Is this possible? - Page 1
Forum section: electronics / projects

author: Dukot
author_rank: Newbie
timestamp: December 04, 2021, 08:25:21 am
content: Hi i would like to ask if this is circuit possible to solve? Thanks.

author: Someone
author_rank: Super Contributor
timestamp: December 04, 2021, 08:28:33 am
content: Quote from: Dukot on December 04, 2021, 08:25:21 amYes, it is possible.

author: Dukot
author_rank: Newbie
timestamp: December 04, 2021, 08:30:07 am
content: how though? im so confused

author: Someone
author_rank: Super Contributor
timestamp: December 04, 2021, 10:11:09 am
content: Understanding Basic Analog – Ideal Op AmpsSLAA068Bhttps://www.ti.com/lit/an/slaa068b/slaa068b.pdfCalculate 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.

author: tooki
author_rank: Super Contributor
timestamp: December 04, 2021, 06:48:36 pm
content: 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.

author: rstofer
author_rank: Super Contributor
timestamp: December 04, 2021, 07:13:53 pm
content: I get the equivalent voltage at the positive input = 0.333...Volts using nodal analysisThe 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)/RThe center resistor (V - (-1)) / RThe right resistor V/RThen 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 = 03V - 1 = 03V = 1V = 1/3 voltsFortunately, LTspice agrees with me. See attachedNegative 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 voltageV(vout): 0.333333 voltage <====V(n002): -1 voltageI(R3): 0.333333 device_currentI(R2): 1.33333 device_currentI(R1): -1.66667 device_currentI(V2): -1.33333 device_currentI(V1): -1.66667 device_currentNow 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.

author: fourfathom
author_rank: Super Contributor
timestamp: December 04, 2021, 07:40:09 pm
content: Quote from: rstofer on December 04, 2021, 07:13:53 pmReally? I get 1V at the non-inverting input.[edit - wrong. I missed the polarity]

author: IanB
author_rank: Super Contributor
timestamp: December 04, 2021, 07:47:38 pm
content: Quote from: fourfathom on December 04, 2021, 07:40:09 pmI just checked and got one third of a volt.Did you notice the polarity of the voltage sources?

author: fourfathom
author_rank: Super Contributor
timestamp: December 04, 2021, 07:50:06 pm
content: Quote from: IanB on December 04, 2021, 07:47:38 pmOops!

author: rstofer
author_rank: Super Contributor
timestamp: December 04, 2021, 08:21:49 pm
content: 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 voltageV(vnode): 0.333333 voltage <====V(n002): -1 voltageV(v-): 0.333332 voltageV(vout): 0.999998 voltage <====V(+15): 15 voltageV(-15): -15 voltageI(R5): 0.000333333 device_currentI(R4): 0.000333332 device_currentI(R3): 0.000333333 device_currentI(R2): 0.00133333 device_currentI(R1): -0.00166667 device_currentI(V4): -0.00133754 device_currentI(V3): -0.00167086 device_currentI(V2): -0.00133333 device_currentI(V1): -0.00166667 device_currentIx(u1:1): 6.66666e-010 subckt_currentIx(u1:2): 6.66664e-010 subckt_currentIx(u1:3): 0.00167086 subckt_currentIx(u1:4): -0.00133754 subckt_currentIx(u1:5): -0.000333333 subckt_currentI 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.

author: gcewing
author_rank: Regular Contributor
timestamp: December 05, 2021, 05:57:35 am
content: Quote from: rstofer on December 04, 2021, 08:21:49 pm10k ohms seems realistic enough to me. Did you read it as 10 ohms?

author: fourfathom
author_rank: Super Contributor
timestamp: December 05, 2021, 03:51:14 pm
content: 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).

author: 741
author_rank: Frequent Contributor
timestamp: December 05, 2021, 04:35:39 pm
content: 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) = 2V+ = I.R = 2*0.5 = 1VNow, The top part is simply a non-inverting amplifier, gain = 1 + (2/1) = 3Finally, Vo = gain * V+ = 3*1 = 3V

author: fourfathom
author_rank: Super Contributor
timestamp: December 05, 2021, 04:46:26 pm
content: Quote from: 741 on December 05, 2021, 04:35:39 pmI see that you also missed the polarity in that "1V" source. I did too.

author: rstofer
author_rank: Super Contributor
timestamp: December 05, 2021, 05:33:05 pm
content: Quote from: gcewing on December 05, 2021, 05:57:35 amWhen 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.

author: strawberry
author_rank: Super Contributor
timestamp: December 06, 2021, 01:57:48 am
content: Vo=0VOP not connected power pins