USB Adapter seems to supply 0A current?

Other Hardware General Electronics
1

Hello,

I want to use a USB adapter and USB cable to supply power to an LED strip.

I am using an AC to USB adapter that provides 5V, 1A on one USB port, 2.1A on another USB port.

adapter
adapter1597×1400 244 KB

I have cut open a USB cable and exposed the inside wires, which include a red, black, white, green, and an uncovered copper wire.

USB cable
USB cable1024×768 133 KB

For power (I don't need data over USB), I think I just need the red and black wires, as shown in this video.

To verify things, I connected multimeter red probe to red wire, black probe to black wire. I measured about 5V voltage as expected.

voltage
voltage1024×768 310 KB

However, with the same multimeter circuit, I measured 0A where I expected 1A.

current
current1024×768 297 KB

What am I doing wrong?

2

You only measure current by placing the amp meter in series with an actual load.

It appears you are shorting out the supply by connecting the amp meter to the power supply.

Use common and 10A terminals.
The internal fuse in the DMM may have burned out.

3

Plug the red lead into the 10A socket when using the 10A range? I would also put a low value resistor in series rather than a short circuit into the meter.

4

The power supply might have "fold back" current limiting.

5

It sure does!

It "folds back" all the way to zero. :grimacing:

Precisely as it should. :+1: With a bit of luck it may not have blown the fuse.

The fuse was probably not in circuit anyway on the 10 Amp range and the power supply would not actually be shorted.

OK, you seriously have a lot to learn about electricity. :astonished:

And multimeters. :roll_eyes:

Perhaps research "How to use a multimeter". :sunglasses:

6

Hi,
As explained you are shorting the 5V 1A supply out.
To test if it can cut the mustard, you need to measure the current with a load that passes 1A

Using Ohms Law;

V = I * R
re-arrange
R = V / I
So;
R = 5 / 1 = 5 Ohms
You need a 5 ohms load.
Power dissipation.
P = V * I
P = 5 * 1 = 5 Watts

So a 5 Ohm 5 W resistor would do as a load, a lower wattage may be okay if you only apply power for a short time.

Use Ohms Law to calculate for lower or high current loads.

Tom... :grinning: :+1: :coffee: :australia:

Paul_B 7


8

Thank you for the explanation. However, I don't understand why this load is necessary. I thought the 1A on the adapter means it will output at most 1A, so if there is no load, it should just be at 1A?

9

Hi,

You need to understand, the power supply puts out a potential of 5V, the current is dependent on the load.
The load provides the circuit for the current to flow.

As in Ohms Law.

If no load (resistance infinit) the current out of the power supply will be zero, as there is not circuit for the current to flow in.

If you short the power supply output with zero Ohms, you have a circuit, but the current that would theoretically flow is;

I = V / I = 5 / 0 which is not mathematically workable, but assume infinite current.
The power supply cannot supply this amount of current, so it shuts down to protect itself.
(Not all supplies have this facility )

If you have a load of 10 Ohms.
The current out of the power supply will be;

I = V / R = 5 / 10 = 0.5A, which the powersupply can produce.

I hope that helps, just finishing first coffee of the day.

Tom... :grinning: :+1: :coffee: :australia:

10

Usually, a generic power supply rated for 1A means that above 1A you will start to lose voltage regulation, and/or it will shut down either via semiconductor magic or by virtual of having a fuse of some kind. Most USB-style supplies are likely to a "polyfuse" a type of fuse that will open when there is too much current, and then close again after cooling off.

A supply that will limit the current to a particular value into a near-short circuit (such as a meter in AMP mode across its output) is usually referred to as a "lab constant-voltage constant-current dc power supply" - it was much different output circuitry.

11

See #5 :rofl: :

That is the salient point. Current only flows when there is a circuit for it to flow through.

The significance of Ohm's Law is that how much current flows depends on how easily it is for the current to flow, which is inverse to how much resistance the circuit poses.

And note - the current rating of a power supply is not how much current it will be able to provide to a circuit, but how much it can safely provide.

If you had connected your multimeter in such a way that it could measure the short-circuit current of the power supply which according to the picture you fortunately did not (does the meter still work? Did you change the meter setting with the power supply turned off?) then it might have temporarily indicated a much higher current - or most likely the power supply would have simply and safely shut down.