Hi,

This seems like a simple question but I've had a surprisingly hard time finding answers from googling.

I have a situation where I need a power supply that outputs the following voltages (in order to replace a legacy power supply that provides said voltages):
16V
12V
5V
-5V

In the past I've always just grabbed an off-the-shelf solution that provided the voltages that I needed, but the closest that I think I'll be able to find for the above voltages would be something like this: https://www.amazon.com/gp/product/B005T9PSKO/

I'm wondering if I could just take a separate 5V wall wart, reverse the polarity on it and get -5V. I'm assuming there is probably a reason this wouldn't work, but I'm not sure what it would be. Can anyone help educate me?

 

if I could just take a separate 5V wall wart, reverse the polarity on it and get -5V

Yes.
The outputs of a wall wart are isolated, so either output can be connected to your circuit ground.

 

A simple LapTop-Computer-Brick would be a good starting point.

You didn't state how much Current You need.

 

After some study of the circuit, I believe the current requirements for the -5V are pretty small. Maybe even less than 10mA. i'm not sure exactly.

 

Each individual Regulated-Voltage-Output needs to have a Current Requirement specified.

 

You could generate -5V using an MC34063; those (or copies) are commonly found in old lighter jack cellphone chargers. The application note for the MC34063 includes an example circuit (and layout) for -5V. It may be feasible to modify a charger board to get -5V by cutting and jumpering. (The older version of the app note has single-sided board layouts that are easily duplicated with a hobby knife. (I used packing tape as resist material))

 

I would recommend bassbindevil's approach in post #6.

Calculators like this makes the design of a power supply easy (plus this saves a wall-ward).
http://www.nomad.ee/micros/mc34063a/index.shtml

 

How much current do you need from the -5V supply?
The devil is in the details.

 

I would recommend bassbindevil's approach in post #6.

Calculators like this makes the design of a power supply easy (plus this saves a wall-ward).
http://www.nomad.ee/micros/mc34063a/index.shtml

Thanks for the link to the calculator!

I tried to use the calculator and I have a few questions.
For Vin I put 5V, for Vout I put -5V and for Iout I put 100 mA.
But I don't understand what Vripple or Fmin mean. I am speculating that the output will have a frequency and so these values determine what the frequency and amplitude of output wave will be. So I put 5mV for Vripple and Fmin for 10 kHz just to put something there so I could move forward. Have I inferred the meaning of Vripple and Fmin correctly? (EDIT: Nevermind, I found this explained in the datasheet here: https://www.ti.com/lit/ds/symlink/mc34063a.pdf?ts=1638894167096 )

The calculated output gives me these values:

Ct=2298 pF
Ipk=470 mA
Rsc=0.638 Ohm
Lmin=489 uH
Co=10340 uF
R1=1k R2=3k (5V)

I am not sure what I am supposed to do with Ipk because it looks like that is something internal which I don't actually control. Is this referring to how much current my 5V input needs to be able to supply?

Does Lmin refer to the minimum value of L? I can't think of what else it could mean.

And what is Ct for? That looks like another internal value but perhaps it is the value of the capacitor between points 3 and 4?

Thanks for teaching me about this new and interesting tool!

 

A simple LapTop-Computer-Brick would be a good starting point.

You didn't state how much Current You need, ( per Output ).

Simply use 4 dead-simple Linear-Voltage-Regulators.
.
.
.

 

If you need no more than 200μA you can use ICL7660 or LMC7660 or similar IC.

 

mpownby wrote:
I am not sure what I am supposed to do with Ipk because it looks like that is something internal which I don't actually control. Is this referring to how much current my 5V input needs to be able to supply?

Yes, kind of. That can be supplied by a large capacitor on the input. The reason Ipk is mentioned is that it is the peak current in the inductor. You need an inductor that can handle currents of Ipk or higher, otherwise it may saturate (be reduced in inductance by the current, which could result in excessive peak currents higher than Ipk).


mpownby wrote:
Does Lmin refer to the minimum value of L? I can't think of what else it could mean.

Yes, your inductor must be that value or larger.


mpownby wrote:
And what is Ct for? That looks like another internal value but perhaps it is the value of the capacitor between points 3 and 4?
Yes.

By the way, the ICL7660 mentioned by MrChips is an excellent solution when the current requirements for your negative supply are low. Easy to use, no calculations necessary! I have used several successfully and are my go-to when I don't need much current.