Dont forget only half the of the pot element is used, the other half is bridged by the brass, so I would think the highest value would be best if you wanted to end up with a 10k log pot using only half the element.
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So far as I can make out from the schematic the pot's value is given as 10k, I take that to mean the full track value ie: end to end value.
You could experiment first if you have a spare lin pot hanging around, mark it at 180 degrees rotation then try it with different values of taper resistor, I modified a 50k lin pot for anti log it was for a screen grid controlled regen det (wiper to the valves screen grid) One end of the pot was to chassis the other end was to HT via a limiting resistor, the taper resistor was connected between the wiper and the end that was connected to the limiting resistor, in the end I settled for a 6.8K taper resistor so approx 7 to 1 ratio. It worked very well. Lawrence.
I hadn't taken on board the brass plate.
So we're looking at about 180° max travel. The 25k pot with a 27k from slider to the end of the track will give you 10k max but the law will be a lot smoother than in the previous graph. I'll tweak it and post it again. A 50k pot and 15k resistor would be better but if you cant find one, you have to do the best with what you've got! EDIT Having looked at the circuit to see where the 'hot' end of the pot goes, I see it is to the GB battery, so its actual value doesn't really matter. A 6k8 resistor and 25k pot will give a similar law to the 50k/15k plot above..
14-02-2017, 08:27 PM
Thanks Terry.
By the time I have fitted the new track into the pot only half the track will be effective so the actual value of the pot will be about 12.5k. Am I right in thinking that minimum volume was intended to be when the pot wiper is at the hot end (more -ve bias) and the condenser is at minimum capacity. I guess the 25k pot and 6.8k resistor combination would just consume a little more current from the bias battery. Please can you just confirm where the resistor would go, is it between wiper and hot end? thanks Mike
14-02-2017, 08:43 PM
VI is a vari mu so more -ve = less gain.
Less regen capacitance = less gain for V2 Lawrence.
thanks Lawrence, just as I thought.
Mike
14-02-2017, 09:11 PM
Looking at the picture of the resistance former removed from the pot with the resistance wire wrapped around it (some of it broken) Which end of that former (as per picture) was connected to the grid bias battery, the near end or the far end?
Just trying to get the full picture in my head as to what's what. Lawrence.
Hi Lawrence
The close wound end which is on the left is connected to GB2, remember the lower half as seen in that photo has no winding on it as it is bridged by a curved brass plate. Do you think the fact that the resistance wire was spaced widely contributed to its failure. I have found this online and have emailed Blore Edwards to see if they can advise. https://media.wix.com/ugd/1636bd_6a64f4c...b51537.pdf thanks Mike
14-02-2017, 09:37 PM
(14-02-2017, 08:27 PM)Crackle Wrote: Am I right in thinking that minimum volume was intended to be when the pot wiper is at the hot end (more -ve bias) and the condenser is at minimum capacity. Well, you will know how the pot was wired but, yes, minimum capacity = minimum reaction = minimum volume. (14-02-2017, 08:27 PM)Crackle Wrote: I guess the 25k pot and 6.8k resistor combination would just consume a little more current from the bias battery. Yes. Because the resistor shunts the top end of the track it plays the greatest part in determining the value and rate of change of that section whereas the value of the bottom section will increase normally and, therefore, more rapidly. With those values, the value of the bottom section will increase from zero to 17% of maximum in the first 10° of rotation; by 13% in the next 10° and 11% in the next. The rate of increase will continue to fall with further rotation: 9%,7%, 5% dropping to an average of 2% per 10° for the last 50° of travel.
14-02-2017, 09:50 PM
Hi Terry
I am being very dim here, but which is the bottom end of the track and the top end? Mike |
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