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Simple Preamp Input Question

1/17/2000 11:21 PM
Randal J
Simple Preamp Input Question
In the classic Fender two input configuration, one of the inputs goes through a 68k resistor straight into the input of the tube, but the other input adds a 1M resistor in parallel with the same arrangement. Am I correct that the 1M resistor only affects the second input? If so, how does this affect the sound versus the other input? Or does the 1M actually factor into both inputs?  
Thanks in advance....
1/18/2000 1:18 AM

If you read the old Fender literature you find that the second input was for a microphone. It's got a little less gain up front and gives more control over the volume.  
If I'm not mistaken though that 1 meg resistor goes to ground, not in parallel.  
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1/18/2000 1:24 AM
John Stokes
Winnie is correct. The 1 meg resistor goes to ground in all cases. The #2 input uses the 68k resistors as an attenuator to reduce the gain on that input.
1/18/2000 1:43 PM
Randal J

You are right, I just meant parallel, with the rest of the circuit, to ground.  
1/18/2000 2:43 AM

Hi, Randal. If you look carefully at the  
schematic, you'll see that when a plug is  
inserted into what you are calling the "second"  
input (actually called input 1 on the Fender  
schems.) the two 68k resistors are in parallel  
because of the connection thru the switch  
circuit of the other input. This input is  
therefore a little hotter, having only 34k  
in front of the grid.  
The 1meg resistor to ground doesn't have much to  
do with setting the sensitivity of the input  
because it is so large a value, rather,  
it's a necessary component of the gain stage  
itself. If you look downstream in the schematic  
you'll find that at each gain stage there is  
a DC path to ground from its grid. Sometimes it's  
thru a pot, sometimes it's a fixed resistor,  
but it's always there. The circuit won't work  
properly without it. Sometimes the value of this  
resistor is smaller to effect a little interstage  
attenuation, but usually it's fairly large.  
Regards, Steve F
1/18/2000 1:47 PM
Randal J

OK, so the 1M is not a "signal path", or at least not one of any consequence? What flows through it is mostly just bias current?  
1/18/2000 1:55 PM

Sorta - the 1M is to reference the grid to ground for DC purposes - the cathode resistor's voltage drop puts the cathode positive to ground, hence the grid is negative with respect to the cathode. There is, for all practical purposes, no grid current flow in this stage.  
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