Please help me identify this component

[iandusud]

Hi I'm new here. I want to replace a burnt out component on a rectifier circuit on a 24V charger for a cordless drill. The component in question is in series on the +ve output. I suspect it is a fuse but don't know. Attached is a photo. Any help very much appreciated.

Ian

[alanb]

It looks like a metal film resistor that has lost its covering. Have you checked it for continuity? Is it open circuit?

[iandusud]

Hi and thanks for your reply. There's no continuity. I did wonder if it was a resistor to regulate the current?? If it is how will I identify the value???

Cheers, Ian

[alanb]

The break may be under one of the end caps rather than on the track that you can see. If you have a multimeter you may be able to find the value by measuring between two points on the resistive track. If you measure near the ends you should be fairly close to the correct value.

[iandusud]

I've done as you suggest but I can only get a reading part-way along the track. However I've measured the current across the output at 2.5A. The output for the charger is rated at 200-240mA so can we come up with a suitable value from that info? My schoolboy physics tells me that R=V/I so if V is 25 (that's the reading I get) and I is .240 the we get a value of about 100 Ohms. Does this sound right?

[Zero999]

I've done as you suggest but I can only get a reading part-way along the track. However I've measured the current across the output at 2.5A. The output for the charger is rated at 200-240mA so can we come up with a suitable value from that info?

Oh no, you didn't just put the meter on the current setting and stick it across the output did you? In that case you've measured the short circuit current, not the charging current. You're lucky the internal resistance was high enough to limit the current otherwise it would've blown the fuse inside your meter. 

If it's just a rectifier and transformer type of charger, the more sensible approach is to measure the open circuit voltage, then work out the required resistor value to put 240mA into a discharged battery.

[iandusud]

Hi Hero999, Don't worry I set my meter can handle 10A! I've edited my last post. The voltage of the output measures 25V. Thanks for your help. Ian

[Zero999]

Hi Hero999, Don't worry I set my meter can handle 10A! I've edited my last post. The voltage of the output measures 25V. Thanks for your help. Ian

The trouble at the time of measuring the short circuit current, you didn't know how big it could be. 2.5A flowed because the power supply has an internal resistance of 10 Ohms but what if it had an internal resistance of 1 Ohm? Another problem is the power dissipation. When the power supply was short circuited, it would have dissipated 62.5W which would have caused it to overheat if it had been short circuited for long enough. It's best to avoid short circuits whenever possible and not connect your meter in current mode across a voltage power supply.

I've done as you suggest but I can only get a reading part-way along the track. However I've measured the current across the output at 2.5A. The output for the charger is rated at 200-240mA so can we come up with a suitable value from that info? My schoolboy physics tells me that R=V/I so if V is 25 (that's the reading I get) and I is .240 the we get a value of about 100 Ohms. Does this sound right?

100Ohms sounds sensible but it depends on the voltage of a typical flat battery. If the battery voltage is above 0V (it should be otherwise it's completely dead) then the current will be much lower, still too low is much safer than too high.

[alanb]

I've done as you suggest but I can only get a reading part-way along the track.

Can you measure from one end to the half way point then double the reading? If so is this consistent with your calculated value?

[iandusud]

I can't get a consistant reading but get a value of anything from 25-60 Ohms over approx half the length of the track. I think I'll try a 100 Ohm resistor and see what happens. What sort of wattage should I use?

[alanb]

using Ohms law again if the voltage drop across the resistor is 25 Volts on a 100 ohm resistor the dissipation will be 25x25/100 = 6.25W. alternatively 240ma through 100 ohms gives a dissipation of .24x.24 x100 = 5.76w. I would go for the next size above 5 watt which is probably 10 watt. Hope this helps

[iandusud]

Hi alanb, thanks for your thoughts. It has been pointed out to me elsewhere that when the charger is connected up to a battery then the voltage across the resistor is in fact 25V - the voltage of the battery (say 20V) = 5V. This changes things considerably and explains why the original resistor is not a big wire wound one. 5V²/100ohms = 0.6125W. Of course as I have a completely dead battery reading 0V the the voltage would have been the full 25V hence the burnt out resistor. Anyway I have now fitted a 100ohm 3W resistor and with the one healthy battery I have it is working fine. I'm now trying to coax the "dead" battery back to life.