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When using a split plate load to lower gain into the next stage, why can't u just leave the resistor (in my case 100k) in place and just use another between the plate and the next stage instead of splitting the 100k into 2 resistors that equal 100k? You would still have to 100k plate resistor but the path to the next stage has the resistance between the next stage and previous plate just like a split load would accomplish. What am i missing? I guess theres more to it than just putting a resistance between the plate and next stage?
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A split plate resistor of say 50k+50k is a well defined voltage divider, lowering output by 50%.
A 100k build-out resistor at the plate isn't a voltage divider by itself and needs to be loaded by another 100k to ground for 50% attenuation.
In other words the attenuation depends on the input impedance of the following circuit/stage.- Own Opinions Only - -
Thanks. Thats what i figured because as discussed before a grid resistor doesn't lower gain. But then it;s after DC has been blocked and i wasn't sure f that made some sort of difference. So you're saying with a 100k load that using 2 resistors directly correlates to the gain droppage given the ratio? In other words. if i used a 33k off the plate and a 68k to the B+ it would be about a 33% gain reduction?Comment
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Thanks HH. Just wanna have a switchable gain reduction if i find preamp gain too much when in a in band context where it's cranked up quite high. I'll have caps coming off both the plate and after the 33k and each cap then to a switch with the common to the grid. I already trid a 50k/50k but it might have been a bit much. Also considering several resistors with a rotary switch at the AC end of the caps. Maybe overkill but it could be useful, at least till i decide on the one or 2 most useful ratios.Comment
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Don't forget that switching between different DC voltages will pop.- Own Opinions Only -Comment
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I'll be switching the AC side of the caps tho. Will that still pop? Not a big deal anyways, as it's something i won't be doing when playing, only at the start to see which works best and i can lower the master when i do.Comment
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Well, right now i have a 3 position rotary so it will be three 33k resistors in series for the plate load and a cap attached to the plate and 2 more between the 33k junctions, then the AC side of the caps to the rotary and the common to the grid.Comment
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I prefer to see a drawing.- Own Opinions Only -Comment
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Here ya go. Don't know if i will be keeping it like this so this is just a quick redraw where i have yet to decide how to illustrate the switch, but A B and C are the 3 switched poles on the rotary and the common is noted at the grid of V2A. Oh, and forgot to remove the switchable cathode R in the drawing that i was using for gain reduction but if the split load works that ill be gone and just a single 820R
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Ok, though the circuit is not complete.
At first turn-on the 3 coupling caps can be assumed to be discharged.
A discharged cap has 0 DCV between its ends, meaning that the voltage to ground at both its ends must be the same.
So switching between different caps will pop because right side ends are at different DC voltages.
To avoid popping you need to ensure that the right side ends of the caps are always at zero DCV.
This is best accomplished with high value (1M or higher) resistors from right side ends of caps to ground.
The resistors make sure that the coupling caps are always fully charged to the left side DC voltage so the right side voltage is zero.
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Will the 1M resistors affect tone at all ? I'm no tech but i do have some ears. Are we talking seriously loud pop? Like i said, i can turn the master down when switching.Comment
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A charging resistor of 1M means a slight additional load which will lower gain by less than 4%.- Own Opinions Only -Comment
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