Author Topic: 1N4148 and 10R resistor in parallel, any way to test diode?  (Read 2629 times)

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Offline Chris WilsonTopic starter

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1N4148 and 10R resistor in parallel, any way to test diode?
« on: December 15, 2018, 09:28:51 pm »
I was wanting to check some out of circuit diodes, but they are each paralleled with a 10 ohm resistor. Is there any fairly straightforward way to test if the diode is fine without splitting it from the 10R resistor? And to check the 10R hasn't shifted much? Thanks
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Offline capt bullshot

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Re: 1N4148 and 10R resistor in parallel, any way to test diode?
« Reply #1 on: December 15, 2018, 09:39:10 pm »
You can check the 10R resistor using a common Ohmmeter, I suppose it'll apply a rather low current (like 1mA to 10mA). So the voltage across the resistor stays well below the diode forward voltage.
For testing the diode, you could apply a constant current of 100mA that would cause 1V drop across the resistor. With diode in parallel, the voltage drop will be lower, approximately the diode forward voltage at 60mA ... 70mA (0.6V ... 0.7V).
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Offline Brumby

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Re: 1N4148 and 10R resistor in parallel, any way to test diode?
« Reply #2 on: December 16, 2018, 12:00:19 am »
.... and the reverse voltage will be 1.0 V

The 1N4148 is a signal diode, not a power one, so it's important to keep the current down.  100mA is ideal.
 

Offline spec

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Re: 1N4148 and 10R resistor in parallel, any way to test diode?
« Reply #3 on: December 17, 2018, 08:29:27 am »
You can check the 10R resistor using a common Ohmmeter, I suppose it'll apply a rather low current (like 1mA to 10mA). So the voltage across the resistor stays well below the diode forward voltage.
For testing the diode, you could apply a constant current of 100mA that would cause 1V drop across the resistor. With diode in parallel, the voltage drop will be lower, approximately the diode forward voltage at 60mA ... 70mA (0.6V ... 0.7V).
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Offline Chris WilsonTopic starter

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Re: 1N4148 and 10R resistor in parallel, any way to test diode?
« Reply #4 on: December 18, 2018, 02:08:56 pm »
That works fine and will save me a lot of time, many thanks :)
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