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Philco 41-kr chassis questions - Printable Version

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Philco 41-kr chassis questions - OldRestorer - 06-07-2014

I am working on re-capping this tiny, tight, cramped chassis and one if the caps (.25) is wrapped in wire and then connected.

Why and do I repeat this?

http://www.nostalgiaair.org/PagesByModel/468/M0013468.pdf


RE: Philco 41-kr chassis questions - Radioroslyn - 06-07-2014

No you don't have to. It's a wave trap for the IF stage.
Terry


RE: Philco 41-kr chassis questions - morzh - 06-07-2014

The cap is AC link between the Chassis and the Negative Common.
It is present in all AC/DC sets. Without it you will hear ghosts, motorboating and howling.
Why a choke....I do not know for sure but as this is a series resonant tank they want zero impedance between the two commons at a certain frequency. Why, don't know, some sort of noise reduction I guess.

Could be a wavetrap as Terry said. Although a cap is kinda large, but then maybe the choke is real smll.


RE: Philco 41-kr chassis questions - Ron Ramirez - 06-07-2014

This page from Chuck's site explains it very well.

http://www.philcorepairbench.com/tips/svctip37.htm


RE: Philco 41-kr chassis questions - Arran - 06-08-2014

Some people remove it entirely, others like myself melt out the capacitor tube and rewind it with a more modern wire of the same gauge. Given Philco's knack for cost cutting they must have been needed at some point. The capacitor needs to be there, the coil probably doesn't, the coil is in series with the capacitor and is used to couple the chassis to B-, it blocks AC voltage but allows RF to pass.
Regards
Arran


RE: Philco 41-kr chassis questions - OldRestorer - 06-08-2014

Thanks guys,
I think I will replace it and wrap it like original


RE: Philco 41-kr chassis questions - morzh - 06-08-2014

I am not fully sure about the explanation on Chuck's site, but I think the representation of the equivalent circuit for the displaced lead is off.
It has no practical ramifications or Kirk, so, Kirk, you may disregard this post.

However, speaking of the circuit, by the same token (1/3 and 2/3 inductances in parallel serialized with the full cap value) if we keep the cap with the lead barely displaced (say 1/100 and 99/100 length division), then we will have reduced the inductance by 99%, creating practically no ESL cap. Better yet, no displacement at all. Leave the pin at the beginning.

I think, a still crude equivalent should have each L (1/3 and 2/3) inuctor connected with the respective partial cap in series and then two of them paralleled. This will take some calcs to figure the resonances and impednces, but will be closer to the truth.