Help: Input Impedance

[burningman]

Should I always place a 1meg tie down resistor at the input of my effects? What, if any, are the drawbacks to doing this?
Thanks,
Ian

[DougH]

OK help me understand. The title of the thread is "Input Impedance" but in your post you never mention input impedance.

Then there are some vague notions in the post about putting a 1meg "tie down" resistor on the input and whether there's a drawback to doing it.

What is it exactly you want to know?

Guys, look... A web forum is a text based medium. That means if you have a question, you need to type some text. Be specific. Don't be afraid to use words.


I'll give your post points for its Haiku-like charm, but it appears to be 2 sentences randomly pulled from a stream-of-conscious electronics rant with no apparent relation to the thread subject and not enough context to decipher what you are really asking.

But I'll give it a shot:

Should I always place a 1meg tie down resistor at the input of my effects?

No.

What, if any, are the drawbacks to doing this?

None.

[Minion]

In 1M Resistor to Ground at the input will help with Clicks and Pops if you plug in yer guitar with the Power engaged on the pedal ,It will also act with the Input cap to form a Filter, It also acts at a High Impedance load on yer guitar.......

I hope this helps a bit.....

[PerroGrande]

Always?  No.  Always is too inclusive, as there are examples where the answer might be no.

On the other hand, let's think a bit about what the purpose of the resistor might be.

If your circuit is AC coupled (i.e. series capacitor on the input line), a large resistor to ground (on the input-side of the capacitor) usually serves the purpose to avoid pops when doing true bypass switching.  If the resistor of which you are speaking is on the circuit-side of the coupling capacitor, then it serves one of several purposes -- typically biasing or dc return (devices like FETs when AC-coupled require a DC return path on the signal line. 

Does the presence of this/these resistor/resistors impact input impedance?  Of course.  They contribute a resistive component which typically ends up being in parallel with the other resistive to-ground components on the input side of the circuit.  Resistances in parallel end up producing a reduced effective resistance, and thus in turn reduce the overall impedance accordingly.  This is one of the reasons that the resistances shown are typically large (several meg).

So if you've got a jfet based op-amp circuit with a gajillion-ohm input impedance and you drop a 1M resistor to ground, your overall input impedance (all other things equal) will end up being dominated by the 1M resistor. 

Another consideration that was mentioned previously, is the filter effect produced by having resistances and capacitances playing together.  A capacitor in series with a resistor to ground forms essentially a frequency-dependent voltage divider where high frequencies are given a free pass and low frequencies are hassled (a.k.a. a high-pass filter).  One must be certain that the resistive path to ground doesn't roll off low frequencies in the range that the circuit is intended to operate in.  In the case of pedals, this is the audio range, so we want to set the low frequency roll-off point *below* 20Hz.  Again, this requires the product of R and C to be large, so larger values of R show up again here.

So -- they don't *always* have to be present, but they are *common* for a variety of reasons.

Drawbacks?  Well in some cases, the circuit may not work without them (some jfet configurations).  You *may* lower the input impedance more than you want by arbitrarily slapping resistors all over the place.  Depending on where you're slapping said resistors, you could end up messing up the biasing of a circuit (mosfet/bjt).

So -- that was a realllly long way to say "it depends on the circuit."

[burningman]

Sorry about the vague post. Thanks for the advice everyone.
Cheers.
Ian.

[brett]

Hi
Interesting question and replies.
re:

So if you've got a jfet based op-amp circuit with a gajillion-ohm input impedance and you drop a 1M resistor to ground, your overall input impedance (all other things equal) will end up being dominated by the 1M resistor.

Hmmmm..... We rarely, if ever, see such things, because the 10^12 ohm input impedance of JFET op-amps would have big noise problems and be affected by static charge, etc.  Op-amps, JFET input or otherwise, are usually given single supplies in our circuits, and the input is biased to half supply via a 1 M or 470k resistor, which effectively "sets" the input impedance to a workable value.  Adding a 1 M pull-down resistor to these, and other, circuits (Zin of 20k to 1M) which will have minimal impact on the circuits' performance.
cheers

[PerroGrande]

Brett,

I was a little unclear on what I wrote about the "gajillion ohm input impedance."   

But consider two cases for a moment:

1) The op-amp has a high input impedance and is powered by a true bipolar supply.  In a non-inverting configuration, the input impedance of this type of configuration can be quite high.  Even high enough to be dominated by a 1M pulldown.  Will this adversely effect most audio/effects circuits?  No -- you're absolutely correct.

2) A non-inverting op-amp running from a single supply that is biased to mid-voltage using a bypassed voltage divider and a 1Meg pull-up resistor to create the biasing network.  Arbitrarily adding a pulldown resistor to this circuit will change its bias point (by creating another voltage divider), so this isn't something that is desirable.

That was where I was trying to go with that.

[brett]

Does (2) not have an input capacitor?

[DougH]

The general idea behind the "1M pulldown resistor" is to avoid switch pop due to a DC voltage left on a coupling capacitor. Basic rule of thumb- don't leave coupling capacitor leads on the input/output "floating" i.e. ungrounded. Always provide a DC path to ground for them. This keeps them at 0VDC and eliminates the pop due to a voltage differential between the input and bypass circuits.

This will essentially "compromise" the Zin of an effect somewhat. But how much? That's the question. IMO, 470k and greater is a very reasonable Zin load for most passive guitar pickups. You could even argue that as low as 100k is reasonable, depending on what kind of interaction you want. As in a fuzz face, sometimes Zin is purposely low to achieve some sort of effect through interaction with a passive guitar pickup. So a 1M pulldown and 1M "biasing" resistor for an op amp circuit gives you essentially 1M || 1M = 500k Zin for the AC input signal- pretty reasonable IMO. Of course, if you are that worried about it, you can use a 10M pulldown to minimize the effect on Zin.

Of course, all this switch pop stuff is documented on RG's site somewhere. I think he was the first one to write about it in these circles.

[PerroGrande]

Hi Brett,

In 2, I'm talking about a pulldown resistor added on the circuit-side ("inside") of a coupling cap.  Of course, comment 2 applies also if the circuit is DC coupled.

[Electric_Death]

Should I always place a 1meg tie down resistor at the input of my effects? What, if any, are the drawbacks to doing this?
Thanks,
Ian

Good question because as I've found out, you really can't get around using this filtering resistor but, the value you'll want will depend on the circuit, particularly if you are designing it from scratch. 1 meg totally killed the overdrive I engineered from scratch, the gain was basically nothing afterwards.
What did I learn?
Yes you need it but the value needed may be anywhere from 1k to 1m.

Side note- I was getting away without using it for a bit because I had enough gain that I could set my control pots low and the hum/noises didn't sneak through.

[DougH]

Good question because as I've found out, you really can't get around using this filtering resistor but, the value you'll want will depend on the circuit, particularly if you are designing it from scratch. 1 meg totally killed the overdrive I engineered from scratch, the gain was basically nothing afterwards.
What did I learn?
Yes you need it but the value needed may be anywhere from 1k to 1m.

Side note- I was getting away without using it for a bit because I had enough gain that I could set my control pots low and the hum/noises didn't sneak through.

Huh?!?

Are you referring to the anti-switch-pop pulldown resistor or something else?

I'm confused...

[PerroGrande]

Please tell me you didn't put the "filtering resistor" in series with the input...   

That *would* be a great way to kill gain and increase noise, etc...

OTOH, it would raise the input impedance.

[aron]

>Are you referring to the anti-switch-pop pulldown resistor or something else?

Come on Doug! It's the filtering resistor!!!!!! :-) The reason you don't know is because the design is not anything out there. It's original remember?

[Mark Hammer]

As I have understood it, the terminating resistor on the input can be any of a number of values.  One often sees values of 2.2meg and higher.  The "ideal" input resistor is one that permits any charge stored in the input cap to drain off.  To some extent, the resistor value selected should permit all of that charge to drain in a reasonable period of time without loading down the input too much.  Consequently, as the input cap gets larger in value, and there is more potential charge to drain off, you want a smaller resistor value to make sure that charge is all drained off between the time you switch the effect to bypass and when you next engage it.  If the input resistor is, say, 10uf and you have a 3.3meg resistor to ground, and it has only been about 15 seconds since you switched the effect to bypass, there is a reasonable chance there will still be charge stored in the cap such that when whatever the terminating resistance is of the thing before that pedal is now connected to the input of the pedal in question, that remaining charge will want to rush out the low-impedance path to ground, resulting in the risk of audible pop.

Of course, if you used a 47k resistor to ground on the input, you could be pretty much absolutely assured that the cap had drained off before you could re-press the stompswitch to engage the effect again, but the down-side would be that it would seriously load down the input.  This is why you tend to see 1M resistors on the input.  They are a) not so large in value that charge does not completely drain from the average input cap, b) not so small in value that the input is loaded down, and c) they are a pretty common value in most parts drawers.

The other historical aspect to consider is that there was a period, long before 3PDT switches, and even before DPDT switches, when effects used SPDT switches and the input of the circuit was always connected to the input jack.  Assuming we had 3 pedals in series, each with a 470k resistor on the front, and they were all "off", even if you thought your guitar would be plugged directly into the amp, it had 3 x 470k in parallel hanging off the hot lead, or a load of 156k.  This is where all those legends of "tone-sucking" came from and why people made such a big to-do about using "true-bypass" when DPDT switches became more available and cheaper***.  Clearly, under those conditions, having a higher-value terminating resistor on the front end made a lot of sense.  Although to be fair about it, many of those SPDT-switched pedals would often have NO terminating resistor on the input cap because if the input cap would always remain connected to the input jack, there was obviously no risk of pop if that cap was never left "hanging".

Long and the short of it is that if you had a .01uf input cap, and ran out of 1M values, then you could pretty much confidently stick in a 1.5M or 2.2M in its place.  If you knew the output impedance of the preceding pedal/stage was pretty low (let's say <2k) then a 470k might do you just fine.  If the input cap is larger, err a bit on the lower side, and if it's smaller, you can go higher in resistance.

(***Though "cheap" is relative.  I bought my Shin-Ei FY-2 in 1992 for $20.  Why did I buy it?  Not because I knew anything about it.  I bought it because it had a DPDT stompswitch and I figured that if a DPDT was going to cost me $14 anyway, that I could get some pots and a chassis for a few bucks more.)

[soulsonic]

I havent used any input anti-pop pulldown resistors since I started doing the kind of bypassing that grounds the effect input when bypassed. Seems to give me satisfactory results - I still use one on the output IF there isnt a volume control "last in line".

[DougH]

>Are you referring to the anti-switch-pop pulldown resistor or something else?

Come on Doug! It's the filtering resistor!!!!!! :-) The reason you don't know is because the design is not anything out there. It's original remember?

Oh yeah, the filtering resistor. Right next to the flux capacitor. Got it...

[JDoyle]

If the resistor of which you are speaking is on the circuit-side of the coupling capacitor, then it serves one of several purposes -- typically biasing or dc return (devices like FETs when AC-coupled require a DC return path on the signal line.

These are the same thing. If you don't have a capacitor between a resistor to ground (or any DC source) and the input of an active device that requires biasing, you have a resistor that is affecting the biasing of that device.

Come on Doug, we all know that the 'filter resistor' is what sets our own personal impedence for when the output impedence of our fingers interacts with the input impedence of the guitar. Obviously, the higher, the better, because less of your talent will be shaved off from the voltage division between your fingers and the guitar. It's really straightforward Doug. But there are a couple of difficuties involved with changing its value. First, it is very difficult to locate and you have to find it by first searching for your talent. Second, it is located in your ass. So if you don't find it when you go looking, your talent is too far up your ass to do you any good and adjusting a single resistor surely isn't going to help.

[DougH]

Yikes... I think I need a beer...

[brett]

Of course, comment 2 applies also if the circuit is DC coupled.

Attention all hypo-sub-sub-bass players....