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How to reduce voltage?

Power supply too high

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Gary Murphy 115/03/2019 15:01:48
464 forum posts
22 photos

I have JP "ultra power" 12v power supply. Well it says 12v on the box BUT when you read the manual it says 14v!

All is well apart from one peak detect charger I use for a few Nimh packs. It shuts down with over voltage warning.

How can I reduce the voltage slightly? Just for this charger,i would like to be able to have a "plug" in device so it still runs in standard form for other devices.

I know you can put a 12 v bulb in line BUT wanted something a bit neater?

Frank Skilbeck15/03/2019 15:21:07
4824 forum posts
107 photos

Is your charger rated for a 12v battery, it's a bit surprising it's shutting down on 14v a fully charged car battery can be over 13v. Also putting a resistor in series (e.g. car headlight bulb) would only work when the charger is on load. the voltage drop across a resistor = current x resistance.

You would need a 12v regulator not a resistor.

Gary Murphy 115/03/2019 15:43:53
464 forum posts
22 photos

The charger manual says 12v pb battery OR power supply.

The annoying thong with the power supply is the box says 12v BUT the manual says upto 15v when on load! average 14v.

Martin Harris15/03/2019 15:49:30
9505 forum posts
256 photos

A simple way to drop a small amount of voltage would be to put a silicon diode (suitably current rated for the charger's input) in series which will knock 0.6V or slightly more off the power supply's output irrespective of the load.

Denis Watkins15/03/2019 16:26:03
4634 forum posts
129 photos

These chargers are a multipurpose compromise, and work well

Your 12v car regulator for the alternator puts out 14v, as your Pb battery would be in a poor state with anything less

Just add a 12v regulator, as suggested for your one off delta peak charger, to allow this to work

Geoff S15/03/2019 16:28:25
3780 forum posts
39 photos

A silicon diode will certainly drop 0.6/0.7 volts. Just make sure it's got a high enough current rating as Martin writes. However, I notice you just mention what the power supply manual says the voltage spec is. Have you actually measured it? It may be well over 14v - especially off-load. In which case the diode may not be enough and you may need 2 in series.


GrahamWh15/03/2019 16:47:07
356 forum posts
53 photos


Also if using a diode, consider the heat it will produce.

Test it carefully and use a heat sink for it if need be.

For example, if using one diode with a voltage drop of 0.6v, if the current going through it (and onwards through the charger) is say 5 amps, then the power turned into heat in the diode itself will be (voltage drop across diode) x current =0.6 x 5=3 watts. This could get the thing quite hot. A higher current will produce more heat. You can get heat sinks that will dissipate the heat produced and connect to the diode.

Heat sinks are rated at temperature increase per watt. Aim to stay below a suitable temperature, say 50 degrees centigrate above room temperature, so a heatsink for 3 watts needs to be 16.7 degrees per watt or better. Of course the diode itself will dissipate its heat to the air naturally without a heatsink, but may get to an unacceptably high temperature to do so.

(It''ll get hotter and hotter until the rate of dissipation (which itself increases with temperature) matches the rate of heat production, ie. 3 watts.)

In our example the heat sink rated at 16.7 'c/watt would get rid of 3 watts by the time it got to 50 degrees above room temperature. Without a heat sink, the diode may get so hot it burns out or causes damage to whatever it touches. When people say they use a diode in a model with high power servos to reduce receiver battery voltage for example, I always wonder how hot that diode is getting .

Edited By GrahamWh on 15/03/2019 16:49:10

Edited By GrahamWh on 15/03/2019 17:01:24

Cuban815/03/2019 18:22:30
3035 forum posts
1 photos

Just had a quick look at the instruction manual, and the device is behaving exactly as you'd expect an unregulated power supply to behave. 15V no load, dropping to 12V at the max rated current. For thirty quid I guess that's all you're going to get. To be fair, the manual warns of potential problems when used with some chargers and that's to be expected. The diode trick will work against you when drawing higher currents.

Edited By Cuban8 on 15/03/2019 18:25:01

Don Fry15/03/2019 18:39:05
4557 forum posts
54 photos

How much is the 'I use for a few NiMh packs' charger worth? If you need it particularly, I might be less stressful, and little more expensive to replace it with a modern unit less fussy about input voltage.

Kevin Wilson15/03/2019 21:06:14
399 forum posts
13 photos

I used a pair of silicon rectifier diodes to drop a bit of voltage off a PSU as suggested.

They do get hot, even just pulling 3A or so.

PatMc15/03/2019 21:55:46
4473 forum posts
548 photos

If you connected another load to the power supply in parallel with the charger this would reduce the voltage. From the OP it sounds like you have another charger, possibly this would provide enough load to drop the voltage below the limit required.

Edited By PatMc on 15/03/2019 21:56:30

fly boy315/03/2019 22:46:06
3757 forum posts
22 photos

Slightly off topic, would a diode do the same thing as long (dangerous) leads used to reduce volts from 2v cell to suitable glow volts for plug ? Thanks

Gary Murphy 115/03/2019 23:15:45
464 forum posts
22 photos

Thanks guys.Looking at the options I might as well buy another power supply.

I will check this one supplies no more than 13.8v

It beats me how the JP supply is a 12v power supply BUT delivers 15v ! Why put 12v on the box when it never supplies 12v.

Peter Jenkins16/03/2019 00:17:26
1644 forum posts
305 photos

If I were you Gary, I'd keep the PSU and change the charger. As has already been mentioned, lead acid batteries when fully charged put out around 14 v. So if you connected your charger to a fully charged 12 v lead acid battery it wouldn't work either. So, change the charger to one that will cope with a supply that goes up to 15 volts.

Cuban816/03/2019 08:58:39
3035 forum posts
1 photos
Posted by Gary Murphy 1 on 15/03/2019 23:15:45:

Thanks guys.Looking at the options I might as well buy another power supply.

I will check this one supplies no more than 13.8v

It beats me how the JP supply is a 12v power supply BUT delivers 15v ! Why put 12v on the box when it never supplies 12v.

They are correctly describing their product. It is designed to provide 12V only at its max rated current, but being an inexpensive unregulated supply it will float up to several volts higher with a smaller load. A commercial regulated psu will be considerably more expensive than what you have at the moment. 

Edited By Cuban8 on 16/03/2019 09:12:15

GrahamWh16/03/2019 11:40:26
356 forum posts
53 photos
Posted by fly boy3 on 15/03/2019 22:46:06:

Slightly off topic, would a diode do the same thing as long (dangerous) leads used to reduce volts from 2v cell to suitable glow volts for plug ? Thanks

It would, but the diode may get really hot - probably burn out. Buy 2 or 3 diodes and test it, and get a heatsink if need be.

Edited By GrahamWh on 16/03/2019 11:40:44

Martin Harris16/03/2019 13:07:35
9505 forum posts
256 photos

Lots of comments about diodes getting too hot but (although I haven't done any testing to prove it) surely a diode rated at (say) 5A carrying 2 or 3 amps average current should remain comfortably within its normal heat range.

Edited By Martin Harris on 16/03/2019 13:08:10

MAD Dave16/03/2019 13:20:19
92 forum posts
9 photos

Place some diodes in parallel to share the load, ie, heat. Recommend the same spec. devices.

PatMc16/03/2019 14:31:19
4473 forum posts
548 photos

Each diode connected in parallel would reduce the level of voltage drop.

John Rudd16/03/2019 14:37:38
96 forum posts
2 photos

Are these power supplies linear transformer based or switch mode?

Either way, there ought to be a means of reducing the output voltage to a more acceptable level.....

Failing that a shunt regulator could be constructed and used to power the more voltage sensitive charger...( more robust than using diodes...)

Edited By John Rudd on 16/03/2019 14:39:47

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