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Re: 68k input resistor?

2/6/1998 6:22 AM
Dave H
Re: 68k input resistor?
I think you have got it back to front. The 68Ks and 1M are on the tube input so they are not a load for the tube they are a load for the guitar.  
If you plug into the hot input the guitar is loaded by 1M and there is 34K (68K // 68K ) in series with the tube grid. If you plug into the cool input the guitar load is 136K (68K + 68K) and the signal the tube grid sees is attenuated by a factor of 2 (6dB) by the 68Ks.  
I donít think that the difference between 68K and 33K stoppers in series with the grid will be audible. I think it is a myth which could have come about because the hot input with a 33K input resistor has more treble than the cool input with a 68K input resistor. The real reason that the hot input has more treble is because it only loads the guitar pu with 1M whereas the cool input loads it with 136K.  
Dave H.
2/6/1998 2:01 PM

I understand the #1 input: 34K in series with the tube grid and in parallel with a load of 1M. But I don't understand the signal path that leads to 136K. Wouldn't the signal path be 68K in series with the tube grid and in parallel with a load of approximately 68K (1M // 68K)? Of course I may not understand the schematic yet.  
Still learning,  
2/7/1998 7:31 AM
with nothing plugged into input #1 (hot), the 1M is SHUNTED (bypassed) by the #1 jack -- therefore, the two 68K are in series to ground (136K total). they are simply acting as a voltage divider (i.e. like a 136K linear pot turned to halfway, feeding the tube).
2/5/1998 10:37 AM
Jack Orman

the grid stopper resistor of 68k acts with the internal tube capacitance to form a low pass filter. This prevents high freq oscillation (as previously noted)... probably doesn't have much effect of audible freq response or noise since the Fc is probably 200K Hz or more.  
For best results the 68k should be connected directly to the tube socket (or as close as possible).  
regards, Jack
2/5/1998 2:20 PM

Thanks to all of you, some of this is starting to make sense.  
2/6/1998 10:30 AM
Looking at the Tweed Deluxe circuit on this site,  
it looks to me like one jack gives you 68k in  
parallel with 68k between guitar and grid,  
the other gives you the two resistors as a  
divider, i.e. the guitar sees 136k to ground  
and the grid sees half the guitar signal...  
hope I'm reading this right!  
2/11/1998 7:38 PM
bob sigafoos

Speaking of the 2-68k input resistors, i've got a newbie type question. I've recently aquired a blackface vibro-champ and sent away for the torres "tweed like" kit to see what it might sound like. The directions with the kit are straight forward enough with the exception of how to install 2-33k resistors to replace the existing 2-66k input resistors. Should I try to remove the angle bracket that holds all the pots together in order to solder the resistors in place or use some type of forceps and pretend I'm a micro surgeon. Any tips or suggestions?

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