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How to reduce voltage?


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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?

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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.

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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.

Geoff

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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

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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

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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

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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.

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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

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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

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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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Slight differences in diodes' forward voltage characteristics will cause at least one of them to overheat....

Adding a small resistor to each will lead to better current sharing.....( similar to adding small resistances to transistors in series pass voltage regulator circuits where the series element  comprises two or more devices...)

Edited By John Rudd on 16/03/2019 14:46:13

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Another alternative trick might be to use say a 20A diode in a through hole mounting TO- 220 package. If necessary this could then be bolted onto small piece of ali plate for a bit of extra heat sinking although with a max junction temperature of up to around 200 degrees C I think that it’s never really going to get hot enough to be a visible problem anyway. I’d have thought one of these would carry up to say 5 amps ok without getting overly excited and a heatsink just makes it more bulky. A typical forward voltage is 0.84V at 5A; 0.96V at 20A. Probably the most difficult exercise is actually buying a single unit, these types of components usually come in a minimum of 5; ebay may do singles perhaps? But even in a bunch they are not expensive, postage could still be the highest cost. One advantage of having more than one, if the volts drop across one diode was still insufficient it would be a very simple matter to place another in series with it, total drop then about 1.2V, that must surely be enough.

One disadvantage of putting a bulb in the line is that when the PSU output lead is not connected to anything, (no load), there is still a standing15 volts at it’s end connection point; therefore when the charger is connected it may well detect this and shutdown before any load can be started to drop the volts; whereas a silicon diode is a constant 0.6V drop or more in all situations.

For anyone dabbling in a bit of electric flying I’d have thought that a DMM, (Digital MultiMeter) would be another rather a useful accessory, at least it’s then maybe possible to get some sort of handle as to what is going on when the predicament gets a bit mystifying. Again, doesn’t always need to be very expensive, from £3 upwards and with free postage and nectar points into the bargain…

Happy electrickyting…

PB

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