Electrical Engineering - Basic Op-Amp question

Author

Basic Op-Amp question

Peter

2007-02-11, 1:25 pm

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
numbers:

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?

MassiveProng

2007-02-11, 8:25 pm

On Sun, 11 Feb 2007 12:55:56 -0600, Peter <private@private.com> Gave
us:

>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
>numbers:
>
>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?

http://www.ecircuitcenter.com/Circu...l1/opmodel1.htm

daestrom

2007-02-11, 8:25 pm

"Peter" <private@private.com> wrote in message
news:TtSdndtXw4sh-1LYnZ2dnUVZ_vamnZ2d@comcast.com...
> 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
> numbers:
>
> 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?

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...

daestrom

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