Basic Op-Amp question

Assuming I have an 'ideal' op-amp.
If I have an Rf and Rin with a voltage into the inverting input and the
non-inverting is grounded: my basic formulas is (-Rf/Rin) * Vin
What if I put a voltage directly into the non-inverting input with the
above conditions?
I understand everything except getting my signs correct. To use real
Rf 10k
Rin 5k
Vin1 2 volts
Vin2 1 volts (this is the voltage on the non-inverting input).
My calculations will be:
The current across Rin will be the difference between 2 volts and 1 volt
because the op-amp will make the inverting equal to the non-inverting input
(in this case 1 volt).
This difference in voltage is divided by Rin to get the current
The current is then multiplied by Rf to get the voltage drop across Rf.
Now my problem is: I have all the correct voltages and currents, however,
my signs get confusing and I'm not sure what to get for the op-amp's outptu
voltage especially when I start changing the input voltages (i.e. making
one higher than the other, making one negative, and one positive, vice
versa, etc..)
Is there is an easier way to do this, a formula, or am I not thinking about
it correctly?
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On Sun, 11 Feb 2007 12:55:56 -0600, Peter Gave us:
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Keep in mind that the typical 'inverting' connection (Rf to the inverting input along with 'signal' to the inverting input via Rin) will try to keep the *difference* between non-inverting and inverting input as close to zero as possible (within the limits of 'real' op-amps).
So when the non-inverting input is tied to ground, the inverting input is kept at a 'virtual' ground. So the solution is, "What output, applied to one end of the voltage divider Rf + Rin, while the input signal is connected to the other end of the voltage divider, will result in the mid-point being at ground?" That's what the whole formula is based on.
Now, you have raised the non-inverting input to +1V. So, "What output, applied.....will result in the mid-point being at +1V??"
It works out to (-Rf/Rin)* (Vin-1V) + 1V.
As you surmise, the current Iin is (Vin - 1V) / Rin. For an ideal op-amp, the current through Rf must be equal and opposite sign (i.e. Kirchoff's current law says that (Iin + If) = 0 ) So, since If = (Vout -1V)/Rf and If = -In, we have.... (Vout - 1V)/Rf = -(Vin-1V)/Rin.
Vout = (-Rf/Rin)*(Vin - 1V) + 1V
Your 'signage' problem can probably be traced back to the 'assumed' directions of current flow in the summing junction. If you 'assume' that all currents flow toward the junction, then Kirchoff's law holds as I stated above. But if you 'assumed' the If flows out of the junction, then Kirchoff's law would look like (Iin - If) = 0 and you get a sign reversal.
Hope this helps...
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