# More current draw from a power supply based on a voltage divider

I've built a circuit that needs a power supply that can provide -12V, 5V and 12V. I've built the circuit below as a proof of concept, being aware that I might need to build something that allows for more current draw once I'm done with testing.

I've used some off the shelf components to get 24V DC, and then built the usual voltage divider with an op-amp keeping the virtual ground in the center. For the 5 volt rail, I simply added an LM7805 regulator. While this technically produces the right voltages, it does not let me draw enough current for my finished circuit and the op amp gets very hot. I need to drive an LCD panel (~100 mA) and a Raspberry Pi Pico (~25 mA) off of the 5 volt rail, which is the majority of the current requirement, the rest of the circuit draws around 20 mA from the +12 and -12 V rails.

Is there any reasonable way to salvage this design or would I be better off designing an SMPS from the ground up that gets its negative rail from AC directly (something like what the OP proposed here) ? Any other smarter design that could safely supply enough current for my circuit?

Edit: I've accepted the answer that adapted my design, but if anyone is in a situation of having to build a PSU with negative voltages, look for better alternatives. If you happen to build the same, use power transistors (the BD139 and BD140 pair worked fine, and make sure you attach both your LM7805 and PNP transistors to a heatsink. That way they run stable and don't get overheated.

You can follow the opamp with a transistor buffer as seen here. This will take the load off of the opamp and increase the available current. You'll probably want a circuit that includes the transistors in the feedback loop.

These buffer circuits can go from very simple to quite complex as seen here.

Here's one I threw together quick in LTspice:

• I’ll have to try this in simulation first to make sure I understand the idea here. Commented May 29 at 21:14
• I've prototyped this on a breadboard and I am able to draw way more current! In fact, I can draw enough that the circuit works as intended. Unfortunately the PNP transistor needs to be something beefier, the 2N3906 lets out the magic smoke after a few minutes (which is probably fine, I'll need to check if I have some stronger replacement). Commented May 30 at 19:57
• @PeterLenkefi Yeah, I just used the generic 2N390x transistors for the example, you'll have to get something that will handle a bit more power. The dissipation for each transistor will be 12 V times whatever current goes through it, so at 100mA that's 1.2 W. The more balanced the positive and negative loads are the less current through the transistors, with equal loads the transistors would be doing pretty much nothing, but typically the positive load is higher than the negative and that extra current goes through the PNP. Commented May 30 at 21:15
• @PeterLenkefi Also note you don't have to use the opamp I used in the example either. If I get a bit of time I'll try to update the example. Commented May 30 at 21:24
• I've built it, they work fine and heat-wise seems all ok after attaching both the PNP transistor and the LM7805 to a heatsink. Thank you, while this design might not be the best approach for a PSU, this was a really neat and easy way to fix what I already had. Commented Jun 1 at 12:01

The VGND needs the ability to sink as much current as the sources produce (if the load currents are not being drawn symmetrically or near symeterically)

For analog electronics most of the supplies draw current near symmetrically and with lowish currents (10's of mA), so this scheme works.

For adding a load on one side the LM358 will need to be able to control the output with as much current as it can source which is ~20mA plus or minus. Your loads are much more than 20mA. In addition there is a 47Ω current limiting resistor and fast changes from digital loads will be a problem.

So what to do? Parallel lots of LM358's to cover the current or find another way go generate power. I would use two DC/DC's to generate the ±12V and another regulator to generate the 5V. I prefer these: https://www.cui.com/product/resource/vxo78-1000.pdf

What's the boost converter for? You already have +12V. Buck that down to +5V. And use an appropriate switcher (inverting charge-pump is probably best) to invert the +12V rail. No opamp or virtual ground necessary.

simulate this circuit – Schematic created using CircuitLab

• The boost was to go 24 Volts, which I could halve easily. I have never built an inverting charge-pump. Could you provide some source or schematics please? Commented May 29 at 21:12
• @PeterLenkefi I understand what a boost converter does. The question is why is it relevant to the circuit in question? As for the charge pump, I'm not going to post any schematics - you can buy complete charge pump IC's like this one: analog.com/media/en/technical-documentation/data-sheets/… Commented May 29 at 21:20
• @PeterLenkefi MOSFET's point is that you can just buy a module that makes -12V from 12V. No need to double and then halve. Commented May 30 at 12:35
• @QuittingDueToAntisemitism I understand that, but then it's hard to answer the "why". This is how I solved it and had no knowledge of inverter ICs like that. Questioning it's relevancy in the circuit is just plain rude, one could just point it out that this is excessive and there are smarter ways to utilize my components, which I'm entirely open to. Commented May 30 at 14:37
• @PeterLenkefi No one's being rude. Questioning the relevancy is paramount to providing a sensible solution. The criticism comes from the fact that your approach is poor. Commented May 30 at 16:26