Function & purpose of "pull down" resistors

[Jittery]

Hi everyone,

i'm a newie on here, a musician who has recently resumed an interest in electronics projects.  Just built a "Ruby" headphone amp and am hoping to get on with some stompboxes soon!

My first question is about "pull down" resistors.  Many preamp and effect circuits have a large-value resisitor (eg. 1Mohm) to ground from the input of the circuit.

What's the function of this resisitor?  I understand that it has little effect on the input impedance of the unit, since that is mainly determined by the first transistor or JFET and associated source/emitter resistors.

Is it something to do with the DC-blocking input capacitor? Sometimes it (the resisitor) is shown before the input cap, sometimes it is shown afterwards.

Thanks & best wishes,
Bruce

[zyxwyvu]

That resistor can serve two different purposes, depending on where it is. It is occasionally used in both places.

If it is before the input cap, it is to drain any voltage from the input cap while the pedal is bypassed. This is to prevent popping when the pedal is switched back in.

If it is after the input cap, it is there to bias the signal. In the picture you included, the JFET needs a signal centered around 0v, so it is biased to ground.

[GibsonGM]

+1, zyx.....I believe that the pulldown is also considered to be in parallel with the input impedance of the FET, BJT, opamp, or what have you.  So its value does affect the input impedance (lowering it), along with the other resistances present. 

[Sir H C]

Imagine if that resistor was not there.  What would set the DC level on the input of the FET?  Nothing, the cap blocks all previous stages, so that sets the DC to pretty much ground.

[Jittery]

Hi all!

Thanks for your replies zyx, Gibson and Sir HC.  I had forgotten about the "popping" thing, that makes more sense now.

I can see that the pull down resisitor will affect the input impedance.  However from what i've been told, it is effectively in parallel with the "reflected" resistance of transistor's source resistor (ie. source resistor x transistor gain), which is smaller and will thus have more effect on the total impedance.  In the example circuit, that would be 1k ohm x gain (say 200) giving 200k, and the total resistance would be about 165k ... am i on the right track here?

Also ... I forgot to thank runoffgroove.com, for providing that little section of the circuit schematic  .  (i think it was from the "Odie" pedal if anyone is interested.)

cheers
jittery / bruce

[QSQCaito]

May I suggesst a search. You'll find very detailed infromation about this topic, which probably won't be repeated here.

Bye!


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