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# A small problem question about opamps

#### circuit975

##### New Member

Vo = -R13 / R14 x Vx + (R16 / R15 + R16 x R14 + R13 / R14) x Vy
Vo = 4 + 4,5 = 8,5V

hello everyone

In this case, in order to find Vo, the equivalent of Vo comes in this way.
But I understood everything until the -R13/R14 x Vx part. But I couldn't understand how to do the inside of the next parenthesis, it sounded like a voltage divider, but I think it won't work like that. How can we explain the second part of the summation?

There are all kinds of shortcuts to get at this.

But easy analytical path is to use superposition. Evaluate the contribution
individually of each input and summing results.

Regards, Dana.

If Vy=6v and no current flows into the non-inverting input of the op amp, you should have 3v at that non-inverting input.

An op amp adjust its output to try to keep its two inputs at the same voltage (for properly designed circuits). From that, let's assume the circuit is reasonably designed (and it is). So, the inverting input should also be at 3v.

To get the inverting input to 3v, the output has to be at...
If you calculate the current through R14 (2k ohms), 11v/2000= 5.5mA
If R14 passes 5.5mA, R13 also passes 5.5mA since no current flows in/out of the inverting input. Therefore, the voltage drop across R13 is 5.5mA x 1000 ohms = 5.5v. Then the input is 3v + 5.5v = 8.5v.

ZZO's technique is probably the easiest way to calculate the output of that op amp circuit.

The gain at the Non Inverting input is (1 + Gain at the Inverting input).
Gain is (R13/R14) = 0.5

Vo = [ -(R13/R14) * Vx + (1+R13/R14) * Vy *R16/(R15 + R16)]
= -.5 * (-8.0) + 1.5 * 6 * 0.5 = 4 + 4.5 = 8.5

I agree.

Intuitively Vin+ must be 50% or 3V and thus with linear feedback Vin- must also be 3V.

Now it should be intuitive that if R14 has -8V - 3V = 11V across 2k then the same current across R13=1k must yield a proportionally lower voltage or half so 11V/2.

Assumptions
1. A virtual null 0V between Vin+,Vin- when not saturated on the output.
2. I(R14)=I(R13)
3. I(R15)=I(R16)
4. Input bias current is far less so ignored.
Result
The output must be 5.5V +3V.
Verify I(R14)=11V/2k Vo= I(R14) *R13= 11V/2k * 1k = 5.5V added to Vin-=3V.

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