# How to determine maximum load?

#### Chillum

Joined Nov 13, 2014
546
I bought cheap (http://www.dx.com/p/universal-eu-plug-dual-usb-car-charger-for-iphone-ipad-ipod-5v-2a-238142) 2.80USD ea for to be a power supply for 5v boards, now I wanna see if they deliver, they say 2000mA, but my phone max out at 1000mA (I DO know that load determines current and not theoretical maximum). Now before I overload and fuse it by shorting live and ground with my multimeter, I wanna know: is there a correct way to do this without fusing the power supply?

Joined Jan 15, 2015
5,499
5 Volts @ 2.0 Amps means it should deliver 2 amps into a 2.5 Ohm load. So I would series 3 one ohm resistors in series and measure the voltage across the resistors. If the voltage maintains at 5 volts into the 3 Ohm load you know the current should be about 1.666 amps less any lead resistance. That or make a good 2.5 Ohm load and drive it measuring the voltage at the load. You want to test a power supply, then load it, do not just short it using a meter on a high current range.

Ron

#### Papabravo

Joined Feb 24, 2006
13,921
The precision way to do this is with a variable load. This will let you determine the point at which the supply drops out of regulation. What should happen is that as the load is reduced the output voltage should drop to key the power output of 10 watts. So for 2.5 Amps you should be at 4 volts and so on. It really depends on the regulator inside the black box.

#### Chillum

Joined Nov 13, 2014
546
got lots of smoke, what wattage? P=V*I=5*2=10watts, don't have 10W resistors

#### Chillum

Joined Nov 13, 2014
546
You're in luck. They are readily available from Digi-Key and other distributors.
These for example
https://www.seielect.com/catalog/SEI-KAL.pdf
unfortuantely not, I live in Cape Town, South Africa and suppliers here don't have the types of parts you could easily get elsewhere in the world... I'm going to check my suppliers www.fort777.co.za <-sucks: have minimal part selection; www.mantech.co.za <-- have most parts I want, and then there's widest range http://www.communica.co.za/ which I avoid because high courier fees and pertrol to get there is about the same, I'd have to have an order of R1000 or more at communica to make it worth my while (1x 0.1uF ceramic disc cost 0.33ZAR)

#### Chillum

Joined Nov 13, 2014
546

#### Papabravo

Joined Feb 24, 2006
13,921
Well you got the idea. You look around and use what you've got. Sorry about the smoke, but we've all done it.

#### Chillum

Joined Nov 13, 2014
546
can you use a breadboard to connect these parts, and will normal jumper wire handle 5W, sorry about the relatively noob questions, but when in doubt, ask for help!

#### Chillum

Joined Nov 13, 2014
546
what does 270E mean?

#### bertus

Joined Apr 5, 2008
20,566
Hello,

That could be 270 Ω.
Some schematic use the E for 10^0 and K for 10^3 and M for 10^6.

Bertus

#### Chillum

Joined Nov 13, 2014
546
0E27 would be 0.27R? I need 2.6R? or 2E6?

#### Chillum

Joined Nov 13, 2014
546
5 WATT ROUND METAL OXIDE RESISTOR / 2E7 OHMS
looks like the best I can get, will there be major problems with this one, except for the always want an error buffer problem (10W for 5W application)

#### AnalogKid

Joined Aug 1, 2013
8,487
If you're only option is 5 W parts, get four 10.0 ohm resistors and wire them in parallel. You could do it with two 5 ohm parts in parallel, but 5 W resistors dissipating the full 5 W will raise blisters if you touch it. Spread the heat around a little, and your parts and fingers will survive. Plus, you now have four load levels.

ak

#### Chillum

Joined Nov 13, 2014
546
If you're only option is 5 W parts, get four 10.0 ohm resistors and wire them in parallel. You could do it with two 5 ohm parts in parallel, but 5 W resistors dissipating the full 5 W will raise blisters if you touch it. Spread the heat around a little, and your parts and fingers will survive. Plus, you now have four load levels.

ak
ok, but then I'm getting 10W 10Rs, price still very low

#### Chillum

Joined Nov 13, 2014
546

#### Papabravo

Joined Feb 24, 2006
13,921
The wires are not subject to the 5 watts. A short piece of wire will have a resistance of a few milliohms. Two amps squared is 4 times 10 milliohms is 40 milliwatts. No problem for lets say a 20 guage hookup wire.