpose of this
capacitor? Is it just to filter out noise? How do you calculate the value
of the required capacitor (does some magic formula exist)?

A capacitor in series with the signal path acts as a high pass filter,
blocking any DC in the signal from passing through while letting
frequencies of interest an easy path.

--
John Popelish

 

[Bob Masta]

On Wed, 28 Jul 2004 17:16:00 -0400, John Popelish <jpopelish@rica.net>
wrote:

Ed wrote:

I've noticed on some circuits that op-amps sometimes have a capacitor on the
output (in series). For example, the circuit I'm looking at has an LM741
op-amp with a 0.02uF capacitor on the output. What is the purpose of this
capacitor? Is it just to filter out noise? How do you calculate the value
of the required capacitor (does some magic formula exist)?

A capacitor in series with the signal path acts as a high pass filter,
blocking any DC in the signal from passing through while letting
frequencies of interest an easy path.

The value of the capacitor depends upon the input impedance of
the circuit that follows it, and the frequencies that you want to
allow to pass. Normally, we consider the "cutoff" frequency to
be that where the output is reduced by 3 dB (0.707). At lower
frequencies, the output is reduced even more, proportional
to the frequency: Every halving of the frequency below cutoff
results in a halving of the output level (-6 dB / octave).

The formula for the cutoff is F3 = 1 / (2 * pi * R * C)
where F3 is the cutoff frequency in Hertz,
R is the following stage input impedance
in ohms, and C is the capacitance in *Farads*.
You can rearrange this to C = 1 / (2 * pi * F3 * C)
when you are trying to pick a C from design
considerations. In your example case, if the
following stage happens to have an input
impedance of 100K the F3 would be
F3 = 1 / (2 * pi * 100000 * 0.02x10^-6) = 79.6 Hz.

Hope this helps!





Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[Activ8]

On Thu, 29 Jul 2004 12:24:21 GMT, Bob Masta wrote:

On Wed, 28 Jul 2004 17:16:00 -0400, John Popelish <jpopelish@rica.net
wrote:

Ed wrote:

I've noticed on some circuits that op-amps sometimes have a capacitor on the
output (in series). For example, the circuit I'm looking at has an LM741
op-amp with a 0.02uF capacitor on the output. What is the purpose of this
capacitor? Is it just to filter out noise? How do you calculate the value
of the required capacitor (does some magic formula exist)?

A capacitor in series with the signal path acts as a high pass filter,
blocking any DC in the signal from passing through while letting
frequencies of interest an easy path.


The value of the capacitor depends upon the input impedance of
the circuit that follows it, and the frequencies that you want to
allow to pass. Normally, we consider the "cutoff" frequency to
be that where the output is reduced by 3 dB (0.707). At lower
frequencies, the output is reduced even more, proportional
to the frequency: Every halving of the frequency below cutoff
results in a halving of the output level (-6 dB / octave).

The formula for the cutoff is F3 = 1 / (2 * pi * R * C)
where F3 is the cutoff frequency in Hertz,
R is the following stage input impedance
in ohms, and C is the capacitance in *Farads*.
You can rearrange this to C = 1 / (2 * pi * F3 * C)
^^^^^^^^^^^^^^^^^^^^^^^^^^

C = 1 / (2 * pi * F3 * R)
^^^

when you are trying to pick a C from design
considerations. In your example case, if the
following stage happens to have an input
impedance of 100K the F3 would be
F3 = 1 / (2 * pi * 100000 * 0.02x10^-6) = 79.6 Hz.

Hope this helps!

Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

--
Best Regards,
Mike