Can u identify the value of this mica cap?..i have a bunch given to me
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help..can you tell me the value of this capacitor?
- Thread starter groovecraft
- Start date
hevans1944
Hop - AC8NS
Looks like a 165 picofarad dipped mica, 1% tolerance, 500 V rating.
Thats the million dollar question..i have 2 pounds of these...does anyone know without a doubt?thanks btw for posting...any help is great
I have a mulimeter...im a beginner..i couldnt figure it out..any help is appreciated
hevans1944
Hop - AC8NS
Either value is within the measuring range of common multimeters with a capacitance measuring function. Why not measure one? Images on the web lend credence to 1650 pF OR 165 pF.
Umm, how much do you want for two pounds of these capacitors? Just asking. I don't have any use in mind.
Umm, how much do you want for two pounds of these capacitors? Just asking. I don't have any use in mind.
hevans1944
Hop - AC8NS
If your multimeter does not measure capacitance, you could measure the capacitive reactance using s 60 Hz low voltage transformer and a resistive voltage divider with your multimeter. A 165 pF capacitor has 16 MΩ reactance at 60 Hz. A 1650 pF capacitor has 1.6 MΩ reactance at 60 Hz. If you connect ten of the "unknown" capacitors in parallel, these reactances will drop to 1.6 MΩ for ten 165 pF capacitors in parallel and to 160 kΩ for ten 1650 pF capacitors in parallel. That gets the capacitive reactance low enough that you can construct a resistive/capacitive voltage divider without the multimeter presenting an excessively low impedance during voltage measurements performed on the divider components.
So, the procedure is this: Connect a 100 kΩ resistor in series with the ten paralleled capacitors and apply low-voltage AC across the series combination. Measure the applied AC voltage and the voltage across the 100 kΩ resistor. If the "unknown" capacitor is 1650 pF the parallel combination will drop about 36% of the AC voltage across the resistor and the remaining 64% across the paralleled capacitors. If instead of 1650 pF capacitors your "unknown" capacitors are only 165 pF, then most of the AC excitation voltage will appear across the paralleled capacitors instead of across the resistor. The problem is to measure this without excessively loading the resistor and obtaining unreliable values.
A typical digital multimeter has an input impedance of one to ten megohms, so if you use the multimeter to measure the AC voltage across the 100 kΩ resistor, it will present an impedance slightly less than 100 kΩ (1 MΩ in parallel with 100 kΩ is 90909 ohms). When you measure the voltage across the 100 kΩ resistor, it will be about 36% of the AC excitation voltage if the ten "unknown" capacitors are each 1650 pF, presenting a reactance of 160 kΩ at 60 Hz AC. In other words you have a voltage divider with 90.9 kΩ resistance (which includes the multimeter input impedance) in series with 160 kΩ of capacitive reactance. The divider ratio (across the resistor) is 90.9/(160+90,9) = 0.36.
If the ten "unknown" capacitors are each 165 pF, the voltage across the 100 kΩ resistor will be much less than the AC excitation voltage and most of the excitation voltage will appear across the ten paralleled 165 pF capacitors. It should be fairly easy to determine which of these two instances is the correct one.
So, the procedure is this: Connect a 100 kΩ resistor in series with the ten paralleled capacitors and apply low-voltage AC across the series combination. Measure the applied AC voltage and the voltage across the 100 kΩ resistor. If the "unknown" capacitor is 1650 pF the parallel combination will drop about 36% of the AC voltage across the resistor and the remaining 64% across the paralleled capacitors. If instead of 1650 pF capacitors your "unknown" capacitors are only 165 pF, then most of the AC excitation voltage will appear across the paralleled capacitors instead of across the resistor. The problem is to measure this without excessively loading the resistor and obtaining unreliable values.
A typical digital multimeter has an input impedance of one to ten megohms, so if you use the multimeter to measure the AC voltage across the 100 kΩ resistor, it will present an impedance slightly less than 100 kΩ (1 MΩ in parallel with 100 kΩ is 90909 ohms). When you measure the voltage across the 100 kΩ resistor, it will be about 36% of the AC excitation voltage if the ten "unknown" capacitors are each 1650 pF, presenting a reactance of 160 kΩ at 60 Hz AC. In other words you have a voltage divider with 90.9 kΩ resistance (which includes the multimeter input impedance) in series with 160 kΩ of capacitive reactance. The divider ratio (across the resistor) is 90.9/(160+90,9) = 0.36.
If the ten "unknown" capacitors are each 165 pF, the voltage across the 100 kΩ resistor will be much less than the AC excitation voltage and most of the excitation voltage will appear across the ten paralleled 165 pF capacitors. It should be fairly easy to determine which of these two instances is the correct one.
Who..i can learn alot from this procedure...i will attempt it...ill have to reread ur post alot ...lol..im not selling the caps but if you give me ur address ill send you 25 -30 ...for taking the time to help me... these r the size of a nickel maybe a little bigger
hevans1944
Hop - AC8NS
Woo woo! That big, they probably are 1650 pF = 0.00165 μF!
These were probably intended for use as resonating or RF by-pass capacitors in some sort of power electronics equipment, like an ultrasonic cleaner maybe. Dipped micas are premium capacitors, once favored by hams who built tube-type transmitters that needed the high-voltage capability. My last tube-type transmitter was something I put together in 1966 during my Novice year in amateur radio. Today everything is solid-state and low voltage.
73 de AC8NS
Hop
For those people not in the US, it means they're a little smaller than an average wallaby dropping.
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