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I need to make a -10 V dc power supply (I_max = 250 mA) to implement in a system where +-10 will be used for powering opamps and a push-pull.

I have a 24 V dc available as power supply.

Regarding the +10 V, I could do it with a switching regulator (buck converter). However, I cannot find an IC that can output the -10 V without having a very high current (~500-1000 mA) through the inductor (which would probably make it explode).

What DC-DC converter and configuration might I use for this purpose?

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  • \$\begingroup\$ See this: electronics.stackexchange.com/q/219827/40609 \$\endgroup\$ – Cornelius Apr 15 '16 at 21:01
  • \$\begingroup\$ @Cornelius Do you think that can work even if I have different loads between GND and +V and between GND and -V? \$\endgroup\$ – cinico Apr 15 '16 at 21:12
  • \$\begingroup\$ This is not an option for +24 V (to GND) and 250 mA load current. The circuit is not suitable and generates too much heat. \$\endgroup\$ – Master Apr 15 '16 at 21:24
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First, there are plenty of inductors capable of several amperes. So - do not be afraid of 500 mA.

Second, you can use the following designs:

  1. SEPIC converter (two inductors and two "power" capacitors)
  2. BUCK (Step-down) converter with special connection (one inductor one cap)
  3. Fly back converter (transformer and one capacitor)
  4. Half bridge on 24 V with transformer and rectifier

I would suggest using option 4 - if the transformer is not a problem for you. You can use PCB transformers (they are fine on 4+ layer PCBs). You get both +10 V and -10 V from the same simple circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The P/N and values of capacitors have no meaning. They must be selected based on switching frequency etc.

There are plenty of IC able to control gates of two MOSFETs in such configuration. I suggest switching at 250-500 kHz.

The disadvantage is: the output voltages are not regulated, they are proportional to the input power supply 24 V.

Alternatively, you can use design 2: you design the common step down converter on one IC. However, its positive output is connected to Ground, while its "Ground" is used as a -10 V output.

schematic

simulate this circuit

Again, values of components have no meaning.

You can use ready made ICs with integrated power MOSFETs in both cases.

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  • \$\begingroup\$ As per my understanding, output voltage calculation for push pull topology is as follows, Vout = Vin * (Ns/Np) * D * 2. This is the general output equation figured after googling. Any link for the derivation of output voltage or duty cycle? \$\endgroup\$ – vt673 Jul 4 '18 at 7:09

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