Questions about a single-to-dual power supply method - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 07, 2024, 05:38:29 pm
content: I recently saw a post (somewhere else, not the forum) about someone using a single DC supply to generate a split-voltage power supply. For number's sake, let's use +24 volts to make +12 and -12 V for a circuit. It seemed like an interesting idea, but I was wondering what some pros and cons of it were? I built some power supplies before, and this seemed like a good idea (certainly, it is different than how I went about it), but I wanted to ask about it here. Alas, I can't find the source again, and I'm a bit stuck on how to conceptualize the negative voltage.I don't remember exactly how it was done, but the idea was that you do something to generate a +12V supply (probably through a linear regulator), and I think you use that as the ground reference for the rest of the circuit (after the voltage supplies). So, when you take the voltage between +24 V and that +12 V reference, you get 12 volts. What gets tricky here is how to get the -12V. How would you generate that negative voltage when you use the "ground reference" for the rest of the circuit? That's where it gets a bit fuzzy for me.I did something like this without the ICs before, but it was on a breadboard. It was meant to power an op-amp for quick testing, and it didn't seem practical in the long term given its design. However, the idea above seemed to take this idea and make it practical with linear regulators. This appears different than how I built power supplies before (using a buck and a inverting buck-boost converter so everything is referenced to one ground, not where they select another ground reference for the rest of the circuit). I'm curious about how practical this idea is.

author: Buriedcode
author_rank: Super Contributor
timestamp: June 07, 2024, 07:33:16 pm
content: Look at the values of R1 and R2. Calculate the current available at your virtual ground - or even better, model it in Spice. What happens when you apply a load to your rails (+12V and -12V)?

author: Rick Law
author_rank: Super Contributor
timestamp: June 07, 2024, 07:40:42 pm
content: It only works if your load is drawing only a very very small current. I am not sure you already know ohm's law or not, and I am not sure if you know how to evaluate resistors running in series or parallel. I am just going to assume you known neither and bring up the important points.Ohm's Law: V = I * R, that is voltage=current*resistance. Mathematical equivalence for V=IR are I=V/R and R=V/IResistors:In Series, the result is their simply addition:Rresult= R1+R2when R1and R2are in series.In Parallel, the result is a bit complex, it is theinverseof thesum of the two inverses:1/Rresult= 1/R1+ 1/R2. Mathematical equivalence is Rresult= 1/((1/R1) + (1/R2))With that in mind, now look at the circuit. Let's removing C2and C3from the circuit for now leaving just the power supply plus R1and R2.R1and R2in series acts as a voltage divider. Current flowing into and out of R1has only one place to go: into and out of R2. So using Ohm's law, V=I*R, both I and R are the same value upper half or lower half. Current are the same, R1and R2are equal so the V1and V2must also be equal: voltage is divided equally, so 12V upper half and 12V lower half.Now imagine, across upper resistor R1side (the presumed +12V half). You give it a load of a third resistor R3, let's say R3is the same value as R1and R2so math is simple. The result of putting R3paralleling R1results in the different resistance. Let's call that R13. You can evaluate R13using the inverse of the sum of the inverses:Result R13= 1/((1/R1) + (1/R3)) = 1/(1/500K + 1/500K) = 250K.Now your voltage divider is out of balance. It is no longer equally dividing between a pair of 500K resistors (ie: R1and R2) in series, but instead the you have R13and R2in series. That is, you have250K and 500K in series. The upper part (+12V part) has only 250K, 1/2 that of of the bottom part of 500K. With V=I*R, and R in R13being half of R2means it's voltage V1is also half of V2. End result would be8V on the upper half, and 16V on the lower halffrom that 24V total.Now if you repeat the mathwith R1and R2each being 5 Ohm, but with the load R3being 500K Ohm, you will see the voltage change to be a lot smaller, R13= 1/(1/5 + 1/500000) = 4.99995. Dividing the voltage between 4.99995 Ohms upper and 5.0000 Ohms lower. The actual voltage would be 11.99994V upper and 12.00006V lower(see EDIT below), you need a really really good DMM to discern that difference.A simple divider works -- as long as the current diverted to the load is small. When the load varies, the voltage divider's division also varies along with it. The more current going through the divider, the better it work, but of course less efficient as less of the total current are actually used for your load. So you can used the voltage divider method with in mind about the load.The capacitor C1 and C2 are for stability (slowing the the change when current drawn is changing). It doesn't change the divider characteristics from division stand point. It may however introduce oscillation in the circuit. That is too complex to discuss in detail for the time being.EDIT (adding this just to be complete for the "if you repeat the math" paragraph):Vupper= 24 * Rupper/ (Rupper+ Rlower)= 24 * 4.99995/(4.99995+5) = 11.99994Vlower= 24 * Rlower/ (Rupper+ Rlower)= 24 *5/(4.99995+5) = 12.00006

author: TimFox
author_rank: Super Contributor
timestamp: June 07, 2024, 08:54:51 pm
content: The previous post details the deficiencies of a resistor divider for that application.Another method, using active components, can be used if the current drains on the positive and negative side are reasonably close: look up "rail splitter".A single op-amp with a divider at its input can be used, where the op amp output requirement is to pass a current equal to the difference between the positive and negative loads.