Pulldown resistor value choices

[dennism]

Really just a hypothetical question for learning purposes.   In many fuzz circuits, the pulldown resistor from input to ground is 1M and sometimes I see 1.5M as a value.   Does it matter much what value is here?  Or does it just need to be large?    What would happen if this value was even bigger?   Say 10M for instance...  How about smaller?  470K for instance...

[CynicalMan]

The value doesn't matter too much. A 10M resistor would be fine. However, if you go too low, then the input impedance will be lowered too much. In a fuzz that wouldn't matter much, but with cleaner pedals you'd hear a loss of treble if your guitar was plugged directly into the pedals. I wouldn't use a resistor lower than 1M, and I personally use 2.2M to 4.7M.

[R.G.]

Really just a hypothetical question for learning purposes.   In many fuzz circuits, the pulldown resistor from input to ground is 1M and sometimes I see 1.5M as a value.   Does it matter much what value is here?  Or does it just need to be large?    What would happen if this value was even bigger?   Say 10M for instance...  How about smaller?  470K for instance...

As Alex said, the exact value doesn't matter much.

There are competing issues hiding underneath this, though. The reason there is a resistor there at all is that the input and output capacitors on an effect circuit are not perfect. They leak DC, a very small amount. When the effect is bypassed, especially with "true bypass", they are also disconnected on one end. The capacitor is charged to the DC voltage it maintains in operation, often the 4.5V bias voltage of many effects. With one end of the capacitor simply disconnected, the internal DC leakage lets this voltage leak down towards zero. When the cap is re-connected by stomping the switch, the capacitor needs to re-charge instantly. So there is a pop as it re-charges. Any voltage change of more than about 5-10mV is clearly audible as a click or pop. The pulldown resistor prevents this by literally keeping the otherwise opened up end of the capacitor pulled down to ground through the pulldown resistor. This keeps it charged to the right voltage, so no pop on reconnection.

What really happens is that the capacitor voltage pulls down to the voltage that's caused by the voltage divider of its internal leakage resistance and the pulldown resistor. Since the internal resistance to leakage is as large as the manufacturer can make it, the pulldown resistor keeps the voltage change down to a few millivolts. That means the capacitor internal resistor is hundreds or thousands of megohms.  So whether you use 1M or 2.2M makes very little difference to the popping. Nor would using 100K, 10K, or ten ohms.

The competing issue is what the resistance causes to the effect. On the input end of the effect, the input pulldown resistor loads the guitar signal. This can cause the treble loss that guitarists so affectionately call "tone sucking" if they hate it and "brown sound" if they like it. Anything much less than 1M causes it detectably, although down to 100K is not too horribly bad. The lower the resistance, the more treble loss. However, there is no point in making the pulldown resistor hugely big for treble loss reasons if the effect circuit itself will cause treble loss. 

The exact value of the pulldown is then very foggily determined by these two competing effects. 1M is kinda in the middle. Lower than 100K is pretty sure to give treble loss, more than a few M is getting into pop territory with leaky electrolytic caps as input caps. In the middle, it doesn't matter much.

[Earthscum]

Thanks for the info, R.G... I read the same thing by you in another thread and couldn't find it.

I was actually wondering about all this myself when I was building the Nurse Quacky. The effective pull-down resistance is about 33k (47k and the 100k pot). I slapped a buffer in front and all of a sudden the thing 'came alive' (I have some seriously dead strings right now). Is this because of the impedance? My GF's bass with new strings (pickups are split-P compared to my pair of J's) didn't have such a noticeable effect, which doesn't make sense because her strings are brighter and should make the effect more noticeable, but I guess that her pickups are lower impedance than mine.

[dennism]

Thanks very much to all who replied, that is very helpful info indeed.

[bluesdevil]

Would wiring the footswitch to ground the effect's input when bypassed a better idea? Then you could just do away with the pull down resistor?

[Brossman]

I believe if that were the case, you might still have to put a resistor in there.  Look at Ohm's Law; V=IR If there is effectively no resistance in that part of the circuit, you'd have 0V... There has to be some resistance to create a potential from one section of the circuit to the other (voltage is this potential).

But, I may be wrong...

[R.G.]

Would wiring the footswitch to ground the effect's input when bypassed a better idea? Then you could just do away with the pull down resistor?

I think that would work, if I understand what you mean - that being the effect input is the pole of the input half of a footswitch, and it is then switched to either ground or the input jack.

It's not a bad idea, as grounding the input can help in some effects that go crazy when their input is left open. Of course, all you save is a $0.02 resistor, and you still have to do something about pulling down the output capacitor. And if you use that switch configuration, then you can't use the Millenium bypass and its variants, so it pushes you into using a 3PDT if you want an LED indicator too.

I guess like in all the other instances, it could be better or worse, depending on how you measure "better".