# Buffered power supply with op amps?

I dont quite understand how this circuit works. I tried to simulate it on multisim 14 and got even more confused! Can anyone help me analize this circuit?

I only understand the second stage (the inverting output) but i have no clue about the left part

• what do you understand about the circuit ... what is confusing you? Apr 3 at 2:29
• I mean I only understand the second stage (the inverting output) but i have no clue about the left part Apr 3 at 3:01
• please add that information to your question ... this site is not a forum ... all info must be inside the question Apr 3 at 3:04
• Post the source of the schematic. If this is your own design, what do you expect it to do? Apr 3 at 3:48
• The original schematic is in the b/w picture. It was handed to us by our professor as an assignment. I dont ask for the solution, I'm asking for help with the first part, though. Apr 3 at 3:54

# Building the circuit

Here is a possible scenario for solving the professor's "riddle". To be clear, I am just analyzing what the op-amps do, but for now I am not commenting on why they do it, what is the point of it. My idea is to build the circuit step by step and revealing the idea at each step.

## Introducing negative feedback

If we simply connect the op-amp output to its inverting input (keeping the non-inverting input grounded), the op-amp will adjust its output voltage (almost) equal to zero (the potential of the non-inverting input). So this circuit will be a zero-voltage follower.

simulate this circuit – Schematic created using CircuitLab

Now imagine that for some reason (unknown to us) we have to introduce positive feedback as well. The first thing that comes to mind is to connect the op-amp output to the non-inverting input as we did above with the inverting input. But for the circuit to be stable (not to become a latch) the negative feedback must dominate over the positive. So we connect a voltage divider R1-R3 in the positive feedback to reduce it. The output voltage is still zero; the op-amp "sees" no reason to change it since the potential difference between the two inputs is (almost) zero.

simulate this circuit

Such a mixed feedback is used, for example, in negative impedance converters (NIC).

## Applying input voltage

Now we need to connect the input voltage source. Imagine again that, for some unknown reason, the source is "floating" (ungrounded), which gives us freedom of action. But what will this so-called "input source" be? Figuratively speaking, it will be a disturbance to the op-amp with negative feedback that will cause it to change its output voltage so as to compensate for this "disturbance". Well, let's then insert a 5 V "input" voltage source V1 in the positive feedback loop (with the same success we could insert it in the negative feedback loop).

simulate this circuit

The op-amp will react to this "intervention" by increasing its output voltage hoping to restore the zero voltage difference between its inputs. Note that, for this purpose, it compares the input voltage at the non-inverting input and the output voltage both referenced to ground. Unfortunately, the negative terminal of the input source is not rigidly fixed and also starts to "move" up, albeit at a slower rate. As a result, the op-amp stops (reaches the equilibrium) later, when VA = 10 V.

## Current source

simulate this circuit

## Circuit doubling

And again, for some reason, imagine that we have to double this circuit like in the famous "long-tailed pair". As a result, the common current through R3 and the output voltages increase.

simulate this circuit

## Virtual grounding

Nothing will change if we connect R3 to virtual ground instead of real ground, but it remains unclear why this is done...

simulate this circuit

# A guess

## Left part

One could assume that the left part - V1, R1 and OA1 (V2, R2 and OA2), is a constant current source because OA1 (OA2) keeps up a 5 V constant voltage across the R1 (R2) constant resistor. This current IL = V1/R1 is passed through the R3 load and then flows through R4.

simulate this circuit

## Right part

It can then be assumed that the task of the right part is to control the current through/voltage across the R3 load. OA3 and R4 can be considered as a current-to-voltage converter (transimpedance amplifier) which controls the LED indicator R5, LED1 and LED2 (the two LEDs can be controlled by only one resistor).

As per homework policy, I will give only some hints.

Your simulation has all the op-amps railed so it's not really "working" whatever that means in this case.

Try simulating with much lower Ve1+Ve2.

The 2.5kΩ resistor in your schematic has no obvious purpose, however the problem set schematic marks it "$$\R_L\$$" (suggesting it is a load resistor) so perhaps the current through that resistor is part of the purpose of this circuit. Try to calculate that current as a function of Ve1,Ve2, and R.

Remember that each op-amp's inverting and non-inverting inputs should be at the same potential if the op-amps are functioning as typically intended. So you can write down several voltages/voltage differences immediately and check against your simulation.

The top opamp’s non-inverting input is Ve1 higher than Vx, and this is presented at the opamp buffer output. So the R connected to it just sees Ve1 across it (Ve1+Vx on one side and Vx on the other).

Likewise, the bottom opamp presents Ve1 across its R.

Those currents sum (Ve1/R + Ve2/R) and it all flows into RL to the second opamp – no current flows through the batteries because it would have to flow into an opamp input.

I don’t see any reason for the addition of the left opamps: seems the circuit would work the same if the batteries connected directly to the Rs. I suspect they’re there just to befuddle you. It looks like it certainly befuddled Multisim. Or maybe I'm befuddled.