Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => AmpTools/Tech Tips => Topic started by: PRNDL on May 01, 2009, 01:18:04 pm
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I'm wondering if it is possible to modify a socket saver for use in a bias meter.
It would involve putting in a 1 ohm resistor on one pin, and drilling a hole for wires.
Has anyone tried this?
Is it sturdy enough?
Here's a photo
(http://www.tubesandmore.com/cemirror/inv/P-ST9-900.GIF)
Thanks
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Hi
I never do that but why not?
is there enough space ?
yes ?
do it
see Bias Rite Head kit instructions here
https://taweber.powweb.com/store/instruction.jpg
and the photo here
http://taweber.powweb.com/store/bias_rite01.jpg
Kagliostro
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That's a 9-pin (preamp tube) socket saver.
You don't often see octal socket-savers, though I'm sure they exist. It's just as easy to buy the plug and socket, and make your own. Pre-made socket savers are useful if you have a tube tester (or other situation where you'd be inserting and removing a tube hundreds of times), and they are generally expensive. I wouldn't waste the 2 I have turning them into a bias tool.
But I would spend the few bucks to make my own bias tool from a socket and plug.
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Thanks Kagliostro and HotBlues
Actually, my bias meter is having problems with the 9-pin circuit board pins - they don't seem to be sturdy enough for the job.
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I see...
Why use the bias meter on the 9-pin tubes? Or do you have a check you want to perform without opening the amp?
Right now I'm in one location and my tube tester and socket savers are in another. The body of those 9-pin socket savers is only about 3/4" to 1" tall (if that; I'm working from memory). I never tried to open one up, but there is a screw through the top center of the socket saver that should allow it. My fear is that the pins extending through the bottom might be welded to the socket contacts. However, I don't know the actual method of connection.
You'd probably need to use a small resistor like a 1/8w to fit inside the shell. 1/4w might fit. In the end you'd have to perform surgery to know for sure. This should not pose a problem since the current through and the voltage across the resistor (and therefore the power dissipated by the resistor) should be quite small.