Changing the Output Voltage of a Power Supply with a Diode?

Don't buy a charging power supply that produces the wrong voltage.
Get the right stuff from the get go.

Each silicone diode will drop .6 volts across itself, be sure it can handle the current. :)

It sounds like your power supply may have foldback current limiting versus straight line current limiting, based on your description of the issue. If this is true diodes won't help. Best to use a power supply without holdback limiting.

Straight line limiting is when the current stays essentially constant but the output voltage drops when in current limiting. Holdback limiting is when the voltage and the current get reduced to zero when a short is applied.

Hi, Salvo,

The issue is, the turnigy does not have constant current limiting. Therefore, when I try to charge my bank with it, the current spike, at power on, causes it to go into over current protection (ocp). If I charge the bank for a few minutes with another charger, until the voltage gets above the turnigy's 13.62v lower limit, and then turn on the turnigy, it powers on, but with no current flow. But once the little bit of surface charge quickly burns off, and the voltage drops back down to the 13.62 mark, the turnigy starts outputting current, starting from 0 amps up to whatever the batteries will accept. So, a soft start of sorts, and the power supply does not go into ocp.

So, I'm attempting to lower the lower end of its output voltage range to below 12.2 volts, (50% soc) so that I can set the adjustable voltage to say 12v, power on the unit, and then slowly raise the voltage until it matches that of the batteries, and initiate a soft start with no current spike. The pot requires about 5 full turns to gain only 1 volt, so turning it up slowly will not be a problem.

Adding a diode is not an elegant way to reduce voltage. What's your objective? How many amps at what voltage? Does it need regulation?

Thanks, Mex,

That particular rectifier is said to have a MAX forward voltage of 3.2v.

(or is that referring to it being in series with additional diodes?)

And it gives a temperature rating of ... Tc = 25C ? ... I'm not sure what Tc refers to? Perhaps it is the same as Tj (junction temperature)?

This and all other rectifiers need an adequate size aluminum heat sink. The heat sink to handle .7 volt drop at a 70 amp load would be EXACTLY the same as what is found in a 70 amp battery isolator. You would do well to consider using such an isolator in fact.

One rectifier will drop voltage .7 - .9 volts. If the power supply is set to 15.00 volts, the rectifier will output 14.1 to 14.3 depending on load. Two silicon rectifiers in series will usually drop voltage 1.5 volts. Your 15.00 volts will drop to 13.5 The voltage drop is even from low to high current, and low to high voltage.

EXAMPLE: I am using (3) 1N4003 diodes in series to drop 12.1 volts power supply output to 9.9 volts for a single 10 watt, 10 volt LED device.

But .7 to .9 volt per series rectifier is all you're going to get and even then you are going to need an impressive heat sink.

Your issue is you are dealing with magnitudes greater current.