# Opamp summing amplifier

1. Mar 3, 2016

### gazp1988

1. The problem statement, all variables and given/known data

i know that it is a summing amplifier and i have vout at 2v, however in the question it asks for I1,I2,I3 and V2(vin)
i have I1 and I3 but having trouble working out I2 and V2
2. Relevant equations
Vo = - RF ( V1 / R1 + V2 / R2 + V3 / R3)

3. The attempt at a solution
when i use this equation and input v2 as 1v i dont get anything close to 2v (vout)

2. Mar 3, 2016

### Merlin3189

Putting aside equations, what do you know about op-amps? In particular, the two inputs?

3. Mar 3, 2016

### gazp1988

the two inputs are the sum of vout, as vout =2v i assume v1=1v and v2=1v but when i input these values into my calculator i get -15

4. Mar 3, 2016

### Merlin3189

If I understand you correctly, I disagree.
Take any op-amp circuit, not your particular circuit and by "input" I'm talking about the IN+ and IN- terminals of the op-amp itself. You know something about the currents and voltages at these inputs.

Edit: Whoops! I don't mean "any op-amp circuit". Sorry, I'm referring to negative feedback amp circuit.

5. Mar 3, 2016

### Merlin3189

If you forget your inputs for the moment, can you calculate Vin- , Vin+ and I3 here?

6. Mar 3, 2016

### gazp1988

they are 0 at the negative input because of the virtual earth.

7. Mar 3, 2016

I3=0.0002A
vin- =0
vin+=0?

8. Mar 3, 2016

### Merlin3189

That's what I think. No current flows into or out of the op-amp inputs. The high gain and negative fb ensure that there is a very small difference between their potentials. Since Vin+ is grounded, then Vin- is a "virtual ground".
So you calculated I3, you can calculate I1 the same way, then you know I2 and can calculate V2.

The net result is the same as given by your formula, if applied correctly.

9. Mar 3, 2016

### gazp1988

i calculated I1 as 0.001A but having problems calculating I2 as i only know R2

10. Mar 3, 2016

### Merlin3189

Right. So I3= 0.0002 A and I1= 0.001 A and no current flows into the op amp, so it must go through the 2k resistor.
Or, use Kirchoff's current law at the junction to find I2. Either way be careful with the sign(!)

11. Mar 3, 2016

### gazp1988

so my calculations are giving me 2v for V2??

12. Mar 3, 2016

### Merlin3189

Do you want to show this in detail please?

13. Mar 3, 2016

### gazp1988

if no current passes through the node then it will pass back through R2, i think, so i worked out that 0.001Ax2000ohm=2v

14. Mar 3, 2016

### Merlin3189

$I_1 + I_2 + I_3 = 0$ because they all flow into the junction and have nowhere to go.
Otherwise you could say $I_3$ and $I_1$ flow into the junction , so must go out through R2 and $I_2 = -(I_3 + I_1)$
It all boils down to the same thing. So $I_2 = -(0.0002 + 0.001)$ or $I_1 + 0.0002 + 0.001 = 0$

15. Mar 3, 2016

### gazp1988

sorry to go over the same question because im confusing myself. how do i get V2 then if the current is 0??

16. Mar 3, 2016

### Merlin3189

Well, if the voltage were 0V then the current would be 0 A and that would be ok.
But it isn't, because I2 is not 0.
$I_1+I_2+I_3\ =\ 0.001 +I_2 + 0.0002 =0$
So $I_2 = - 0.001 - 0.0002 = - 0.0012$

17. Mar 3, 2016

### gazp1988

so to work out the voltage we use ohms law that gives me -12v which doesnt seem right

18. Mar 3, 2016

### Merlin3189

Yes.
I must say, I wish you wouldn't keep just giving answers without your working.
If $I_2$ is -0.0012 A, then Ohm's law gives $V_2 = I_2 \times R_2 = -0.0012 \times 2k \ \$ which is not -12V
BUT there is no reason why it should not seem right. The summing op amp can add both positive and negative signals.

19. Mar 3, 2016

### gazp1988

-0.0012 x 2000 = -2.4= v=-2.4v
Vo = - RF ( V1 / R1 + V2 / R2 + V3 / R3)
vo= -10000(1/1000+-2.4/2000)= 2

2v =vo so it proves that the formula is correct i hope

20. Mar 3, 2016

### Merlin3189

Yes, that's right.
Your original formula is ok, but refers to a 3 input summing circuit. And RF is your R3.

$V_o = -R_F(\frac{V_1}{R_1} + \frac{V_2}{R_2})\ \$ is the correct formula to use here
So $2 = -10k(\frac{1}{1k} + \frac{V_2}{2k}) \ \ =\ -10(1 + \frac{V_2}{2}) \ =\ -10 -5V_2$
So $2 = -10 -5V_2$ and $12=-5V_2$ and $V_2 = -\frac{12}{5}\ \ = -2.4$

Your original error was in assuming that $V_2$ was 2V. There was no reason for that.
I'm not familiar with the formula, so I just looked at the circuit and saw what the currents were. Only when I worked it through for you, did I realise that the formula was applicable if you interpreted it correctly.
If you understand how this formula is derived and used, then it is fine to use it. But, if you have understood the steps we worked through, perhaps you'll have a better understanding of the circuit than you would get from just using the formula.

Edit: And BTW, I think you should try harder to set down your detailed working in your posts. It will help us to help you better.