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Jiaqi
Guest
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 Posted:
Sat Feb 05, 2005 10:57 am Post subject:
A basic question about current-to-voltage converters |
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I have a basic question seeming quite simple:
An op amp with an open loop gain A, and a feedback resistor Rf, are
connected as a current-to-voltage converter which can be found in many
common applications. If the input impedance of the op amp is very large
and can be regarded ideally "infinite" (as the case of FET-input op amp),
the input resistance of the current-to-voltage converter can be estimated
as Rf/A (Why?). With this presumption, can I estimate the voltage at the
virtual ground pin as I*Rf/A (I denotes the input signal current)? How can
I estimate the maximum feedback resistance Rf which the op amp can load?
Does that value depend on the open loop gain A? If yes, please tell me the
way or where I can find a definite answer.
Thanks.
Jiaqi
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Jonathan Kirwan
Guest
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| Posted:
Sat Feb 05, 2005 1:55 pm Post subject:
Re: A basic question about current-to-voltage converters |
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On Fri, 04 Feb 2005 23:57:59 -0500, "Jiaqi" <sjq1@citiz.net> wrote:
| Quote: | An op amp with an open loop gain A, and a feedback resistor Rf, are
connected as a current-to-voltage converter which can be found in many
common applications. If the input impedance of the op amp is very large
and can be regarded ideally "infinite" (as the case of FET-input op amp),
the input resistance of the current-to-voltage converter can be estimated
as Rf/A (Why?). With this presumption, can I estimate the voltage at the
virtual ground pin as I*Rf/A (I denotes the input signal current)? How can
I estimate the maximum feedback resistance Rf which the op amp can load?
Does that value depend on the open loop gain A? If yes, please tell me the
way or where I can find a definite answer.
|
With the non-inverting input at zero volts and your inverting input at some
value, V, then your output voltage is V*A, right? This produces a current
through Rf of (V*A - V) / Rf, the voltage difference between the output and the
inverting input divided by the feedback resistor. So:
I = (V*A-V)/Rf = V*(A-1)/Rf, so V = I*Rf/(A-1)
So I think the voltage at the virtual ground node (the inverting input) is
I*Rf/(A-1), not I*Rf/A. This makes the input impedance Rf/(A-1), I think.
For DC considerations, your Rf depends on your maximum allowed output voltage
and your maximum input current. If you have a maximum output of 10V and your
maximum current will be 1mA, then your Rf is limited to 10V/1mA*[(A-1)/A] or
about 10k Ohm.
Jon |
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Jiaqi
Guest
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| Posted:
Sat Feb 05, 2005 7:37 pm Post subject:
Re: A basic question about current-to-voltage converters |
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On Fri, 04 Feb 2005 07:55:24, "Jonathan Kirwan" wrote:
| Quote: | With the non-inverting input at zero volts and >your inverting input at
some value, V, then your output voltage is V*A, >right? This produces a
current
through Rf of (V*A - V) / Rf, the voltage >difference between the output
and the
inverting input divided by the feedback resistor. So:
I = (V*A-V)/Rf = V*(A-1)/Rf, so V = I*Rf/(A-1)
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I think the output voltage should be -V*A rather than V*A, and the input
current I should be expressed as (V-(-V*A))/Rf=V*(A+1)/Rf. So the input
voltage V is I*Rf/(A+1) rather than I*Rf/(A-1). The input impedance is
Ri=Rf/(A+1) rather than Rf/(A-1). Am I right, Jon?
Jiaqi
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Ban
Guest
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| Posted:
Sat Feb 05, 2005 8:21 pm Post subject:
Re: A basic question about current-to-voltage converters |
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Jiaqi wrote:
| Quote: | I have a basic question seeming quite simple:
An op amp with an open loop gain A, and a feedback resistor Rf, are
connected as a current-to-voltage converter which can be found in many
common applications. If the input impedance of the op amp is very
large and can be regarded ideally "infinite" (as the case of
FET-input op amp), the input resistance of the current-to-voltage
converter can be estimated as Rf/A (Why?).
|
The input impedance (Rf/A) is almost zero, because the opamp tries to keep
the inverting input at gnd potential by means of the output voltage through
the feedback resistor. With a smaller resistor the compensation current will
need less voltage swing on the O/P.
With this presumption, can
| Quote: | I estimate the voltage at the virtual ground pin as I*Rf/A (I denotes
the input signal current)? How can I estimate the maximum feedback
resistance Rf which the op amp can load? Does that value depend on
the open loop gain A? If yes, please tell me the way or where I can
find a definite answer.
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Your are right with the voltage at the inverting input, additionally is
superimposed the offset voltage.
Rf is (almost) independent of the open loop gain. It is only dependent on
the max output voltage swing. If your input is +1uA, with 1Meg Rf the O/P
voltage will be -1V, so with a voltage swing of +/-13V the max. Rf to
prevent saturation would be 13Meg.
| Quote: | Thanks.
Jiaqi
welcome |
--
ciao Ban
Bordighera, Italy |
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Jonathan Kirwan
Guest
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| Posted:
Sat Feb 05, 2005 9:39 pm Post subject:
Re: A basic question about current-to-voltage converters |
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On Sat, 05 Feb 2005 08:37:44 -0500, "Jiaqi" <sjq1@citiz.net> wrote:
| Quote: | I think the output voltage should be -V*A rather than V*A, and the input
current I should be expressed as (V-(-V*A))/Rf=V*(A+1)/Rf. So the input
voltage V is I*Rf/(A+1) rather than I*Rf/(A-1). The input impedance is
Ri=Rf/(A+1) rather than Rf/(A-1). Am I right, Jon?
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I'm a hobbyist and get my signs wrong! Yes, I like that better.
The output would indeed be -V*A.
Jon |
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