2
\$\begingroup\$

I'm designing an audio DSP and DAC board, and in the analog section, I'm using opamps that take in differential audio signal and turn it into single ended output. Because the signals are AC, they obviously swing below zero, so I will need a plus/minus power supply. Keep in mind, I'm a software guy, and I'm only doing this as a hobby -- not a pro hardware designer.

I've been researching the internet for about a week, looking for solutions. So far, I've found:

Charge pumps: These are awesome if it wasn't for the fact that they only supply maximum of 100mA. Driving 16 audio opamps will need more current than that.

Buck/Boost inverter These suck. They require complex circuitry and can barely supply higher current output than the charge pumps. To me, this is the big and ugly solution.

Cuk topology I don't have enough background in power electronics to fully understand the mechanisms for this one, but it still requires several inductors and some mess of external circuitry. They are also limited in their current output for the most part.

Power modules This would be a plug-and-chug solution, but it takes the fun out of DIY, not to mention they are quite huge and take up lots of board space.

I've seen some configurations where there is a sort of "half-rectifier" diodes arrangement used for generating negative supply, but I don't understand it.

If someone with enough electronics background can explain to me an easy to implement and understand solution to generating a negative voltage supply, it would be fantastic. Here are my constrains:

  • My input voltage is 12V.
  • I want my output voltage to be between -12 to -15v (any value there will do)
  • I want 1 A current output.
  • I want minimal external components. A few is okay, max one inductor.
  • Don't care about ripple, noise, isolation, temperature characteristics, or anything else.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
19
  • 7
    \$\begingroup\$ You ask for the most painless way, then, when you mention modules, you say it takes out the fun of DIY. You have to choose... \$\endgroup\$
    – dim
    Apr 27, 2018 at 4:34
  • 2
    \$\begingroup\$ @darksky they are expensive because they often have an integrated transformer and flyback topology, which is needed to isolate and create a negative rail at such high currents, sadly. I understand what you are asking for - you don't want a 1-part solution which takes no effort at all, but you also don't want to spend heaps and heaps of time designing a complicated power supply setup. Sadly, you can only really pick one of those extremes for such a high power negative rail :( \$\endgroup\$
    – DSWG
    Apr 27, 2018 at 4:56
  • 1
    \$\begingroup\$ All right, what you could do for a start is to try using something like the TI webench tool. You enter your specs and it spits out a lot of designs with TI chips. Some of them may be quite simple. Other manufacturers may have equivalent tools, e.g Linear Technology. Bu the TI one is very good and let you compare estimated cost/part count/PCB area of each solution, provides simulation plots, etc... \$\endgroup\$
    – dim
    Apr 27, 2018 at 5:02
  • 3
    \$\begingroup\$ The easiest solution might be to shift the signal to be always positive, rather than to generate a negative supply voltage. \$\endgroup\$
    – The Photon
    Apr 27, 2018 at 5:46
  • 2
    \$\begingroup\$ Why not bias all your signals to half your main voltage? Or stated differently, use an op-amp buffer to generate half your main voltage, and then use the output of that buffer as circuit ground. \$\endgroup\$
    – user253751
    Apr 27, 2018 at 9:45

6 Answers 6

Reset to default
10
\$\begingroup\$

The inverting SMPS would seem to be the most appropriate

schematic

simulate this circuit – Schematic created using CircuitLab

When Q1 is on, current into L1 builds, storing energy in it. When Q1 turns off, current continues to flow into L1, and it has to draw this current from D1, pulling charge out of C1, so pulling it to a negative voltage. L1 current falls during this phase, as the voltage across it is negative.

This sort of circuit will happily deliver amps, even 10s of amps. You can use a p-BJT or a p-FET for Q1.

Control consists of detecting the output voltage and changing/stopping the power switch drive.

The usual suspects, Analog Devices, Texas, Linear and Maxim, all have integrated power supply control parts that can be configured to work in this topology, if you don't want to build it from scratch.

For a particular high load or low ripple application, you can parallel two of these converters driven in antiphase, or even more converters driven in different phases, to get a smoother current delivery waveform into the output filter.

\$\endgroup\$
10
  • 1
    \$\begingroup\$ I think your advice would result in most painful way to create 1-A negative rail. \$\endgroup\$
    – Ale..chenski
    Apr 27, 2018 at 5:40
  • \$\begingroup\$ @HarrySvensson No, the feedback isn't on a pulse by pulse basis. There are two ways to do it, voltage control and current control. In voltage control, easier to understand but more difficult to stabilise, a VCO operates the PWM at the transistor switch, and the feedback controls the PWM duty cycle, which controls the nominal output voltage. Current control involves sensing the current flowing in the inductor, turning off the switch when it's more than a threshold, and using the feedback to control the threshhold. A fixed 50% duty cycle is more painless, but less accurate. \$\endgroup\$
    – Neil_UK
    Apr 27, 2018 at 7:10
  • \$\begingroup\$ Thank you very much for explaining that! I will read more about this particular configuration soon. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 7:27
  • 2
    \$\begingroup\$ The first is completely painless, to use a fixed duty cycle, like 50%. This will mean the nominal output voltage is equal to the input voltage. Not true for light loads unless you plan on using a synchronous MOSFET in place of the diode. On a light load you will get a big nasty negative voltage that is un-regulated. \$\endgroup\$
    – Andy aka
    Apr 27, 2018 at 8:36
  • \$\begingroup\$ @Neil_UK agreed. Which is why I deleted my previous comment. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 8:36
3
\$\begingroup\$

You could use an old-fashioned LM2596 adjustable to make a -12V supply, with at least 10V in you should be able to get about 1A @12V out (5V version schematic shown).

enter image description here

Follow the design information in the datasheet to the letter and you should be okay. In particular make sure the inductor meets the requirements and that the PCB layout recommendations are followed.

Protection for an automotive electrical system source is additional, and I'm not covering that here.

Personally I would definitely consider using an isolated 12-15W DC-DC converter for the negative rail. Just wire the output to get negative polarity (and put a diode across it to prevent reverse biasing as D3 above).

\$\endgroup\$
8
  • \$\begingroup\$ Thanks. Is it enough to adjust R1 and R2 in the above design to change the output to -12? I suppose the inductor value should change also. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 12:05
  • \$\begingroup\$ Also since this is a buck converter, does it not mean that the output should always be less than the input? In that case, does it not mean if drop out voltage is 1V, I can only get -11V using this? \$\endgroup\$
    – darksky
    Apr 27, 2018 at 12:14
  • 1
    \$\begingroup\$ @darksky What part of "read the datasheet" was unclear? \$\endgroup\$
    – Spehro Pefhany
    Apr 27, 2018 at 12:14
  • \$\begingroup\$ Yes, it's a buck converter. -12V is less than +10V. \$\endgroup\$
    – Spehro Pefhany
    Apr 27, 2018 at 12:22
  • \$\begingroup\$ The datasheet only gives one example with negative output, and their resistor calculation is for positive output. This is what's confusing. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 12:43
2
\$\begingroup\$

The least painful way to power analog electronics with +-15 V rails at 1 A is to get a DC-DC converter into +-15 V, 30 W overall power. Isolated converters are fine, you need to check for ripples, or filter them out with additional LDO. Try Digi-Key, something like "Mean Well" DKA30A-15, or many similar.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ "Power modules This would be a plug-and-chug solution, but it takes the fun out of DIY, not to mention they are quite huge and take up lots of board space." - Nice. \$\endgroup\$
    – Harry Svensson
    Apr 27, 2018 at 5:05
  • 4
    \$\begingroup\$ @HarrySvensson These modules are designed and produced by professionals. There is certain currently-achieved limit on power density of DC-DC converters. There is no "but". DYI can't possibly match this, and can't make them smaller. That's why this sentiment makes no sense. Either you want fun and make kludges, or use quality components and solve the problem in most optimal way. \$\endgroup\$
    – Ale..chenski
    Apr 27, 2018 at 5:32
1
\$\begingroup\$

From your 12v create a half rail- simplest, 2 resistors. Capacitively couple the input to remove the DC swing below 0. Bias up the input of the opamp (after the cap) to 6v. Your output will swing 0-12 around the 6v point.

\$\endgroup\$
0
\$\begingroup\$

If you really ONLY use the -12V and ground on your device (and not the other +12V too) you can simply use the device -12V, connect it to ground and the device ground connected to +12V. So you get inverted power. But be sure: this only works if you need -12V and ground and NOTHING MORE. Extra: No additional components needed.

\$\endgroup\$
1
  • \$\begingroup\$ I will most certainly need the +12V as well the -12V. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 12:08
-1
\$\begingroup\$

If you had a 24V supply you could use a center tap or voltage divider to get 0V, 12V and 24V rails that you rename to -12V, 0V and 12V.

You can also do this using 2 independent isolated 12V supplies and connecting the + of one to the - of the other:

schematic

simulate this circuit – Schematic created using CircuitLab

This allows you to size each supply accordingly should the 0 to 12V rail need more power than the -12 to 12V rail.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Unfortunately, I only have one 12V power supply. \$\endgroup\$
    – darksky
    Apr 27, 2018 at 8:44
  • \$\begingroup\$ Could you use this method with two dc to dc converters with isolated outputs? \$\endgroup\$
    – Willtech
    Apr 27, 2018 at 12:44
  • \$\begingroup\$ @Willtech they need to be isolated so you don't create a short. \$\endgroup\$
    – ratchet freak
    Apr 27, 2018 at 12:48
  • \$\begingroup\$ Ah, I edited and then your comment showed. Thank you. \$\endgroup\$
    – Willtech
    Apr 27, 2018 at 12:49
  • \$\begingroup\$ ping @darksky does this help? \$\endgroup\$
    – Willtech
    Apr 27, 2018 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.