# How to handle four resistors in a square (OP-amp)?

I want to analyze this circuit and find $$\\frac{V_{out}}{V_{in}}\$$. But how should I handle the four resistors in the square?

simulate this circuit – Schematic created using CircuitLab

Can I redraw the circuit as following?

simulate this circuit

Are now $$\R1\$$, $$\R3\$$ (and $$\R2\$$, $$\R4\$$) in series? I assume it is not the case because I end up with:

simulate this circuit

So my question is: How should I handle the four resistors?

• 1. to 2. schematic: that's not even a redrawing, that's just changing a few purely visual angles. Yes you can do that: 2. to 3.: This makes no sense at all. Whatever the intention was, you can't do this. This is against all rules of linear circuit analysis. Commented Jul 24, 2021 at 11:16
• 2nd schematic is same as 1st like @MarcusMüller mentioned. But that can give you a clearer picture Commented Jul 24, 2021 at 11:20
• What are you trying to find? The DC gain? Are you allowed to use software to solve systems of equations? Do you have to use pencil and paper?
– Carl
Commented Jul 24, 2021 at 11:22
• @Carl $\frac{V_{out}}{V_{in}}$ with pencil and paper. Commented Jul 24, 2021 at 11:27
• Useful search term : Wheatstone bridge.
– user16324
Commented Jul 24, 2021 at 12:47

Just look at this circuit. what I have done is that since both R3-R1 and R4-R2 are going to ground , I split it into two separate path flowing to ground. Now you can just re draw it and observe the working of this circuit. You can further reduce it by finding R1/R5.

EDIT: you can also calculate the Thevenin equivalent and find the voltage at +ve terminal of Opamp.

• You can also find the Thevenin equivalent for Vin, R2 and R4. (Thanks to @jonathan 's explanation in his answer) Commented Jul 24, 2021 at 11:46

Your first transformation is correct (it's exactly the same circuit, nothing changed), your second is not. In general, electronic components are only "in series" when there's nothing connected between them.

R5 and R1 both go from V2 to ground, so they're in parallel and you can simplify them to a single resistor (let's call it R51). Next, R3 and R51 form a voltage divider. You can replace it with its Thevenin equivalent.

R4 and R2 also form a voltage divider that you can replace with its Thevenin equivalent. This means that the Thevenin equivalent voltage source is Vin*R2/(R2+R4) and its impedance is R2||R4. It connects to the inverting input of the OpAmp. R6 stays where it is because you can't include it in the Thevenin calculation due to it being involved in the feedback path.

After these transformations, you're left with a voltage source connected directly to the non-inverting input of the OpAmp. The OpAmp itself is connected as an inverting amplifier.

• I was confused by 'R4 and R2 form a voltage divider'! There is some current flowing through R4 to R6 right ? Since there is a negative feedback , we cant assume it as a normal voltage divider and find its Thevenin right ?@Jonathan Commented Jul 24, 2021 at 11:34
• You can use Thevenin to simplify any subcircuit that consists of passive elements and voltage/current sources. In this case, the Thevenin voltage is Vin*R2/(R2+R4) and the impedance is R2||R4. R6 stays where it is. Commented Jul 24, 2021 at 11:36
• Yeah 'Passive circuits'.Will it be safe to use it here ? since opamp is present in circuit ? @Jonathan Commented Jul 24, 2021 at 11:38
• Yes, it is safe to use as long as you don't include any variable (dependent) voltage sources (like the OpAmp's output voltage). This means you can't fold R6 into the Thevenin equivalent circuit, it has to stay separate. Vin, R2 and R4 are all fixed (independent), though, so you can calculate an equivalent circuit for them. Commented Jul 24, 2021 at 11:40
• The current through R6 will change, but the Thevenin equivalent circuit of Vin, R2 and R4 behaves exactly like the original circuit even when this changing current is applied to it. You just can't include dependent voltage/current sources in the Thevenin calculation. Any subcircuit that only consists of independent elements (independent voltage sources, fixed resistors, etc) is fair game. Vin is independent and R2/R4 are fixed resistors, so this works out just fine. Commented Jul 24, 2021 at 11:43

Keeping it simple: R4 is irrelephant. Ignore/remove it. It just makes extra heat. All it does is change the input impedance which does not matter since the circuit is driven by a voltage source.

• This isn't right. You should look up "Wheatstone Bridge" Commented Jul 25, 2021 at 3:00
• If any of the elements was variable I might agree, but the questioner simply wants to calculate the output voltage of the circuit. For a Wheatstone bridge you would be interested in the volatge across the nodes. In this circuit, due to the op amp, there will be no voltage across the bridge. Commented Jul 25, 2021 at 3:11
• @AuldDuffer I suspect your answer might be taken more seriously if it involved less pachyderms. Commented Jul 25, 2021 at 11:36
• Yes, and the current through R6, and thus the output voltage, depends on R4 Commented Jul 25, 2021 at 15:47
• Apply KCL at the inverting input node, and it's clear. Commented Jul 25, 2021 at 15:48