basic op-amp contradiction

[Joe.T]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

Thanks !

 

[Jim Thompson]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

Thanks !

Well duh! It's unstable... positive feedback.

...Jim Thompson

 

[Genome]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

Thanks !

By not asking stupid twat questions..... Would you care to explain why you
have not asked a stupid twat question? You might start off by giving the
sums you have used to reach your contradiction.

On the other hand you might just float away.

DNA

 

[Joe.T]

I understand,

Thanks alot!

Especially thank you, Genome, for the helpful response.

 

[Ban]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

It will settle right on the rail, since the circuit is useless. Look how the
current flows from the output back into the output, it is independent of the
output voltage, thus there will be always a voltage between +in and -in,
overdriving the opamp. but you can at least know what direction the PD was
connected.

 

[Tom Bruhns]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

Thanks !

Amplifying a little on what Ban wrote...

Notice that nowhere in your circuit do you have a ground
connection--that is, a point of reference for measuring the op amp
output voltage. Since the inputs of your ideal op amp are infinite
resistance, then the current source just pumps current around the loop
R, R1 and R2. The op amp output voltage, no matter what it is (in an
ideal case), can't deliver current to R1 or R2, since there's no path
from it to ground through either of them. What is the voltage drop
across R1+R2? What is the voltage difference between the op amp + and
- inputs? What's the op amp open loop gain?

Cheers,
Tom

 

[Eeyore]

Hi,
Basic question I guess:
consider the following circuit -
A current source connected in serial with a resistor R - both between
the "in -" and "in+" of an opamp.
the "in - " is connected to the opamp "out" via a resistor R1.
the "in + " is connected to the opamp "out" via a resistor R2.

what is the output voltage?

i seem to be reaching a contradiction using the ideal opamp model.
How is this settled?

By not trying to model an nonsense circuit mainly.

Graham