# How do I reduce output voltage from a DC power supply?



## JimH (Jan 12, 2006)

This isn't really a flashlight question, but then it's not really a Cafe question either. I hope no one minds me posting it here.

Here's the situation. I have a 4.5A DC power supply that puts out 16.8 volts. I have a load that can only accept 15.8v - 16.2v. What would be the best way to reduce the output voltage of the power supply.

I was thinking I could just make a small adapter with a resistor in it, but, other than trial and error, I don't know how to figure what size resistor to use. If I use a resistor, does it matter which side of the load I place it on?

I know the CPF talent pool in this area is pretty deep, so I figured this would be the place to come for some help.


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## evan9162 (Jan 12, 2006)

Since you only need to drop < 1V, I would suggest finding a 5A diode and simply place it in series. You can find one that should drop 0.5-1V across the whole range of current.


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## pbarrette (Jan 12, 2006)

Hi Evan, Jim,

I was thinking of the diode myself, and even located a sub $2 part on digikey as an example. The problem is, how well regulated is the power supply and how critical is the input voltage on the load?

Many power supplies, such as a laptop brick, specify a voltage output. However, that voltage output isn't always that tightly regulated. Often, the voltage sags under load or the local line power conditions can alter the voltage. If it is critical to keep the load's input voltage between 15.8v and 16.2v, then the diode will simply drop the voltage based on the diode's Vf. If the power supply fluctuates at all from the 16.8v, the voltage on the load could easily end up above or below the requirements.

pb


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## magic79 (Jan 12, 2006)

evan9162 said:


> Since you only need to drop < 1V, I would suggest finding a 5A diode and simply place it in series. You can find one that should drop 0.5-1V across the whole range of current.


 
Yep, that's the ticket.

Try this one:

http://www.mouser.com/index.cfm?handler=displayproduct&lstdispproductid=280244

6A diode and only a half-a-buck!
.


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## Flash_Gordon (Jan 12, 2006)

Is your power supply precisely regulated at 16.8V? If it is the diode concept should drop the output into your range. If not, the output will sag if your load current is high or line voltage fluctuates.

If it is regulated and you have a schematic, look for a trimmer inside. It might have enough range to drop the .8V that you need.

A resistor is impractical (and very very hot).


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## JimH (Jan 12, 2006)

You guys are great. Thanks for the great answers. Here are the details:

I'm trying to power a Toshiba Tablet PC with a PM118 Li-ion battery pack. The pack has selectable output of 16v or 19v. 

When set to 16v the unloaded voltage is 16.8. As soon as I connect it to the computer, the computer crashes and locks up. The internal battery must be removed from the computer in order to free it up.

I have an older model of the battery pack (slightly less capacity). When the older model is set to 16v, it puts out 16.2 volts unloaded. The older model powers the PC just fine.

So somewhere between 16.2v and 16.8v there is a threshold that causes the PC to go tilt.


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## Flash_Gordon (Jan 12, 2006)

Good to learn the details. I might suggest that you have an issue with your new battery pack. Your computer should not be anywhere near this sensitive to .6V difference. Your standard AC power pack varies at least that much.

I wonder if the voltage is going low when the load is applied causing the problem.


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## MrAl (Jan 12, 2006)

Hi there,

The diode idea sounds like it should work, however there are a couple points to
keep in mind...

1. You may do better using two diodes in series to drop a little more voltage.
Purchase at least two diodes just in case you need a second one.
2. You should also connect at least a 1k resistor as a load so that when the
load device is just powering up if it draws less current it wont be hit with the full
output of the power supply before the diode(s) start to conduct heavily. With
the resistor there will always be a load for the diode so it always drops around
0.5 volts or so. Once the device powers up the diode should drop it's normal
voltage and you should check the output with a meter during various power
modes of the load device.
3. A cap across the output isnt a bad idea either, say 100uf or so, to smooth
short voltage dips and spikes.


Take care,
Al


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## VidPro (Jan 13, 2006)

i would have to read the voltage, under load.
to (only) add to what has been said, it might be the total opposite what your thinking.
the output might be to little, or it might be dirty power, or even (insomuch) ac as opposed to dc.

and also add, that powering a laptop with some OTHER power supply might be a good way to lose more than a power supply or fuse (assuming it HAS one) 

underpowered motors, have to work hard, and laptops like other smaller electronics do not take kindly to large variations, because the parts arent 2X what is needed.

check your power supply under a load, ummm what kinda load i aint telling, cause you might fry it using my stupid methods.

so how about some examples. a cheap arse 12V powersupply could be putting out 14+V without any load, put a load on it, and it might snap down to somewhere close to 12V.

a good computer power supply from turn on to turn off, with no load or huge load, might vary .5V.
BIG huge difference. and when your booting up there is a Large Initilization load that is Attempted to be spread out (cpu then ram then hard drive) but it still isnt spread out much at all, it hits Hard and fast on init.
that COULD be why it acts like its going to go then bails out on you. IF the power supply is going to actually run the computer, replacing it with even a good power supply , might not be sufficient.

Watch it jim  dont be frying that thing.

and for them that dont understand what Big Al said, a diodes drop is also under LOAD, its not such a perfected .7V thing as it SEEMS when its , cold, first hit, over or under rated. still makes a great silicoln resister, but to waste the power it does, it STILL wastes it in heat.
aparentally the solar people have blown quite a few diodes, thinking they would hold up over time, without heatsyncing or cooling, so they do VARY, they do dissipate heat, and they will fail when not cooled or highly overrated, over time.

a diode IS a good solution for a voltage drop, (i am not knocking it) i just like Fat heatsynced bridge casements with 25A heatsynced suckers in them.
even the car chargers use fatty 20-30A ones for 6-10A chargers.


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## MrAl (Jan 13, 2006)

Hello again,


Hey some very good points VidPro! I was assuming he was using a regulated
power supply and just dropping the voltage a little, but if it's an unregulated supply
the peaks could be as high as 1.4 times the nominal output voltage (that's 22 volts
for a 16v nominal supply) which could very well blow out a load designed for
around 16 volts only.

If it was *my* computer i think i would go with either very well regulated output
or take the loss and purchase a power supply made just for that, or else at the very
least make sure the power supply is regulated before dropping the voltage with
diodes, and i'd be very sure to have a resistor tied to the output and measure
the voltage without the load to make sure it isnt too high before connecting it.

BTW, to check the output of a supply with a meter you sometimes need to
connect at least a 100uf cap to the output in case it is full wave rectified
but not filtered. A full wave rect. output will measure a nominal DC voltage
(like 16 volts) even when the peaks are much higher (22 or more volts) but
with the cap connected to the output it will measure the full peak output
with no load (around 22 volts if it's not regulated). If it's got a cap
connected internally then it doesnt matter as you'll always read the peak
with no load. If it's a 'computer type' power supply then it's probably 
decently regulated already.

Be sure before connecting the load or risk blowing out the load device permanently.


Take care,
Al


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## eebowler (Jan 13, 2006)

I have a 12V,10A power supply home which I used to power car audio previously. Maybe the design is different but, inside of mine, there was a flat variable resistor which changed the output voltage noticably. Take a look inside of yours and you may be lucky. Tonight I'll play with it again and give you better numbers in the morning.

*EDIT: Sorry about the delay. Voltage varies from 10.98-13.75V when turning the pot.*


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