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Author Topic:  What is the proper way to measure pickup resistance?
Jack Hanson


From:
San Luis Valley, USA
Post  Posted 22 Apr 2014 3:18 pm    
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I have measured the resistance of the pickups on all of my electric instruments by inserting one of these
into the output jack, turning the pot(s) full on, and measuring with the meter probes at the tip and the sleeve. How accurate is my method? Would it produce an appreciably higher or lower resistance reading?

Is the preferred method of obtaining a totally accurate reading only doable by unsoldering the pickup leads? Or, is my method close enough to be reliable in any knowledgeable discussion on pickup resistance.
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Craig Baker


From:
Eatonton, Georgia, USA - R.I.P.
Post  Posted 22 Apr 2014 4:33 pm    
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Jack,
Your method will show the pickup's resistance in parallel with the instrument's volume control pot. The reading you get will depend on the resistance of the volume pot, but because of the parallel connection, will always be less than the pickup's resistance.

Sincerely,
Craig Baker 706-485-8792

cmbakerelectronics@gmail.com

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P.O. Box 3965
Eatonton, GA 31024
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Stephen Cowell


From:
Round Rock, Texas, USA
Post  Posted 22 Apr 2014 6:58 pm    
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Let's do a little math...

Formula for parallel resistance is (x*y)/(x+y)
Let's try 250K for the pot and 10K for the pickup... that would be (in K's) 2500/260 or 9.16K.

Percent error: error/true *100.
Error is .84K... /10K gives .084, *100 gives 8.4% error.

So your method is plenty good enough for telling whether a pickup is open or not... you'll get within about 10%. No need to get inside and unsolder anything.
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Jack Hanson


From:
San Luis Valley, USA
Post  Posted 27 Apr 2014 9:08 am    
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Thanks for your knowledgeable replies, Craig and Stephen.

In summation, would it be relatively safe to assume that, using my method described above, a closer estimate could be achieved by adding 10% to the meter reading?

For example, my early '50s Fender Champion measured 5.7K, and my late '40s Gibson Century measured 7.3K. Would 6.3K and 8K, respectively, be more realistic figures, or would that be a gross oversimplification?
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Donny Hinson

 

From:
Glen Burnie, Md. U.S.A.
Post  Posted 1 May 2014 2:34 pm    
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That's not an oversimplification, just a false conclusion. For one thing, the lower the resistance or inductance of the pickup, the less it will be affected by the parallel effect of the pot. Also, like a lot of players, you're assuming that if the resistance or inductance is the same, then that pickup will sound the same, and that's simply not the case. There are other factors that enter into what creates a pickups sound, and you can't just ignore them.
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Doug Beaumier


From:
Northampton, MA
Post  Posted 1 May 2014 2:59 pm    
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I wouldn't unsolder anything to measure a pickup's resistance, especially on a vintage steel guitar. Just take the measurement off the output jack. That should be close enough IMO.
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Daryl Thisdelle

 

From:
New Brunswick, Canada
Post  Posted 3 May 2014 11:41 am     Pick Up Resistance
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The only way I ever measured it was once me and the better half were in the music store and I wanted to buy a new pick up and had it on the counter ready to pay for it and the little women came by and grabbed it and put it back in the salesman hand and said , over my dead body. Now that is what I call pick up resistance...

Daryl
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Dickie Whitley

 

Post  Posted 3 May 2014 11:52 am    
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Daryl, I would totally agree...
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Paul Arntson


From:
Washington, USA
Post  Posted 6 May 2014 6:43 am    
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number of slaps
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Tony Glassman


From:
The Great Northwest
Post  Posted 6 May 2014 7:26 am    
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Push really hard on the pick-up and measure how hard it pushes back by using a strain gauge. That should determine it's resistance. Rolling Eyes
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Paul Arntson


From:
Washington, USA
Post  Posted 6 May 2014 8:33 am    
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Seriously, what Stephen said. You can even do a little more algebra and back out the exact value of the pickup if you know the potentiometer value.
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