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Possible Duplicate:
Chosing power supply, how to get the voltage and current ratings?

I've got a broken gadget that is discontinued. The only thing broken is the power supply (verified by my multimeter). I'm confident enough to open up a replacement power supply and solder in the wire I'll salvage from the broken one. What I don't know is what Volt and Amp bounds are acceptable for the replacement.

The broken adapter states it outputs 8V DC 2.6A
The gadget it plugs in to states it expects a 8V input
The battery it goes to charge is 6V

I can't find a power supply that outputs exactly 8V DC 2.6A
Is it safe to change this to 9V DC 1A?
Or perhaps to 9V DC 2A?

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marked as duplicate by Kellenjb, W5VO, Kortuk Jun 29 '12 at 5:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ How do you know the power supply is the only thing broken? \$\endgroup\$ – stevenvh Jun 28 '12 at 18:43
  • \$\begingroup\$ The only thing the power supply does is charge a battery via a docking station. There is still enough juice left in the battery for it to turn on for around 5 seconds before giving up. \$\endgroup\$ – Perrin255 Jun 28 '12 at 18:45
  • \$\begingroup\$ If you've repeatedly run the battery for the few seconds before it gives up, you may need to replace the battery too now. \$\endgroup\$ – Olin Lathrop Jun 28 '12 at 18:48
  • \$\begingroup\$ Notice that most power supplies today are glued tight and it is really difficult to open them (not to mention close them afterwards). \$\endgroup\$ – jippie Jun 28 '12 at 18:48
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    \$\begingroup\$ @Kellenjb I did find that question before I asked mine. I proceeded to ask mine as I wasn't sure if increasing the voltage meant I could lower the amp requirement. My question may be simple however the variety of the answers has already helped me enormously; even if that's just helping me know what to put in a search engine to learn more. If my question was too basic to be on-topic, sorry. I will be back though as I can see that this is a great community and I have a couple of projects I'm taking on as a hobby. \$\endgroup\$ – Perrin255 Jun 28 '12 at 20:50
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I don't know what Volt and Amp bounds are acceptable for the replacement.

Neither will anyone else that isn't familiar with the internal circuit design of the gadget.

Is it safe to change this to 9V DC 1A?
Or perhaps to 9V DC 2A?

No, it's not safe to assume that either of those will work.

In fact, even if you did find a "8V 2.6A" power adapter it might still not be safe unless you also know whether the old power adapter was regulated or not. If it's unregulated, then it's quite possible that the gadget depends on the particular behavior of the adapter, especially when charging the batteries.

Ideally you would disassemble the old power adapter and determine how smart it was - if it's regulated, or even if it has a full charging circuit inside it.

Then you would look for another adapter with those same features.

Anything else is a risky assumption.

However, without additional information, you can almost always replace one power supply with another that has the same voltage output with an equal or greater amperage output.

Given that the output voltage is not standard, though, I suspect the adapter plays a role in charging that might not be replicated by a generic adapter.

Your best bet may simply be to repair the adapter. Consider disassembling it, taking pictures, and posting here to ask for help with troubleshooting steps.

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(simplified terms ahead)

Devices operate on voltage. Your replacement power supply MUST be rated for the same voltage. Going too low can cause the device to not work (or worse, not work some of the time!) whereas going too high on the voltage could cause overheating or damage to electrical components.

When the device is operating at its rated voltage, it has been measured by the manufacturer to "request" a maximum amount of amperage from the power supply. It's akin to going to a gas station and filling your car. If the tank only holds 10 gallons, it doesn't matter that the underground reservoir holds 3000 gallons. Your car only needed 10, so it only took 10. Amperage works the same way. If the device is rated to draw 1A at 12 volts, it will try its hardest to pull 1A (and no more) at 12V.

It's impossible for a power supply to supply too much current to a device, assuming the device is working correctly. If the original power adapter was 1A, and the replacement is 3A, then the device will draw 1A from it which is less than the maximum the supply can put out. If the replacement supply is too small (say, 0.5A), the device will still try to pull 1A. The supply is too small and may destroy itself or catch fire trying to supply the current that the device wants.

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  • \$\begingroup\$ Nice way of explaining how a device only draws the current it needs, not what the supply is capable of. \$\endgroup\$ – Olin Lathrop Jun 28 '12 at 21:22
  • \$\begingroup\$ @OlinLathrop: thank you :) I'm tempted to edit the OP as well to make it more generic. I've seen this question way too many times now. \$\endgroup\$ – Bryan Boettcher Jun 28 '12 at 21:24
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If the old supply put out 8 V and 2.6 A and the equipment says it wants 8 V, then the obvious answer is a power supply that puts out 8 V and 2.6 or more amps. The equipment might be OK with 9 V, or not. It could operate mostly fine, but in some corner case overheat, suddenly stop operating, or vanish into a greasy black mushroom cloud.

You are right in that 8 V is not a standard value. However, many OEM power supplies can be tweaked a few 10s of percent from their nominal value with a trimpot or something. Manufacturers know that lots of different voltages are required but can't stock a different supply for every weird voltage a customer wants. Take a careful look at data sheets of some "9 V" OEM supplies, and you will probably find most of them can be convinced to do 8 V.

I just wrote a detailed answer about this general topic.

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    \$\begingroup\$ A general tip you are given when being trained in giving presentations is to never use the word obvious, we all let it slip in, but to those that it is not obvious it is belittling and reduces how receptive they are to your advice. That is non-obvious to the general public, only to us. I agree with the answer on the whole, but someone pointed that out to me once and I often find it hard to keep myself from using the word obvious, almost as a way of saying, this is basic enough that explaining the details is not required. \$\endgroup\$ – Kortuk Jun 28 '12 at 19:23
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Given that you stated that it charges a 6V battery, I am guessing it does this directly or almost directly (i.e. through a diode). Given this, higher voltages may cause the battery to die prematurely.

You should be able to find an adjustable voltage supply, although 2.6 Amps is relatively high current as such things go (i.e. expensive).

You should replace it with a power supply rated at or above 2.6 amps and no more than 8v if you want to avoid damaging the battery.

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If the battery is a Ni-MH or Ni-Cd pack, it is probably charged with a resistor and diode in series, which will slowly charge a battery at a fairly constant current.

An increase in voltage would increase charge current slightly, although probably not enough to matter. In the worst case, you may be able to use one or two series 1N5400 diodes to drop about 1V.

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