Hi to all,
I'm using a LT1033 for obtain -3.3V from -12V of an ATX PowerSupply.
It works well but the LT1033 heats up a lot.

I had then thought of using a different method, with a LM2596T-3.3, which should provide the maximum possible current (3A nominal) but I can't find any circuits capable of using it to create a negative voltage from a positive one.

Do you think it's a good solution?
What voltage should I power the LM2596T-3.3?
And finally... do you have any circuits to suggest to me to be able to do this?

Thanks everyone!

 

The LT1033 is linear. It will get hot.
The LM2596T-3.3 will only work on positive voltages. It will not create heat.

 

Hi to all,
I'm using a LT1033 for obtain -3.3V from -12V of an ATX PowerSupply.
It works well but the LT1033 heats up a lot.

Is there a heatsink on the regulator?

 

Is there a heatsink on the regulator?

Yes.

 

[...]The LM2596T-3.3 will only work on positive voltages.[...]

I read about the possibility of doing it and I also saw some circuits. But if I try to simulate them (Multisim) I don't get the results I hoped for.
Circuito di alimentazione usando LM2596 : r/ElectricalEngineering

 

Yes.

Another option is placing a resistor in series with the input of the regulator to transfer some of the dissipation.

 

I have used this part as a simple buck and can get near 3A from it.
I have not tried to get much current out of it in the Inverting mode.
I think your LM2596-3.3 will also work in this mode.
SPICE might not be happy with the circuit. If it does not start, try adding a diode across Cout so Vout does not go above Ground.

 

I have used this part as a simple buck and can get near 3A from it.
I have not tried to get much current out of it in the Inverting mode.
I think your LM2596-3.3 will also work in this mode.
SPICE might not be happy with the circuit. If it does not start, try adding a diode across Cout so Vout does not go above Ground.
View attachment 341448

Thanks for your reply.
I tried it and now it works, although I get a slightly different voltage (about -3.6 volts instead of -3.3 volts):
How can I instead understand with the simulator how much current it will provide?