Buffers and low-impedance rack units

[Dr.Pain]

Hey guys, I have a question regarding low impedance rack units (not quite about building anything, but I will if I have to). I'm looking at getting some Roland rack units, but they expect a low input impedance of 56k. Now I know that my guitar has an output impedance of 500k (volume pot), but I was planning on running the rack unit last in my pedal chain after my Boss RV-3 which according to Boss has a 1k output impedance. This should work fine since the RV-3 will be driving a higher load, right? If not, I'm guessing that I could build a simple buffer to match the impedances, but preferably the above method would be a lot easier. 
Also, the output impedance is stated as between 100-600 ohms, would the be fine going into your average guitar amp? I'm not sure what guitar amps expect to receive up front (high impedance?). If they do, shouldn't that be fine too?
Thanks a bunch.

[GibsonGM]

Amps want low OUTPUT impedance from a guitar/effect, and they tend to have a moderate-to-high INPUT impedance themselves.  This prevents "loading" the guitar or effect, and thereby losing input signal (starting with the highs...there is a filter action which occurs between the output impedance and input capacitance of the amp that bleeds off highs).  So the 100-600 ohm output is GREAT. 

Look at it as "the unit (amp, effect, rack, whatever) having high input impedance will only 'take' a little bit of current to operate, rather than suck all the life out of the pedal before it".  So you get a 'higher fidelity' representation of the signal you were feeding it.

Sounds like your rack units have a ~56k INPUT impedance.  That is sort of medium; many transistor-based effects have a lower input Z than that.    Try it with your guitar and see what it sounds like. If it sounds like it's losing treble, a buffer would certainly help.  Easy build, just a little box with a JFET in it and many plans on here.  A Boss pedal before it will also act like a buffer, even when not turned on.
{If it was an output impedance, it would be considered high, BTW. Good output impedance is like <1k ohms} 

Your guitar has 500k output impedance, which is high, only when the knob is turned fully down - which is why the highs go away as you turn the vol. down.  That rapidly goes to like 100K and then lower as you play w/the volume knob!   Low output Z is good, high input z is good.  Together, they cooperate and transfer your signal efficiently with little loss of treble.  Taken the other way, they start to act like a filter and bleed off treble.

Seems like you should be fine as long as your rack unit expects the guitar signal level.  Start off on a low volume to the unit, see what it sounds like, go from there )

[R.G.]

I'm looking at getting some Roland rack units, but they expect a low input impedance of 56k.

??
Expect or have a low input impedance?

Now I know that my guitar has an output impedance of 500k (volume pot)

Actually it's worse than that. Your guitar has a 500K pot, which has a maximum output impedance of 250K, but what drives it is a guitar pickup winding of 4-18K ohms resistive and an inductance of 2-8 Henries. The inductance makes for an output impedance of the pickup itself of upwards of 0.1-1Mohm at the high end of guitar frequencies. That's the one that really gets you.

but I was planning on running the rack unit last in my pedal chain after my Boss RV-3 which according to Boss has a 1k output impedance. This should work fine since the RV-3 will be driving a higher load, right?

Yes. You should always drive a load with a driver with less than 1/10 of the load's impedance to avoid much level loss. In this case, that's fine.

If not, I'm guessing that I could build a simple buffer to match the impedances

Careful. "Matching" as in "making equal or similar" is not the right word. For voltage based audio you want to MISmatch in the direction of the driver being much lower impedance than the input/load it drives. But yes, what you describe with the RV3 driving will work fine. It's MISmatched correctly.

Also, the output impedance is stated as between 100-600 ohms, would the be fine going into your average guitar amp? I'm not sure what guitar amps expect to receive up front (high impedance?). If they do, shouldn't that be fine too?

You may expect guitar amps to have a high input impedance, usually a bit over 1M, so that they can successfully be driven by impedances up to about 100K. This is about where guitars get in their upper harmonics, roughly.

Any impedance less than that will drive a guitar amp fine.

There is a hidden benefit. Tube guitar amps clip a little more softly if the clipping source is the first tube stage and it's driven by a low impedance. It's a quirk of tubes. But yes, a low input impedance into a guitar amp is fine.

[Dr.Pain]

Thanks for the replies, much appreciated. I'd like to leech some more knowledge from you two if you don't mind. 

??
Expect or have a low input impedance?

Sorry, bad wording on my part. They have a low input impedance.

Actually it's worse than that. Your guitar has a 500K pot, which has a maximum output impedance of 250K, but what drives it is a guitar pickup winding of 4-18K ohms resistive and an inductance of 2-8 Henries. The inductance makes for an output impedance of the pickup itself of upwards of 0.1-1Mohm at the high end of guitar frequencies. That's the one that really gets you.

I'd like to get this quote straightened out for my understanding, I'm an EE student, but I haven't done too much analog stuff yet and I'd like to understand how this works. First of all, why do you say that the 500k pot would only have a max output impedance of 250k? Second, it's a total combination of the pickup winding, the pickup inductance and the volume pot (tone pot is bypassed when "on"), right?

Yes. You should always drive a load with a driver with less than 1/10 of the load's impedance to avoid much level loss. In this case, that's fine.

That's the rule of thumb then for audio equipment then?

Careful. "Matching" as in "making equal or similar" is not the right word. For voltage based audio you want to MISmatch in the direction of the driver being much lower impedance than the input/load it drives. But yes, what you describe with the RV3 driving will work fine. It's MISmatched correctly.

D'oh, my post is a trainwreck. Yeah, I realize that you'd want to drive a load with a a higher current, thus wanting low to high impedance. Sorry, that's what I meant. In any case, a buffer could take care of the improper/poor impedance combinations disregarding levels/etc, right?

You may expect guitar amps to have a high input impedance, usually a bit over 1M, so that they can successfully be driven by impedances up to about 100K. This is about where guitars get in their upper harmonics, roughly.

Any impedance less than that will drive a guitar amp fine.

Again, sorry for the poor wording. So to prevent extreme filtering, use the 1/10th rule of thumb, check. So then a low output impedance from the rack unit (650ohms let's say) would  drive the amp pretty hard, wouldn't it or is it typical considering that even the RV-3 states an output impedance of 1k? Talking about pedals and typical guitar gear here.

Again, thanks for the replies!

[PRR]

> Your guitar has a 500K pot, which has a maximum output impedance of 250K

If one end were a low-Z source, and the other end is ground, the maximum impedance at the wiper is 125K, not 250K. (Wiper at half, 250K each way, 250K||250K is 125K.) Going to zero at full-up or full-down, of course.

> what drives it is a guitar pickup winding of 4-18K ohms resistive and an inductance of 2-8 Henries.

Taking 5K as the bass to midrange value, then worst-case is 126.25K.

The inductance gives a rising impedance. 5H is 31K at 1KHz. While we can't directly add R and L, when the reactance of L is smaller than R we can skip the small angle and say the total around the loop is 531K and the worst-case at the wiper is 132K.

Above 1KHz the capacitances matter. There are two: the coil has hundreds of pFd self-capacitance, and the guitar cable is 30pFd per foot. When pot is full-up, the two caps may be lumped together. They resonate the L at 2KHz-3KHz. The reactance is around 70K, the series resistance is 5K, the shunt resistance is pot plus amp, say 330K. The effective impedance at resonance is a bit less, in the 200K zone. BUT if the pot is not FULL-up this resonance shifts in frequency, shape, and impedance. When pot is "8" to "5" the wiper impedance is around 200K at peak.

The jack impedance tends to stay between 100K and 200K for most usable pot settings below full-up and above full-down. The pot buffers-up the winding's low bass resistance and damps-down the resonance.

> 100-600 ohms, would the be fine going into your average guitar amp? ..... I'm an EE student

There are important exceptions. But it is usually convenient to measure voltage, and to connect loads in parallel, so we design systems for "constant" voltage. (Even when voltage flutters with the music and there is only one load.) The dam for Green Lake Hydro puts out (say) 120V at anything up to 1,200 Amperes. Or at least the delivered voltage is controlled to be 120V whether 3am (light load) or dinner hour on a cold evening (heavy load).

Consider instead a "matched" electric generator. Matched to what load? Well, say we regulate for 120V at no load, and wind the dynamo to match the maximum load. Then when everybody is cooking and heating, or the paper-mill goes back to work, the delivered voltage is 60V. Also the waste heat in the dynamo equals the billable power delivered. That's usually not acceptable in utility work. The economics of copper versus MegaWatts allows a "mismatched" system. The load may be 120V/1,200A= 0.1 ohms, while the dynamo may be 0.01 ohms (will actually be nearer 0.002 ohms).

Electric systems are usually mismatched (or over-regulated) for 2% sag at maximum load. 2% of MegaWatts is a lot of waste heat, a lot of coal (or water over the dam). Sag makes lights flicker and on a shared utility that's annoying. In audio our power is smaller (you can't melt a guitar pickup by loading it) but "expensive" (the pickup power is all from your sore arm), and we usually won't notice a 10% sag. That's a rough 10:1 mis-match.

[R.G.]

Actually it's worse than that. Your guitar has a 500K pot, which has a maximum output impedance of 250K, but what drives it is a guitar pickup winding of 4-18K ohms resistive and an inductance of 2-8 Henries. The inductance makes for an output impedance of the pickup itself of upwards of 0.1-1Mohm at the high end of guitar frequencies. That's the one that really gets you.

I'd like to get this quote straightened out for my understanding, I'm an EE student, but I haven't done too much analog stuff yet and I'd like to understand how this works. First of all, why do you say that the 500k pot would only have a max output impedance of 250k? Second, it's a total combination of the pickup winding, the pickup inductance and the volume pot (tone pot is bypassed when "on"), right?

First of all I told you wrong. It's what I get for typing fast. Paul got it correct. A 500K pot has an output impedance that maxes out at 125K ohms when it's turned to the middle resistance.

If you're an EE student, pay attention to Norton equivalent circuits and Thevenin equivalent circuits and the transformation between them. A voltage source across two resistors in series (potentiometer tries to be this) causes a divided voltage. It also inserts some resistance, as you'd guess. But what resistance?
Saving me typing in a lot of math, the pot inserts the least resistance when it's dead in the middle of its rotation. At the ground end, the impedance is small (connection to ground) and at the top end, the impedance is small because the pot resistors are no longer in the way, they just load the source a bit. So the series resistance that the pot adds peaks somewhere in the middle. The peak is when both resistors are equal. And with some math you can find out that the series equivalent is always the two pars of the pot in parallel.

A 500K pot set to middle gives two halves of 250K. These halves appear as two resistors in a divider to make the pot output, and the equivalent series resistance of the two halves is the two halves in parallel, or half again  - 125K. A pot used as a divider has an output impedance which varies from zero up to 1/4 of the pot value from end to end. This happens at the point where the two resistance halves are equal. That is in the middle for a linear pot, and as little as 10% rotation for log or reverse log pots.

Yes. You should always drive a load with a driver with less than 1/10 of the load's impedance to avoid much level loss. In this case, that's fine.

That's the rule of thumb then for audio equipment then?

It's mine. Well, it's H. Jack Allison's rule of tens, converted to audio. This just states that if you have two things and one's got less than 1/10 of the effect of the other, you can ignore the little-effect one with less than 10% error. So if what you want is audio voltage, and you're worried that the impedance will cause loss, then if the source impedance is less than 10% of the load it drives, you get less than 10% loss of signal voltage.

Careful. "Matching" as in "making equal or similar" is not the right word. For voltage based audio you want to MISmatch in the direction of the driver being much lower impedance than the input/load it drives. But yes, what you describe with the RV3 driving will work fine. It's MISmatched correctly.

D'oh, my post is a trainwreck. Yeah, I realize that you'd want to drive a load with a a higher current, thus wanting low to high impedance. Sorry, that's what I meant. In any case, a buffer could take care of the improper/poor impedance combinations disregarding levels/etc, right?

Yes. A buffer is in fact the standard answer for driving a low impedance load from a high impedance source.

So then a low output impedance from the rack unit (650ohms let's say) would  drive the amp pretty hard, wouldn't it or is it typical considering that even the RV-3 states an output impedance of 1k? Talking about pedals and typical guitar gear here.

The idea is to make the driving impedance be less than 1/10 of the input impedance it drives. A guitar amp at low frequencies has an input impedance of  - typicall - 1M. Some older solid state amps and most stereo hifi amps have input impedances down to about 10K. For the guitar amps, source/driving impedances up to about 100K are fine. For the stereo stuff, you want to get down to 1K or lower, which is one reason a lot of pro audio stuff runs lines at 600 ohms.

Notice that all of this is because we send signals as audio *voltages*. If we used current, not voltage, as a signal standard, we'd want to inverse things, using an input impedance that's *less* than 1/10 the impedance of what drives it so you would not reduce the current (and hence received signal) that flows into the input. Mirror image world, but just as valid.