Is this possible?

[Dukot]

Hi i would like to ask if this is circuit possible to solve? Thanks.

[Someone]

Hi i would like to ask if this is circuit possible to solve?

Yes, it is possible.

[Dukot]

how though? im so confused

[Someone]

Understanding Basic Analog – Ideal Op Amps
SLAA068B
https://www.ti.com/lit/an/slaa068b/slaa068b.pdf

Calculate voltage at + input.
Calculate output voltage to make - input equal the same.

Superposition may not be the best way to describe the method, as there are not multiple variable inputs.

[tooki]

There are multiple sources here (in this case, two voltage sources), so superposition is one way to solve it. Evaluate one at a time with all other sources set to 0, then add up.

[rstofer]

I get the equivalent voltage at the positive input = 0.333...Volts using nodal analysis

The easy way is to take the voltage on that trace headed toward the + input as V and then write the Kirchhoff nodal equations assuming all currents are leaving the node.

The left resistor current flowing out of the node = (V-2)/R
The center resistor (V - (-1)) / R
The right resistor V/R

Then sum them up and since they total 0 and R is in each denominator, it can be dropped and the circuit works for all values of R as long as they are identical

(V-2) + (V-(-1)) + V = 0
3V - 1 = 0
3V = 1
V = 1/3 volts

Fortunately, LTspice agrees with me.  See attached

Negative currents simply means that the assumed direction (leaving the V node) is incorrect and current is flowing into the node through that branch.


       --- Operating Point ---

V(n001):    2    voltage
V(vout):    0.333333    voltage   <====
V(n002):    -1    voltage
I(R3):    0.333333    device_current
I(R2):    1.33333    device_current
I(R1):    -1.66667    device_current
I(V2):    -1.33333    device_current
I(V1):    -1.66667    device_current



Now that you have the voltage at the positive input, you can work out the voltage at the negative input (it must be the same) and from that and the gain equation, you can get the op amp output voltage.

Since this is a non-inverting amplifier, the output voltage should be 3 times the input or 1.0V.  When answers are this neat, they are usually homework type problems and likely to be correct if the numbers are clean.

[fourfathom]

I get the equivalent voltage at the positive input = 0.333...Volts using nodal analysis

Really?  I get 1V at the non-inverting input.

[edit - wrong.  I missed the polarity]

[IanB]

I get the equivalent voltage at the positive input = 0.333...Volts using nodal analysis

Really?  I get 1V at the non-inverting input.

I just checked and got one third of a volt.

Did you notice the polarity of the voltage sources?

[fourfathom]

Did you notice the polarity of the voltage sources?

Oops!

[rstofer]

OK, I'll confess!  I missed the polarity when I first started writing the equations as well.  I knew it was an issue and forgot it anyway.  Old age...  In fact, I wanted to see if LTspice agreed before I dared to post a solution.

I added an op amp to the project and the output is 1.0V.  OK, the gain is 3 and it's non-inverting and the input is 1/3V so that seems fine.


       --- Operating Point ---

V(n001):    2    voltage
V(vnode):    0.333333    voltage   <====
V(n002):    -1    voltage
V(v-):    0.333332    voltage
V(vout):    0.999998    voltage   <====
V(+15):    15    voltage
V(-15):    -15    voltage
I(R5):    0.000333333    device_current
I(R4):    0.000333332    device_current
I(R3):    0.000333333    device_current
I(R2):    0.00133333    device_current
I(R1):    -0.00166667    device_current
I(V4):    -0.00133754    device_current
I(V3):    -0.00167086    device_current
I(V2):    -0.00133333    device_current
I(V1):    -0.00166667    device_current
Ix(u1:1):    6.66666e-010    subckt_current
Ix(u1:2):    6.66664e-010    subckt_current
Ix(u1:3):    0.00167086    subckt_current
Ix(u1:4):    -0.00133754    subckt_current
Ix(u1:5):    -0.000333333    subckt_current


I had to get a little more realistic about the resistors.  The generic op amp can't drive a heavy load!  But as long as the ratios hold the circuit output is independent of the resistor values.

[gcewing]

I had to get a little more realistic about the resistors.

10k ohms seems realistic enough to me. Did you read it as 10 ohms?

[fourfathom]

I did it in my head using 1 Ohm (and 2 Ohm) and an ideal opamp.  As long as you can add, divide by three, and multiply by two, it all works out (if you mind the polarities).

[741]

By superposition, mentally short out one of the signal voltages (thus keeping only the series resistance).

For each signal, the 'bottom' resistor is effectively 0.5R.

Therefore, we can sum the two separate currents, which combine in the "0.5R" effective bottom resistor.

I1 = 1/(1 + 0.5),  I2 = 2/(1 + 0.5), I_SUM = (1+2)/(1.5) = 2

V+ = I.R = 2*0.5 = 1V

Now, The top part is simply a non-inverting amplifier, gain = 1 + (2/1) = 3


Finally, Vo = gain * V+ = 3*1 = 3V

[fourfathom]

Finally, Vo = gain * V+ = 3*1 = 3V

I see that you also missed the polarity in that "1V" source.  I did too.

[rstofer]

I had to get a little more realistic about the resistors.

10k ohms seems realistic enough to me. Did you read it as 10 ohms?

When I solved just the input voltage, I used 1 Ohm.  That's fine because the divider is based on ratios, not values.  1 Ohm, 1k Ohms, same result as long as they're all the same.

When I added the op amp, LTspice didn't think much of 2 Ohms and 1 Ohm (or at least it seemed that way at the time) so I multiplied everything up to 1k and 2k.

The values aren't important other than as a load on the op amp.  It's the ratios that matter.

[strawberry]

Vo=0V
OP not connected power pins