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It says 100J and is much bigger than the other components.

component

It has a resistance of 2ohm when the circuit is without power, and 15 ohm when I attach power to the circuit. It has a 1nf capacitance. So it might be a capacitor, but why is it bigger than the other capacitors, and could it replaced by any 1nf capacitor?

user220980
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  • Could be an inductor – Colin May 02 '19 at 11:04
  • https://electronics.stackexchange.com/questions/334128/how-do-i-identify-smd-components-or-how-do-i-identify-any-component – dim May 02 '19 at 11:08
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    Possibly a transient suppressor. – Peter Smith May 02 '19 at 11:11
  • Thanks for the link – user220980 May 02 '19 at 11:11
  • When you say that it has no resistance, do you mean that it has 0 ohms resistance, or that it is showing open circuit on your multimeter? – HandyHowie May 02 '19 at 11:30
  • "No resistance." Do you mean zero ohms, or do you mean "open circuit?" – JRE May 02 '19 at 11:30
  • @JRE Just beat you :) – HandyHowie May 02 '19 at 11:31
  • I mean it shows 0 ohm on my multimeter, but more precisely it shows a 2 ohm resistance. – user220980 May 02 '19 at 11:32
  • You should update your question with the 2 ohms information. It will make it clearer. – HandyHowie May 02 '19 at 11:33
  • Very likely it is an inductor then. Next time also include information which seems irrelevant to you like, what is this PCB? Where do the wires go. Since I think I see a microphone at the left, the black and red wires could be the supply, then maybe the gray wire is an antenna. Trimmer on the PCB and only a few transistors: probably an FM audio transmitter. The inductor will then be connected between the antenna and ground (black wire) so make a resonator at the output with the antenna. – Bimpelrekkie May 02 '19 at 11:46
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    *and 15 ohm when I attach power to the circuit. It has a 1nf capacitance* **Don't** measure resistance in a powered circuit, you can **damage** your multimeter. Always power off, then measure. The 1 nF means nothing, at this low resistance the capacitance cannot be measured properly. – Bimpelrekkie May 02 '19 at 11:48
  • Thanks a lot everyone. I've learned a lot today. This was my first question on stackexchange, I'll be more clear and precise next time as well. – user220980 May 02 '19 at 11:51

1 Answers1

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100J and a short circuit would be an inductor.

100J is a 10 microhenry inductor. J means it is accurate to 5%

This site explains how to read inductor codes.

It also lists all the precision codes.

Short form:

First 2 digits is the value. The third digit is how many zeroes to put on the end. Final value is in microhenries.

So, 100 means value 10, add no zeroes to the end. So, final value 10 microhenries.

It seems to be OK. Inductors fail by becoming open circuits.

Don't replace it. Look elsewhere for whatever problem made you suspect it.

Do not replace it with a wire. It appears to be part of an RF amplifier. If you put in a piece of wire, you will kill the output transistor.

Do not replace it with a capacitor. The output amplifier won't work then.

In response to comments:

NEVER try to measure resistance with the power on.

Since you say it is only a short circuit when the power is on, it looks like it is dead after all.

You need a 10 microhenry inductor to replace it.

Be careful when playing with that board. Touching the gray wire to ground will kill the inductor - that shorts the power supply to ground through the inductor.

JRE
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  • This would make sense as it is short circuited. Interestingly it's only short circuited when the power is on. – user220980 May 02 '19 at 11:44
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    @user220980 -- measuring resistance with the power on is pretty much meaningless. The meter looks at the voltage drop across the component when it's powered by the meter's battery. Adding another voltage source skews the reading. – Pete Becker May 02 '19 at 11:47
  • Thank you. Very good to know! – user220980 May 02 '19 at 11:55
  • It's a cheap wireless microphone, which the OP might have been kind enough to mention. I doubt if anything would be damaged by shorting the inductor, but it would eliminate any useful RF output at the antenna. – Dave Tweed May 02 '19 at 12:31