Integrator question

J

Jim Miller

Guest
I have been looking at using an integrator to measure a current. The
integrator would be the typical opamp with about 1nf of feedback cap. My
problem is that the current source is a photodiode with about 300pf
capacitance.

Am I correct that the integrator output as a function of time in response to
the current source will be larger due to the "gain" created by the ratio of
the feedback to source capacitance?

thanks

jtm
 
On Tue, 5 Oct 2004 09:20:16 -0400, "Jim Miller"
<jim@removethisjtmiller.com> wrote:

I have been looking at using an integrator to measure a current. The
integrator would be the typical opamp with about 1nf of feedback cap. My
problem is that the current source is a photodiode with about 300pf
capacitance.

Am I correct that the integrator output as a function of time in response to
the current source will be larger due to the "gain" created by the ratio of
the feedback to source capacitance?
Click to expand...
No. The integration rate depends only on the pd current dumped into
the integration cap. A very large pd capacitance can cause ringing or
oscillation in the opamp, but your numbers are probably safe.

John
 
John Larkin wrote:

On Tue, 5 Oct 2004 09:20:16 -0400, "Jim Miller"
jim@removethisjtmiller.com> wrote:


I have been looking at using an integrator to measure a current. The
integrator would be the typical opamp with about 1nf of feedback cap. My
problem is that the current source is a photodiode with about 300pf
capacitance.

Am I correct that the integrator output as a function of time in response to
the current source will be larger due to the "gain" created by the ratio of
the feedback to source capacitance?



No. The integration rate depends only on the pd current dumped into
the integration cap. A very large pd capacitance can cause ringing or
oscillation in the opamp, but your numbers are probably safe.

John

The way to understand this is to add the following rule to your toolkit:
Click to expand...
An op-amp will do everything it can to keep the sum of all of the
currents flowing into the V- node equal to zero. So if the only
connections are the op-amp (assume zero current for now; that's not
always valid but pretend), the feedback cap and the photodiode, then the
output voltage _must_ change at a rate sufficient to keep the capacitor
current equal to the photodiode current. It ends up looking like the
photodiode is charging the cap (which is what you want).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
hi john

so the input capacitance wouldn't act as a divider to reduce the feedback
from the "feedback" cap and thereby require the opamp output to be greater
in order to achieve a null at the summing junction?

i should have added some additional information on the original post: the
photodiode has a gigohm impedence other than the capacitive element if this
has any importance.

if you're right then since the input current is in the picoamp range and the
slew rate of output is consequently in the volt per minute range i probably
shouldn't have any issues with stability.

thanks
jtm

"John Larkin" <jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote in
message news:9rj5m0dbe93evs5bll97fsod29fb2ugrlo@4ax.com...
On Tue, 5 Oct 2004 09:20:16 -0400, "Jim Miller"
<jim@removethisjtmiller.com> wrote:

I have been looking at using an integrator to measure a current. The
integrator would be the typical opamp with about 1nf of feedback cap. My
problem is that the current source is a photodiode with about 300pf
capacitance.

Am I correct that the integrator output as a function of time in response
to
the current source will be larger due to the "gain" created by the ratio of
the feedback to source capacitance?
Click to expand...
No. The integration rate depends only on the pd current dumped into
the integration cap. A very large pd capacitance can cause ringing or
oscillation in the opamp, but your numbers are probably safe.

John
 
On Tue, 5 Oct 2004 13:03:42 -0400, "Jim Miller"
<jim@removethisjtmiller.com> wrote:

hi john

so the input capacitance wouldn't act as a divider to reduce the feedback
from the "feedback" cap and thereby require the opamp output to be greater
in order to achieve a null at the summing junction?
Click to expand...
The opamp forces the voltage across the pd to be zero, or at least to
be constant if the diode is biased. At any rate, photocurrent doesn't
change the potential across the pd, so the pd capacitance is not
charged, so all the photocurrent must flow into, and charge, the opamp
feedback cap. That's the cool thing about an opamp!

i should have added some additional information on the original post: the
photodiode has a gigohm impedence other than the capacitive element if this
has any importance.

if you're right then since the input current is in the picoamp range and the
slew rate of output is consequently in the volt per minute range i probably
shouldn't have any issues with stability.
Click to expand...
The slew rate doesn't factor into the stability analysis... the thing
could oscillate wildly irregardless of the charge rate. You do have a
230 pF load on the opamp output, among more obvious other things that
could make trouble. This loop does rate proper analysis.

John
 
thanks!

jtm

"John Larkin" <jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote in
message news:gbn5m0tqqcm09cn4ilib1be5h2tq1ni7dm@4ax.com...
On Tue, 5 Oct 2004 13:03:42 -0400, "Jim Miller"
<jim@removethisjtmiller.com> wrote:



You do have a 230 pF load on the opamp output, among more obvious other
things that
could make trouble. This loop does rate proper analysis.

John
 
spiced it.

simulation says you were right!

tnx
jtm


"John Larkin" <jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote in
message news:gbn5m0tqqcm09cn4ilib1be5h2tq1ni7dm@4ax.com...
On Tue, 5 Oct 2004 13:03:42 -0400, "Jim Miller"
<jim@removethisjtmiller.com> wrote:


The slew rate doesn't factor into the stability analysis... the thing
could oscillate wildly irregardless of the charge rate. You do have a
230 pF load on the opamp output, among more obvious other things that
could make trouble. This loop does rate proper analysis.

John
 
Jim Miller wrote:
spiced it.

simulation says you were right!
If your light source has no sharp spikes in output you might improve
Click to expand...
stability by putting a resistor in series with the detector, before
the summing junction. This will hurt the high frequency response but
stabilize the opamp. Something on the order of 1k might be enough.

--
John Popelish
 
On Thu, 07 Oct 2004 11:51:51 -0400, John Popelish <jpopelish@rica.net>
wrote:

Jim Miller wrote:

spiced it.

simulation says you were right!
If your light source has no sharp spikes in output you might improve
stability by putting a resistor in series with the detector, before
the summing junction. This will hurt the high frequency response but
stabilize the opamp. Something on the order of 1k might be enough.
Click to expand...
That brings up an interesting point: PDs have some series resistance,
and that may be a major contributor to loop stability.

John
 
Jim Miller wrote:
hi john

so the input capacitance wouldn't act as a divider to reduce the feedback
from the "feedback" cap and thereby require the opamp output to be greater
in order to achieve a null at the summing junction?

i should have added some additional information on the original post: the
photodiode has a gigohm impedence other than the capacitive element if this
has any importance.

if you're right then since the input current is in the picoamp range and the
slew rate of output is consequently in the volt per minute range i probably
shouldn't have any issues with stability.

thanks
jtm
-- snip --
Click to expand...
Because the photodiode's capacitance (and resistance, for that matter)
are in parallel with the apparent current source, and because the opamp
is holding the voltage across the diode constant, the PD capacitance
will affect stability but not DC precision.

These effects do cause issues: the parallel resistance _could_ cause you
problems if you get more dark current from the PD than signal current,
and the parallel capacitance could cause stability problems if you try
to lower the feedback capacitance for better response speed or gain.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
John Larkin wrote:
On Thu, 07 Oct 2004 11:51:51 -0400, John Popelish <jpopelish@rica.net
wrote:

Jim Miller wrote:

spiced it.

simulation says you were right!
If your light source has no sharp spikes in output you might improve
stability by putting a resistor in series with the detector, before
the summing junction. This will hurt the high frequency response but
stabilize the opamp. Something on the order of 1k might be enough.

That brings up an interesting point: PDs have some series resistance,
and that may be a major contributor to loop stability.
Click to expand...
Perhaps. I modeled the ones I am using on a project at the moment
http://www.osram.convergy.de/upload/documents/bulkload/infrared/bpx61.pdf
and the model that matched tested forward drop data indicated about 1
ohm series resistance at DC. I don't know how that would vary at the
instability frequency.

--
John Popelish
 
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