grounding input: is pulldown resistor still necessary?

[hans h]

Dear all,

I have been reading up on good true bypass methods and prevention of switch popping. I understand that using multiple methods at the same time can be beneficial (e.g. ground circuit input upon bypassing + pulldown resistor). However, for simplicity, reduction of the number of components and possibly the tiny loading effect of the 1M pulldown resistor, I would like to use only input grouding and remove the pulldown resistor.

1) Does this adversely affect the input impedance of the circuit (I do not fully understand this yet).
2) Are there any other reasons besides switch pop and possibly input impedance that a pulldown resistor should be used?
3) Is it correct that when you use multiple pedals in series, the 1M pulldown resistors are in parallel of eachother and thus their value is effectively lowered (e.g. 2 pedals => 0.5M)? In this case it would certainly be interesting to omit the pulldown resistor

Any ideas/ thoughts are greatly appreciated,

Hans

[antonis]

1. No..
Input grounding only occurs in by-pass mode..
(circuit out of order..)

2. No..

3. No..
(actually yes , but it stands for previous effect output pull-down resistor in set in parallele with next effect input pull-down one.. In almost every pedal there is an output pot which takes the role of pull down resistor - in the rare case of out pot absence AND output pull-down resistor presence, you're right..)

P.S.
On the contrary of the above, I always place input pull-down resistor AND wire 3PDT switch with Input & Output grounded when effect by-passed..
(it's a strong recomendation of my Zen teacher for peace of mind..)

[hans h]

Hi Antonis,

Thanks for the swift response!

Just to clarify, with question 1) I meant to ask whether removing the 1M pulldown resistor affects the circuit's input impedance (and whether this matters).

I think that you read it as: does input grounding affect the input impedance.

What I got out of your response so far:

Point two => I am probably safe in using only input grounding and omitting the pulldown resistor.

Point three => However, signal quality is probably not really affected by the 1M resistor, even if multiple pedals are on => there is not really a strong case for omitting the 1M resistor

[antonis]

Got it now, Hans.. 

1. Yes, it affects it for sure..
(any parallel impedance removed raises the total one..)

2. Yes..
(except for a veeeeeeeeeeeeeeeeeery slow switch..)

3. Yes..
(but you face it quite wrongly..)

Number of pedals involed doesn't have to do nothing with signal "quality" (from individual input impedance point of view) as far as each effect takes care of its own signal attenuation..

[j_flanders]

The input impedance of a pedal mostly matters to your guitar's passive pickups.
So, it only matters for the first pedal, the one directly connected to your guitar.
It matters very little to the output of a pedal, as that output isn't affected as much by a low(er) input impedance of a next pedal.

As to your question about several pedals in series and their input impedances adding up (in parallel):
-your pickup only sees the volume pot + tone pot + input impedance of the first pedal.
-the output of the first pedal sees only the input impedance of the second pedal, etc.
- when pedals are in bypass, that 1M pull down resistor is out of the signal chain, otherwise it wouldn't be true bypass.
Some (older) pedals leave the input connected and only switch the output. In that case all those input impedances would add up, even in bypass.

Here is what resistive loading does to your passive guitar pickup:
http://buildyourguitar.com/resources/lemme/

As you can see, a lower resistive load lowers the resonance peak (treble content)


Take various guitars as an example to see what load their pickups see and what effect an additional 1M pull down resistor has:
Jazzmaster:
1M vol pot in parallel to 1M tone pot: 500k Ohm load
An additional 1M pull down resistor will make the load/input impedance: 333kOhm load

Les Paul:
500k vol pot in parallel to 500k tone pot: 250k load
An additional 1M pull down resistor will make the load/input impedance: 200kOhm load

Strat:
250k vol pot in parallel to 250k tone pot: 125k load
An additional 1M pull down resistor will make the load/input impedance: 111kOhm load

On a Jazzmaster you'll probably notice the extra parallel 1M, on a Strat probably not.

As a side note, this is what a capacitive load does to the frequency response of a passive pickup.
Instead of lowering the amplitude (Q) of the resonance peak it lowers the frequency of the resonance peak.

[duck_arse]

As you can see, a lower resistive load lowers the resonance peak (treble content)

not quite terminologied correctly. a lower resistive load, that is, less load, resistive, will increase the resonance peak. a lower value of resistance as load will load the pickup more, hence lower the resonance peak.

[j_flanders]

As you can see, a lower resistive load lowers the resonance peak (treble content)

not quite terminologied correctly. a lower resistive load, that is, less load, resistive, will increase the resonance peak. a lower value of resistance as load will load the pickup more, hence lower the resonance peak.

100% correct. Thanks, for clarifying.

I changed my sentence a couple of times while typing but it never came out the way I wanted it to come across, so I just kept the (less confusing?) but wrong way of saying it.

By the way, I often see people struggling with the same concept/terminology when talking about the load for a tube amp. Usually resulting in people discussing, meaning the same thing, but thinking the other person means the opposite of what they're saying.

[antonis]

I often see people struggling with the same concept/terminology when talking about the load for a tube amp. Usually resulting in people discussing, meaning the same thing, but thinking the other person means the opposite of what they're saying.

Probably due to "opposite" meanings of light load vs low value load..

[duck_arse]

I changed my sentence a couple of times while typing but it never came out the way I wanted it to come across, so I just kept the (less confusing?) but wrong way of saying it.

I use the simple method of just continuing to add more and more words until they are all there.

[merlinb]

Probably due to "opposite" meanings of light load vs low value load..

As a beginner I remember being confused by statements like:
"we can ignore this impedance because it is negligible"
because sometimes it meant negligibly small, and sometime negligibly large!

[antonis]

"we can ignore this impedance because it is negligible"

meaning nothing about its actual value as long as it doesn't refer on its specific placement (series or parallel)..