Understanding impedance when it comes to daisy-chaining pedals

[PowerJack]

Assuming I have two pedals and each of them has an op-amp somewhere inside it. As I understand it the input of the op-amp is HIGH impedance, and the output LOW impedance. Not sure exactly what this means since I'm just starting my reading in the topic.

Still if I connect them in series like you usually do with guitar pedals, I would be connecting a LOW impedance output into a HIGH impedance input. In writing this doesn't sound right but in reality there is no problem with this.

Anyone care to explain in layman terms why this works?

[blackieNYC]

http://www.muzique.com/lab/imp.htm
Check this out.
Bigger impedance= bigger resistor= bigger voltage

But I have an additional question on the topic which I'm going to tack on to this one:
If I only have a SPST switch for bypassing an effect, the output will be connected to the input, but if the circuit must be connected to the input or output jack. Is it better to have the circuit input connected to the bypass signal thruput, or the circuit output?  I would think it preferable to have the input connected, since its greatest threat is impedance loading, while a circuits output will be putting out a certain amount of noise.  Especially if I am determining the input impedance with a resistor, which I can increase.  True?

[GibsonGM]

Ideally, you have MATCHING impedances....like, in radios, they will use a transformer to make the relationship nearly 1:1!   Lucky for us, we don't need to do that, ha ha.   Low out to high in is just FINE.   

High impedance essentially means the circuit will not *TAKE* much current from the block before it...and low output impedance means that it won't "resist" that taking.  Basically, the output and input of pedals creates a voltage divider with impedances instead of resistors (AC resistance is all that impedance really is, plus your DC resistance...of course, in some places there ARE resistors in series, too).    Take a good look at a schematic of one pedal's output vs. the input to the next one, and you'll see how that works. 

There isn't much need to give it too much more thought than that, until you decide to 'get into it' and study more about it.


2) I believe the one always connected, before nice bypass switches, was the input, making your statement true.  BUT - please don't do this; the results stink to varying degrees, and the lack of clarity you experience doing it is called "tone sucking".

You can learn as much as you'll ever want here!    http://www.geofex.com/article_folders/bypass/bypass.htm

[PowerJack]

It's all starting to make sense... Thanks!

[R.G.]

Still if I connect them in series like you usually do with guitar pedals, I would be connecting a LOW impedance output into a HIGH impedance input. In writing this doesn't sound right but in reality there is no problem with this.

Anyone care to explain in layman terms why this works?

How you match or mismatch impedances depends on what you want to wind up with - most voltage, most current, or highest power.

Every electrical source, including every circuit output, has some internal resistance (impedance) to current flowing out. If you measure the voltage on the output with no loading, just an AC voltmeter, it will read some voltage.  This is the largest voltage it can put out, because it is not loaded down. If you short the output - that is, apply a very low resistance like a wire, the output voltage across the wire is nearly zero, because the source cannot supply more current than its internal resistance/impedance lets flow.

An open output (that is, an output loaded with a very high resistance/impedance) puts out the most voltage it can. A shorted output puts out the most current it can. There is a very special case where an output is loaded with a resistance/impedance that's equal to the internal impedance - that gets you maximum POWER delivered to the load.

In pedal work and most non-power-amp audio work, you don't want maximum current flow or maximum power flow. You want the most signal voltage to be supplied to the next stage. So you want the output impedance of a circuit to be low(ish) compared to whatever loads it, so the voltage is not dropped by the loading. This is helped even more if everything has a high input resistance/impedance. Again, the idea is to get the most signal VOLTAGE into the next input.

Think of the water system in your house. If you are in the shower and someone flushes a toilet nearby, the water pressure drops noticeably, right? That's because the water system in the house has some internal impedance. When all the valves and faucets are shut, the pressure is highest. If you open the shower valve, you get some water flow, and you can see by the arc of the water an indication of the pressure. Flushing a nearby toilet adds more load to the system by letting more current flow, and the pressure drops.

For audio you almost always want the most voltage(pressure), not the most water flow.

Yep, power amps are like fire hoses... you want the most flow... well, you see. 

[PowerJack]

R.G., I've been keeping this text document on impedance where I copy pasted bits I collected on various posts, forums, etc, with things that I thought would help me understand this concept. Your explanation is by far the one that I understand best. I'm really glad you posted it!

Here are a bunch of snippets from my notes on this topic that have also helped me figure this out (for future readers of this post):

"Even when things are very clear and linear, it's important to note that impedance just describes a ratio-- it doesn't describe the limits to the system, and it's not "bad." You can definitely get as much current/velocity as you want (in an ideal system) by adding more voltage/pushing harder."

"The impedance of a circuit element is the ratio between voltage and current in that element."

"In the case of a high impedance guitar output (7,000 to 15,000 Ohms or more) driving a relatively low impedance input of a mixer (2,000 to 10,000 Ohms), it's like connecting a garden hose to a fire nozzle."

[Digital Larry]

I think of impedances as springs.

A low impedance is a stiff spring.

A high impedance is a... err... not very stiff spring.

If you connect two equally stiff springs together, whether they are stiff or not so stiff, connect the far end of one to "ground", aka some unmovable point, then wiggle the free end of the top one, the middle point will wiggle right the the middle.  Congrats, you've just created a voltage divider.

If you connect a stiff spring to a not so stiff spring, and wiggle the stiff spring while the far end of the not so stiff spring is attached to ground, the middle point depends mostly on the stiff spring and not so much on the other one.  Switch the ground and source (of wiggling) and the relationship is still true.  The stiff spring (relatively low impedance) dominates the proceedings.

[R.G.]

Understanding impedance is easiest with three tools: Ohm's Law, and Thevenin/Norton equivalent circuits. Well, OK, also the concepts of an ideal voltage source and an ideal current source.

Ohm's Law is the biggie. Staying with resistors for the beginning understanding, The Law says:

V = I * R, R = V/I, and I = V/R, which are all the same thing viewed from different viewpoints.

An ideal voltage source is an imaginary black box which provides a fixed voltage on its two terminals, regardless of the current involved.  A 10.0V ideal voltage source provides 10.0V whether its output current is zero, or 10,000,000 amperes. Of course - and frankly, thankfully! - none of these exist in the real world.

Every real-world voltage source has some internal impedance. This is most easily thought of as an internal resistor in that black box with the voltage source. If you have an ideal voltage source inside a box, in series with an internal resistor, you have a good model of a real world voltage source. A 9V battery, for instance, when fresh, has an open circuit voltage of 9..4-9.6Vdc. But if you hook a fat copper wire between + and -, you could measure the short-circuit current. For a fresh 9V battery, this will be from one to a few amperes. And from Ohm's law, we could calculate that the internal resistance R is 9.4V/2A = 4.7 ohms (just picking reasonable numbers out of the air for illustration).

As the battery runs down, the chemicals get more resistive inside, so the external voltage goes down because there is less chemical action, and the short circuit current goes up because the internal resistance goes up.

A car battery might have an open circuit voltage of 13.2V and a short circuit current of over 1000A (that's a realistic number), or an internal resistance of 13.2 milli-ohms.

But you get the idea - source resistance is the open circuit voltage divided by the short circuit current.

The model of a voltage source as an ideal voltage source in series with some impedance is the Thevenin Equivalent Circuit. All real world sources can be modeled this way. An alternate is the Norton Equivalent Circuit, which is an ideal current source (any voltage out, even infinite current, to force the actual current to flow) paralleled by a resistor. The resistor in a Norton Equivalent prevents the voltage from rising to infinity; it can be at most I *R. If you short circuit a Norton Equivalent, the current that flows is just the current the source wanted to do all along, but the voltage drops to zero.

Notice that you can change a Thevenin to a Norton; they are both characterized by an open circuit voltage and a short circuit current. But that's for a different lecture. 

For audio signals, we mostly think of sources - such as the collector connection of a bipolar amplifier, or the output of an opamp, as a Thevenin Equivalent. When there is a high enough resistance load on it, you get so close to the open circuit voltage that you can ignore the difference. When you put a resistor across the output, the voltage drops because the internal resistance of the Thevenin circuit acts with the extra external resistor to make a voltage divider. The voltage that appears on the output pin is the voltage which the two resistances, internal and external, let happen, just like a volume control.

If you have a bipolar transistor circuit with an output that is the collector of the transistor through a (for example) 10K resistor to the power supply, the transistor's action is much like a Norton Equivalent. The transistor collector tries to pull a current through the resistor. This current makes a voltage on the resistor/collector, and this is what we see as signal. This produces some open-circuit voltage at the collector. If we put a resistor load on the output (through a capacitor, to block the DC), then the output voltage will sag lower than the open circuit voltage.

For instance, if the output voltage is swinging +/-1V peak with no external load, and we hook a 1M resistor load to it, the voltage drops to 1M/(1M+10K) = 0.990099... of the truly open circuit voltage. That's closer than the accuracy of your meter to no drop at all. If we load it with 100K, the voltage drops to 100K/(110K) = 0.90909... peak. Now we're losing about 9% just because we loaded the output. If we put a 10K on the output, the voltage drops to 10K/(20K) = 0.5V peak. We lost half the signal voltage to the internal resistance!

And that's the essence of why we want low output impedances and high input impedances. Low output impedances, ideally much lower than that 10K example, can drive things connected to them with little noticeable loss of signal voltage. High input impedances can keep nearly all the output voltage from a signal source without having it sapped away by the internal resistances of the signal source.

For audio, then, we want low output impedances, and high input impedances so we don't have to go through the mess of calculating and adjusting for how much gain (amplification) and loss ( sag from loading) the circuits give us, and we can just connect things without the losses.

[ashcat_lt]

Definitive as always, RG.  But, of course, you're talking about pure resistances there.  It talks about broadband attenuation, but doesn't explain "tone suck", which is the main concern for most guitarists investigating "impedance issues".

Luckily, while actual impedance (including both reactance and resistance) can be a complex (as in imaginary numbers!) to work out, it is actually pretty easy to visualize in the same way that RG laid out.  What it takes is to first identify the reactive components and then consider them as though they were frequency dependent resistors.

A guitar pickup is not an ideal voltage source.  There's a lot of very thin wire there, which makes the Norton series resistor pretty big, but also introduces inductance, which basically means that that internal resistance (the top of the voltage divider) is larger for higher frequencies.  When the "top resistor" gets bigger, it drops a larger share of the voltage, leaving less to be sensed across the load.  So higher frequencies are attenuated more - it's a low pass filter.  If you reduce the "bottom resistor", you get more attenuation for a given top resistor, so that lower frequencies are attenuated more than with a bigger load impedance - the cutoff frequency is reduced.  So you get more treble out of the pickup with a higher load impedance.

But that ain't it!  Unless you've got an onboard buffer, the signal passes through a pair of wires very close together and separated by a non-conductive material.  We usually call this a cable, but it's also the definition of a capacitor.  So you've got a capacitor in parallel with the load.  Think of this as though the "bottom resistor" gets smaller with higher frequencies.  That's opposite to the action of the inductance, but it's also in the opposite position on the divider, so again we have the bottom resistor getting proportionately smaller with higher frequencies.  If the source were purely resistive (no inductance), then reducing the load resistor parallel to the capacitance would not affect the cutoff frequency, just create broadband attenuation.  Changing the "top resistor", though, moves the cutoff.  The low-Z (mostly resistive) output of an opamp makes that top resistor smaller enough for reasonable cable capacitance that high frequency attenuation isn't significant until the frequencies get really high.  But remember the guitar pickup, even without inductance, is a really big resistance.

It happens to work out to where the cutoff of the inductive filter tends to get close to the cutoff of the capacitive filter, and where they overlap you normally end up with some resonance - an actual boost at certain frequencies.  This is counterintuitive to what we've discussed so far, and is where your imaginary numbers and phase angles and vector math comes in.  I'm not going there.  The "frequency dependent resistor" gets you close for purposes of conceptualization, just be aware that resonance will modify things a bit.

[PRR]

Read R.G.'s essay.

Read R.G.'s essay.

Read R.G.'s essay.

> Ideally, you have MATCHING impedances

ONLY when you are *desperate* for signal-gain.

When you want Power To A Load, you MUST "mis-match" impedances or drown in your losses.

As for Gain: as soon as you have Amplifiers which don't cost a LOT (around 1920), it is very-very wise to design on NON-Matching principles, using an additional tube (etc) as needed to cover the (often small) signal loss.

Also-- Impedances are rarely "perfect". (If they were, we'd just say "resistance".) A guitar pickup is 5K in bass, 50K midrange, 200K treble-peak, dropping again to 10K at the top of the audio band.... how do you match to that? The little 10K transformers on R.G.'s splitter are really 800 ohms sub-bass, get near 10K in mid-bass, rise over 50K in mid-range, then fall-off again. A tone-stack will usually be Meg-ohms sub-bass and 100K at far-treble. Even a "simple wire", like the power line to my house, is a mix of resistance and reactance.

> In the case of a high impedance guitar output (7,000 to 15,000 Ohms or more) driving a relatively low impedance input of a mixer (2,000 to 10,000 Ohms)

Aside from your water-spray visions.... the guitar pickup evolved to work with a high impedance load (vacuum tube grid). Because it has impedance that varies with frequency, if you connect a low Z load you get a very different frequency response, not what you get with hi-Z loading.

> connecting a LOW impedance output into a HIGH impedance input. ....this doesn't sound right

This fooled wiser men than us back in the 1870s.

[duck_arse]

As the battery runs down, the chemicals get more resistive inside, so the external voltage goes down because there is less chemical action, and the short circuit current goes up because the internal resistance goes up.

does the short circuit current increase with the failing chemicals?

[R.G.]

does the short circuit current increase with the failing chemicals?

DOH!!!

Short circuit current goes DOWN with the failing chemicals.

I knew that. I was just testing you ...

Right...

Good catch. Thank you.

[duck_arse]

we are usually too scared to say "RG, are you just testing us with that last ..... ? etc".