# Power Supply Limits? Do & Don't

Discussion in 'Power Electronics' started by KWXYZ, Apr 14, 2021.

1. ### KWXYZ

12
2
Jan 25, 2021
Howdy I've learned a lot from posting question and getting answers from the Experts!

So AC/DC Power supply 101 (Wall Warts) 9V, !2V, 6V, etc.

So I understand the V need to be exact, do not exceed Voltage.
What about ma what is the threshold for Amperage for devices?

I guess what I"m asking lots of older equipment have various requirements.
Not all new power supplies are made for these tolerances. so you tend to buy used.

Is 10% or 20% of the device stated requirement safe with out frying it.

Thanks.

2. ### Nanren888

469
135
Nov 8, 2015
For most things it tends to work like this.

Needs to match in voltage. The power supply needs to be able to supply what the device needs or more.

The output can be AC or DC - you want to get this matching. A device made to work on one won't like the other.
The voltage is important as you said. Get it just right. Some devices that you might want to power will have a range of voltages that they are happy enough with. If so it will say. If it just says one voltage, get a power supply with that voltage.
.
The current, A, mA, amperage of the power supply should match or be bigger than the one on the device you want to power.
This one is a little more complicated, but basically, at a fixed voltage, the device being powered will draw, pull, use, a certain current and the power supply needs to be able to supply that, or more. If the device being powered does not need more, it won't draw more.

So a power supply 5volts, DC, 300mA is fine for a device marked 5 volts, DC, 300mA (max).

So a power supply 5volts, DC, 2A is fine for a device marked 5 volts, DC, 300mA (max).

So a power supply 5volts, DC, 1A is NOT fine for a device marked 5 volts, DC, 2A. The power supply will be stressed and fail immediately or later, possibly with heat, flame, destruction of property &c.

Sometimes things are rated by power rather than current (I, mA). Then you can use P = I * V watts = volts times Amps.
Again. voltage match and power supply of greater power capability than the device will draw.
.
Most supplies nowadays are "regulated" meaning that the voltage output is steady, controlled, accurate. Sometimes you get one "unregulated" which means that the voltage will be somewhere near the stated value, or perhaps vary as the load varies. So it makes it a little difficult to judge the voltage match, but he same rules apply, so you don't want a supply that goes higher than the device likes & don't want one that can't supply the current the device needs.
.

Hope it helps. Anything unclear, ask more. People will be happy to help & fill in anything that I left out.

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3. ### Martaine2005

3,328
910
May 12, 2015
Just to add to @Nanren888 's advice, one obvious thing that is regularly overlooked is polarity. Always check polarity!.
Also, as a rule of thumb, don't exceed 75% of the adapters power rating.

One very annoying factor when using adapters (walwart) is cable exit location. Sometimes you need two or three adapters plugged in near you on the same extension strip. The cable exiting the adapter doesn't allow for another to be plugged in next to it.
Sounds silly but very annoying sometimes.

Martin

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4. ### KWXYZ

12
2
Jan 25, 2021
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5. ### Martaine2005

3,328
910
May 12, 2015
Anything you are not clear about?
BTW, that's not a bad article you linked to.

Martin

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6. ### KWXYZ

12
2
Jan 25, 2021
Hey Thanks,
Yeah,

So Nanren888; posted Sometimes things are rated by power rather than current (I, mA). Then you can use P = I * V watts = volts times Amps. Again. voltage match and power supply of greater power capability than the device will draw.

I'm a little confused about the PS Wattage? Since it's in the input stage a la, 12W, 24W, 15W etc. how does this affect the performance of said device. So if it's an exact match for the product ( let say it came with it new) no problem, what if you get a PS that is same
V and mA but the W is higher than called for is there a problem? Or is that only for the output stgage?

Thanks.

7. ### hevans1944Hop - AC8NS

4,595
2,149
Jun 21, 2012
A specific example of such a power supply would be helpful, but if the voltage doesn't change under the rated mA load, it makes no difference what the wattage rating of the PS is. Of course no power supply will have a wattage rating less than the product of its maximum voltage and maximum current, but some (inefficient) power supplies might require considerably more power or watts than the PS is capable of delivering. Again, no harm, no foul, if you stay within the rated voltage and current outputs.

A third possibility is a PS whose voltage output "sags" or decreases as the load current increases. Many wall-warts exhibit this behavior, and most of them should be avoided because, un-loaded, their output voltage is higher than their loaded output voltage. This over-voltage, which may decrease to a normal level as the connected device begins to draw current, can damage the electronic device to which such a wall-wart is connected. At the very least, the over-voltage represents an undesirable electrical stress that should be avoided.

I do use over-voltage producing wall-warts from my collection of fine, and antique, junque wall-warts, but then I mostly know WTF I am doing at the time.

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8. ### KWXYZ

12
2
Jan 25, 2021
Ex. 2 power supplies #1 has Input 0.19 A #2 has input of 12 W ?

but thanks I'm clear as to what to do to keep it safe.

9. ### Martaine2005

3,328
910
May 12, 2015
I am a little confused by this.
What are the voltages? eg: 63V x 0.19A = 12 W (almost). And 12W / 63V = 0.19A
Use the same calculations with both of your power supplies to find the difference.
That is of course if I understand you correctly.

Martin

10. ### KWXYZ

12
2
Jan 25, 2021
Hey Martaine2005,
sorry for the confusion maybe this will show.

https://imgur.com/TxUbpA3

2 transformers 1 has W listed the other does not.
Thanks.

11. ### Harald KappModeratorModerator

11,294
2,586
Nov 17, 2011
The relevant input rating is mains voltage, AC 1^20 V in both cases.
The relevant output ratings are 12 V, identical for both and current, which is 1 A and 0.8 A 800 mA) respectively. "+" is tip, "-" is sleeve, identical for boh.
As long as the load is rated 12 V and does not require more than 0.8 A, both wall warts can be used.

Watts can be calculated from P = V × I (power in Watt equals voltage in Volt times current in Ampere). There's no need to write the wattage down on the power supply if the other two parameters are given.

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