Opamp Summing Amplifier: Finding I1, I3 & V2 for Vout = 2V

[gazp1988]

Homework Statement

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

Homework Equations

Vo = - RF ( V1 / R1 + V2 / R2 + V3 / R3)

The Attempt at a Solution

when i use this equation and input v2 as 1v i don't get anything close to 2v (vout)

 

[Merlin3189]

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

 

[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

 

[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.

 

[Merlin3189]

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

 

[gazp1988]

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

 

[gazp1988]

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

 

[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.

 

[gazp1988]

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

 

[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(!)

 

[gazp1988]

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

 

[Merlin3189]

Do you want to show this in detail please?

 

[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

 

[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 $I_1 + 0.0002 + 0.001 = 0$

 

[gazp1988]

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

 

[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 $I_2 = - 0.001 - 0.0002 = - 0.0012$

 

[gazp1988]

so to work out the voltage we use ohms law that gives me -12v which doesn't seem right

 

[Merlin3189]

so to work out the voltage we use ohms law

Yes.

that gives me -12v which doesn't seem right

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.

 

[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
thank you for your time

 

[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 realize 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.

 

[gazp1988]

thank you for your time and patience. I'm pretty new to the forum thing so in the future ill put more information down and show my calculations.
the original formula was one i found in my paper work but looking back, it seems it wasnt the right one to use.

 

[Merlin3189]

the original formula was one i found in my paper work but looking back, it seems it wasn't the right one to use.

The formula is ok provided you understand how to use it
$V_O = - R_F ( \frac{V_1}{R_1} + \frac{V_2}{R_2} + ... )$
is the general formula for a summing op-amp circuit for 2 or more inputs. You can extend is for as many inputs as you like.
The $R_F$ is the negative feedback resistor and the others are the resistors for each input.
You just picked up the formula with 3 inputs and used it in a circuit where $R_F$ is labelled R3, which probably confused you.

The formula rearranges to
$\frac{V_O}{R_F} = - ( \frac{V_1}{R_1} + \frac{V_2}{R_2} + ... )$ which represents
feedback current = - (input 1 current + input 2 current + ...) as I was trying to get you to work out