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For the following schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

Easy circuit - amplifies signal difference 10 times and offsets it 2.5 V.
The gain is said to be:

$$G = 1 + \frac{R_3}{R_2||R_1} = 1 + \frac{4530}{500} = 10.06$$

This is correct, I tested it in LTspice, but I don't understand one thing:

Why in the gain calculations we use \$R_2\$ and \$R_1\$ in parallel? I don't see how we end up thinking they are parallel, they have only one common point A, therefore they are serial to each other, why do we apply parallel calculations here?

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  • \$\begingroup\$ Both ends of a DC supply are considered AC ground. \$\endgroup\$ Aug 29, 2016 at 19:07

4 Answers 4

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Think of that voltage divider in terms of its equivalent Thevenin circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Inside the box is the Thevenin equivalent circuit of your voltage divider. Where

$$ V_{\mathrm{TH}}=\frac{V_{\mathrm{cc}}R_1}{R_1+R_2}=2.5\mathrm{V}$$ and \$R_{\mathrm{TH}}=R_1||R_2\$

So, to the outside world that voltage divider looks like a voltage source with a series resistance, all you need to find is the value of the source and the value of that resistance. From there should be easy to see why you have those two resistors in parallel in your gain equation.

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  • \$\begingroup\$ oh, yeah, Thevenin comes in handy this time, thank you! :) \$\endgroup\$ Aug 29, 2016 at 19:31
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Node A, by virtue of negative feedback, is at the voltage Vin, not, as you think, the voltage dictated by voltage division on the left. The current through R1 is thus Vin/R1, and the current through R2 is (5V-Vin)/R2. Now you can use Kirchoff's Current Law to figure out the current through Rf, and once you know that you can calculate Vout.

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Science Samovar - I think you are not correct in assuming a DC offset of 2.5 volts. Using superposition, Vin=0, the voltage V1=5V will be amplified by the factor A=-4.53. Hence, the DC output will be (theoretically) larger than 20 volts. The resistor R1 appears across the diff. input and does not influence the dc gain.

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An intuitive way to understand why R1 and R2 are in parallel is by altering the voltage at node A.

Let's say the voltage is increased by 1V. Now the current through R1 will increase by 1mA. This current flows from node A to node C. The current through R2 will decrease by 1mA, which is equal to a positive current that flows from A to B.

So the total current is 2mA. Whatever source caused that increase of 1V will observe a change of current of 2mA. Which is equal to 500 ohms or R1 and R2 being in parallel.

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