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I am trying to get a voltage divider which has a positive and negative rail such as this:

schematic

simulate this circuit – Schematic created using CircuitLab

I know with a standard voltage divider the equation for Vo would be (R2/R1+R2)*Vin which is fine.

I know if R1 and R2 are equal, you get half the supply. I know this works with a positive and negative rail also, for example: V+ = +5V V- = -5V R1 = 100R R2 = 100R

Then I know Vo will be 0V as this is the halfway point.

My question then, is what if the value of R1 or R2 changes? I have done simulations, for example, when I make R2 half the value of R1, my simulation tells me the Vo is -1.67V. When R2 is a quarter of the value, then Vo is -3V.

Also, If I change the value of R1, then it gives me the same results, but positive, rather than negative.

I have tried to search for an equation to calculate Vo but cannot find one. All the voltage divider calculators online just assume that V- is GND. I have tried to find some way of calculating it by putting the -5 in the equation somewhere but I never get the same result as the simulation.

Does anyone know if there is an equation for this, and of so, what is it? Would be useful for a project idea I have in mind!

I hope I have worded this well enough and have supplied all the appropriate information, if not, please do not hesitate to ask if I have not communicated something properly.

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    \$\begingroup\$ GND is just an arbitrary point we assign to a circuit, you can set V- at GND and then adjust V+ to be the same distance away. \$\endgroup\$
    – PlasmaHH
    Mar 17, 2017 at 15:20
  • \$\begingroup\$ Your problem is in looking for a formula you can memorize. Once you learned Ohm's Law you knew how to solve it. The current through the two resistors is the voltage across the pair (V+ - V-) divided by their series resistance (R1+R2). The voltage drop across either resistor is that current times the resistor's value. You can find Vo by subtracting the drop across R1 from V+ or add the drop across R2 to V-. Don't just memorize. Stop and think. \$\endgroup\$
    – user128351
    Aug 3, 2017 at 23:16

3 Answers 3

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This is your equation for Vout in terms of V+,V- R1 and R2:

VO = V+ - ((V+ - V-)/(R1+R2))*R1

In short, just find the total voltage, divide by the total resistance to get current, then find the voltage drop across one of the resistors (I used R1) and use that to get VO. If you found the voltage across R1 then do V+ - VR1 or if you found it across R2 then do V- + VR2.

Oh, and don't use a divider like that to power any circuitry, if you actually need the rail buffer with an opamp. Also, don't use that circuit as a voltage reference, with 200 Ohm total R you will be throwing away lots of current.


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  • \$\begingroup\$ Those resistor values were just the default when drawing the circuit and couldn't be bothered to change it! The Vo will be going through a buffer and my resistor values are in the 10's of Kohms \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:28
  • \$\begingroup\$ Cool, just make sure the buffer can handle the current required by the rest of the circuit. :) And remember, don't make the resistors to large, although you save current you make the reference more unstable and susceptible to noise. \$\endgroup\$
    – Tim M
    Mar 17, 2017 at 15:30
  • \$\begingroup\$ I tried your equation, and it still didn't match the simulation results.... I also found the total current (with a difference of 10V between V+ and V-) and used Ohm's law in that way, but it still does not give me the answer I want. For example, lets use the one that gave me -3 which was R1 = 100R and R2 = 25R. With 10V, that gives 80mA (10/125) and times that by R1, you get 8V.... not the -3 the simulation tells me \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:35
  • \$\begingroup\$ Also, my load is in the region of 100-200uA so I'm pretty sure the buffer can handle it! Just to add aswell, I will be wanting a negative voltage out of this. \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:36
  • \$\begingroup\$ Never mind, I was just being silly. I looked at my answer again, then re-read your reply and it's just clicked! Thank you very much for that, I was trying Ohm's Law right at the beginning but it just never clicked into place! \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:40
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You cannot really calculate the voltages across the divider because they depend on the current. This means the voltages depend on the load you plan to connect. This is because of Ohms law (U=I*R), where I is unknown.

In general, a voltage regulator such as the 7805 (positive 5V) or 7905 (negative 5V) is a better idea. Those devices regulate the voltage regardless the current. When you plan a huge load (more than 1 Amp), this will not work anymore. Pleas note that other voltages are available in the 78 series (positive) and the 79 (negative). For example, the 7815 and 7915 are used often in audio applications, where a sinus with a peak to peak voltage of 30V is desired.

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  • \$\begingroup\$ It will be going straight into a buffer so I won't need to think about the load on it. I don't want -5 or +5, I would like to calculate the voltage in a resistor divider with those rails \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:30
  • \$\begingroup\$ Read the answer by Tim Mottram. You can calculate it, I don;t know if you read my question incorrectly or misunderstood it, but unfortunately your reply wasn't really any use as it didn;t have much to do with the question \$\endgroup\$
    – MCG
    Mar 17, 2017 at 15:42
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try this circuit with GND reference:

enter image description here

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    \$\begingroup\$ Sorry but this question has already been answered.... plus, did you actually read it? Your answer has nothing to do with the question asked.... you just copied my circuit but added the voltage supplies.... Literally had nothing to do with what was asked \$\endgroup\$
    – MCG
    Aug 3, 2017 at 22:00
  • \$\begingroup\$ @MCG Sorry but this question has already been answered... so what? People can answer questions years after they've already been answered by others. Also this answer is an attempt to get you to work through the problem yourself some more so that you can arrive at the correct conclusion. \$\endgroup\$
    – Jashaszun
    Aug 3, 2017 at 23:00
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    \$\begingroup\$ @jashaszun The answer isn't am attempt to get me to work through it.... I gave my attempts to work through it in my original question. This answer is literally a copy of my circuit with the power supply shown and says 'try this circuit' which is what I've already done. I had done the practical many times, I was trying to get my head around the theory, which was the question. The reason I gave this -1 is because it adds nothing and answers nothing. It's a copy of the question circuit and that's it \$\endgroup\$
    – MCG
    Aug 4, 2017 at 6:22

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