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#1 |
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diyAudio Member
Join Date: Aug 2007
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Hello
I have a condensator with this information on it: 105J100 ![]() can someone can explain me what is the value please? For me it's a 105nF +-5% 100vdc is it right ? the condensator is in an amplifier classD thanks a lot in advance
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#2 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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the capacitor value is 10 with five zeros added after the ten giving 1000000.
This is the value in pF i.e. 1000000pF To help us read this and remove errors in typing we shorten it to 1000nF. It can be restated as 1uF. I much prefer to use the shortest form when describing values. The J is tolerance and as you correctly identified shows +-5% the final 100 is the rated voltage. Almost certainly 100Vdc. some will add dc or ac after the voltage rating helping to avoid confusion, but as in this case you must assume the safer value of working at <=100Vdc.
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regards Andrew T. |
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#3 |
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diyAudio Member
Join Date: Aug 2007
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ok Thanks a lot AndrewT now i understand
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