0
\$\begingroup\$

I've got an old Sony Ericsson with a charger that's sustained some damage. The cord has been melted and flattened in one place. Neither the charger nor phone has been used in a long time. The battery for the phone is not charged and at right now the charger doesn't seem to charge the phone.

Using a multimeter, I tested the voltage on the pins of the charger. I get the correct 5V readout. This leads me to believe the charger still works. However, when I try to read the amperage from one pin to the other I always get zero. Can anyone tell me why this might be?

\$\endgroup\$
  • \$\begingroup\$ How are you measuring the current? With at least a 10Ω resistor in series? \$\endgroup\$ – Samuel Dec 13 '12 at 21:17
  • 3
    \$\begingroup\$ You've probably blown the fuse in your ammeter. That isn't how you're supposed to do it. \$\endgroup\$ – Dave Tweed Dec 13 '12 at 21:18
  • \$\begingroup\$ @Samuel: Ah yes. That would probably help wouldn't it. It's been a while... \$\endgroup\$ – golmschenk Dec 13 '12 at 21:27
  • \$\begingroup\$ Please be aware that some chargers/charging systems are designed to use a constant current source/charger. That has an implication on how you calculate things and what can be considered as a healthy range of values and response expected of the device in test. You might wonder why the current remains the same despite varying resistive loads. As always, refer to the specific datasheet for guidance. \$\endgroup\$ – shimofuri Dec 14 '12 at 10:38
4
\$\begingroup\$

You probably blew out the fuse on your ammeter. An ideal ammeter has zero resistance, so when you measure current you need a load.

Poorly made picture:

bad picture

Notice the light-bulb providing a load. Try connecting a 100Ω resistor in series when you test, if you're getting 5V out you should get 50 mA. You can try lower resistances, but you'll want to ensure that the resistor you use has a high enough power rating. If you use a 100Ω I'd go with a 1/2 W resistor to be safe. If you only have 1/4 watts lying around then try a larger value resistor.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.