questions about schematics

[beinz]

hi all. i have been spending some time lately looking a bunch of schematics, looking at their differences and similarities and trying to better understand how the different parts interconnect.
(disclaimer) im still a major newb when it comes to electronics, so i have tried to find as many answers as i can by reading on sites like merlin's tone wizard, and aiken's site before coming over here and bothering you fellas. i think im understanding things correctly, but im hoping the experts here can help me out with some questions i have.


1] from what i understand, common hi/lo inputs are basically a voltage divider between 1m to ground and 68k to grid for hi input, and 68k to ground and 68k to grid for lo input.
if you were to use only one input jack, but still wanted hi/lo capability, could you use a switch to jump between 1m, and a resistor matching the resistor at the grid?
is this correct?



2]in the example of the above attachments, one has cap off the plate going to a gain pot to ground, then to a resistor into the grid. the other has a cap off the plate, to a resistor, then to a gain pot to ground connected to the grid.
my question is does it matter what side of the grounded resistor/pot the other resistor is? or, does it matter if the grid sees the grounded resistor? from what i have read, there is a slight drop in gain (minor) when the grounded resistor is on the grid side as opposed to the other side... but otherwise would it have pretty much the same effect regardless of which side?



3] if im understanding this correctly, is the 470k resistor not in parallel with the 500k pot to ground meaning that the pot essentially has a range 0-250k? if so, then why not just use a 250k pot? or is there something else going on there? if not, then what is the purpose of 2 resistors to ground?

thanks for your help guys, any feedback is greatly appreciated.

[Enzo]

The common Fender type hi/low is indeed a voltage divider. Into the high and the two 68k are in parallel through the other jack, and in series with the grid. 1 meg to ground as a grid return. Plug into low and there is a 68k in series with the grid and a 68k to ground, making a 2/1 voltage divider, or a 6db pad if you prefer. Rather than switch resistors like that, wire the 1 meg in and switch the other resistor in or out of parallel with it.

2. They are two different circuits. In the left one the signal comes through a cap to the volume control, which is grounded. The 470k is in series with the grid. It acts as a grid stopper but also the two resistors allow VOlume 1 and Volume 2 to mix if needed. In the second one, the signal comes through a cap (or direct from the low jack) into a 3/2 voltage divider formed by the resistor and pot. The pot 1 meg is 2/3 of the resistor and pot total. The signal is thus reduced by a third before it gets to the volume control. In that example there is no grid stopper.

Does the order matter in the second? Yes indeed. The resistor on top means 1/3 of the signal will drop across the resistor, and the volume will sweep from zero to 2/3 of the signal. If the resistor were on the bottom, then the volume would sweep from 1/3 of the signal up to 100% of it. The volume would never turn all the way off.

3. yes, the resistor and pot are in parallel. COuld you use a 250k pot by itself? Sure, just like they did in the number two question. And like that #2 example, R6 and R7/P1 form a 2/1 voltage divider.

[beinz]

thanks for your reply enzo.

as i re-read #1, i realize i was wrong. the hi jack has 68k+68k in parallel = 34k to grid, 1m to ground, and the lo jack has 68k to grid, 68k to ground. but in any event, are you suggesting (using the 30k in the image instead of 68k) that 30k should be wired parallel to the 1m with a switch, or the 30k to the grid with a switch? if parallel to the 1m with a switch, isnt that pretty close to the same as in the image i attached (1m parallel with 30k = 29.12k vs 30k in the drawing) or is your suggestion to have only 1 resistor attached to a switch instead of 2 for less solder points?

your answers for 2 and 3 seem pretty clear to me, thx. i am curious in regards to question 2, how dual channel volume pots match up or meet up at the next stage. i was reading on a site using a fender tweed as an example of how the interaction between volume pots is a bit funny due to the (in that case) 270k isolation resistors used to separate two 1m volume pots before the next grid. this causes the two pots to hit max volume at "lower than full rotations of the pot". the article said that 270k was too low of a value, in the image i attached, they have used 470k resistors to isolate the two 1m volume pots. is this a better value to let the two volumes act more "normally" or independently?

[eschertron]

thanks for your reply enzo.

as i re-read #1, i realize i was wrong. the hi jack has 68k+68k in parallel = 34k to grid, 1m to ground, and the lo jack has 68k to grid, 68k to ground. but in any event, are you suggesting (using the 30k in the image instead of 68k) that 30k should be wired parallel to the 1m with a switch, or the 30k to the grid with a switch? if parallel to the 1m with a switch, isnt that pretty close to the same as in the image i attached (1m parallel with 30k = 29.12k vs 30k in the drawing) or is your suggestion to have only 1 resistor attached to a switch instead of 2 for less solder points?

I'm not Enzo, but I did stay at a Holiday Inn express...

I suspect that the suggestion to switch in a 30k (or 68k, or whatever) resistor in parallel with the 1Meg grid LEAK resistor is so that if the switch fails 1) the amp is still functional, and 2) the tube doesn't burn up from thermal runaway due to a loss of bias. Search for grid leak and thermal runaway for more info. There are quite a few preamp tube circuits out there, but the Fender input section you show is fairly standard simply because it adheres to the RadioDesignersHandbook.

[beinz]

that makes sense, thx

[Enzo]

Those are good reasons, but my reason was simpler. By leaving the 1meg all the time, you can switch the smaller resistor in or out. If you select one resistor or the other as you originally posted, then each time you flip the switch, there is a brief moment when neither resistor is there. It won't hurt the amp, but it likely will pop out the speaker. By leaving the 1 meg permanent, the grid return is always a complete circuit, so less likely to pop. and the 30k or whatever is not much affected by 1 meg in parallel. Comes to a little more than 29k. Never switch unterminated parts in the signal path.

[beinz]

im not sure if i am overthinking this, but in the example from before, when the pot to ground has the wiper into the grid, as ive read about pots, they are sort of like 2 series resistors in one package with the wiper being the point in between. so, in this example, the pot is essentially a 250k pot. when the pot is fully "on" the pot is sending 250k to ground and 0k to grid. but when the pot is at 50% resistance, it is sending 125k to ground, does that mean 125k is on the grid? and then when the pot is full "off" there is 0k to ground, but is there now 250k to the grid? or am i thinking about pots all wrong, and its it simply only a resistor to ground of varying values? also, this pot combined with the coupling cap effects frequency response. but does this change as the pot value changes, or is it always on the whole pot value, regardless of pot position. this is probably pretty simple, but my mind wants to make it confusing..., thx.

[mhuss]

Your thinking is a little confused. Think of resistance as an obstacle to climb over for electrons. Like said obstacle, resistance doesn't go (or get sent) anywhere. It is useful, as you suggest, to think of a pot as two resistors in series with the tap connected to the middle. At the top of rotation you'll have near-zero in series with 500k, and vice-versa at the bottom.

From the point of view of the driving stage, it looks like 220k in series with (470k in parallel with 500k), no matter where the control is set, because the grid of the following tube looks essentially like an open circuit (*very* high resistance). About half of the driving voltage will be dropped across the series 220k.

[beinz]

thanks for helping me with this. so to use an example with fixed resistor values, theoretically, if you liked the sound of the 500k pot at 5 (aprox 50k on an audio taper pot) could you just use a 47k resistor to ground in place of the pot? would this achieve a similar effect? how would this effect the driving stage?

[eschertron]

thanks for helping me with this. so to use an example with fixed resistor values, theoretically, if you liked the sound of the 500k pot at 5 (aprox 50k on an audio taper pot) could you just use a 47k resistor to ground in place of the pot? would this achieve a similar effect? how would this effect the driving stage?

Thinking of the pot as a voltage divider you'll see that with 47k (approx) from wiper to ground, you'll have the other 90% resistance between the wiper and the audio source. So a 470k (handy value) resistor on top with your 47k resistor on the bottom will simulate a volume pot stuck at 5 on the dial. Looking closer at existing schemtics, there are a lot of places where the designer did exactly that, although not necessarily with the values we're discussing. The Marshall interstage divider that you posted above has a fixed 470k resistor on top, and a pot from grid to ground. Why didn't they simply use a 1M pot (or a 500k pot) and ditch the resistor on top? My guess is that they wanted it that way. So the pot, at full, really only takes the volume up to 7 or so.

Also, taking into consideration the parallel caps, there's a treble "boost" going on, but that's another thread.

[beinz]

thank you, this is what i was trying to get at a couple posts ago. this makes sense!