Is voltage divider applicable on circuits where there is no ground?

Question

For the circuit below, if the bottom terminal is -Vin and not grounded, is applying voltage divider possible at all?

If this is the case, what would the equation be? I never heard of a voltage divider with no reference ground at the bottom.

See How do I use superposition to solve a circuit?. Once you've worked through it once by superposition you'll see it's easy to reduce the solution to a simple formula. — The Photon, Mar 28 '16 at 01:56

score 15 · Accepted Answer · answered Mar 28 '16 at 01:57

15

Yes. Consider a more general case and apply the superposition theorem:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

$$ V_o = V_o(V1) + V_o(V2) = V1 \frac{R2}{R1+R2} + V2 \frac{R1}{R1+R2}$$

answered Mar 28 '16 at 01:57

arielCo

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Peter Bennett · Answer 2 · 2016-03-28T02:36:59.180

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"Ground" is (usually) just a label we put on the point in a circuit that we want to call "Zero Volts" -it has no special magic properties.

A voltage divider will work whether or not one end of it is connected to "Ground". Vout is still proportional to the resistor values, but the total voltage is Vin-Vb (Vb = voltage at the bottom of R2), and will be shifted by the difference in voltage between Vb "ground".

edited Mar 28 '16 at 02:36

answered Mar 28 '16 at 01:50

Peter Bennett

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score 1 · Answer 3 · answered Mar 28 '16 at 01:58

Sure. The analysis is really straightforward. Let's say you have Vin1 (the positive input) and Vin2 (the negative input). The derivation is as follows (with \$i\$ being the current flowing through R1,R2):

$$ i = \frac{V_{in1}-V_{in2}}{R_1 + R_2} \\ V_{OUT} = V_{in_2} + i R_2 = V_{in_1} - i R_1 $$

It's still a voltage divider, but the "lower end" isn't 0V anymore. Still has its uses, but not as common.

Is voltage divider applicable on circuits where there is no ground?

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