Lets talk about input impedance...

[Stickman393]

A'ite,

I'm going to use an example of a pedal that I recently built. I've attached a simplified version of the circuit: this is the "wah" section of the Cosmo King Wah/Ringmod from DEFX:



This was done with the assumption that the wah pot was at it's midpoint. I then modeled the input impedance by plotting V(input)/I(V2):



Assuming that I did all that right, this confirms something that I had suspected: that this design has a quite low input impedance.

If I'm thinking about this correctly: if an effect's input impedance is too low, that reduces the overall peak to peak voltage that a pickup is able to produce because the effect circuit is, for lack of a better phrase, "Current Hungry". Therefore a low input impedance will load down the guitar pickups: any increase in peak to peak voltage will increase current as well. Because the guitar's pickups are limited in how much current they can produce, a low input impedance will sacrifice some voltage for current.

By using a higher input impedance, an increase in signal voltage will result in a smaller increase in signal current. This allows for a wider dynamic range. Did I get that right?

So...what are one's options if one wants to "correct" this in an existing design? I'm familiar with seeing 2.2M resistors tied to ground immediately after an input: doing so in LTspice certainly does the trick. I realize I'm creating a voltage divider at this point if I do so, which will attenuate the signal to some extent. Though how much I can't say: 27K is, what, 1.2% of 2.2M? Realistically speaking, the majority of the current should still be traveling through the 27K resistor.

Are there any considerations there that I'm not seeing?

I suppose another option here would be to add a buffer to the input stage, though that would be more involved.

Any insight y'all could provide would be greatly appreciated. Stickman bang rocks to get er, learn good.

 

[jwin615]

By using a higher input impedance, an increase in signal voltage will result in a smaller increase in signal current.

Not sure that follows
V
IxR

 

[Stickman393]

Lemmie rephrase that to more closely reflect what I mean:

Take two scenarios, one with a smaller impedance, one with a larger impedance. Call it, 1 ohm and 10 ohms.

Example 1) 10V across 1 ohm will yield 10 amps. 20V across 1 ohms will yield 20 amps.

Example 2) 10v across 10 ohms will yield 1 amp. 20V across 10 ohms will yield 2 amps.

Within the same circuit a change in voltage will correlate to a proportional change in current. A circuit with a larger impedance will correlate to a lesser *absolute* change in current, because the denominator has changed.

 

[Stickman393]

I should make clear: I learn best when I'm forced to explain things in my own words and/or work through problems. Preferably at gunpoint. Thats when all the good le'rnin' happens. So I'm talking through this in an effort to try to get closer to the truth. Feel free to call out any fucked up logic on my end, cause I wanna learn and maybe become 3D some day.

That being said: I might have been going about this spice analysis wrong.

I had read that it would be more useful to model this based on a current source, rather than a voltage source. I tried that, and lo and behold, that little 2.2m to ground input resistor does not in fact increase input impedance.

Which, duh. Of course it wouldn't. We're talking about adding another resistance in parallel, which can only decrease the input impedance (1/Rt=1/R1+1/R2). That makes a ton of sense.

With that being the case, it seems to me that simplest way to solve this "problem", without more in-depth knowledge of how the circuit works, would be to add a simple input buffer. Instead of trying to modify the existing circuit to increase it's input impedance, I can tack a buffer with a high input impedance onto the input, which will then be able to provide the current required to drive the input of this circuit.

 

[YourGuitarist]

Low input impedance is IMO best thought of in terms of voltage transfer. The output impedance of your pickups (or preceding pedal) form a voltage divider with the input impedance of the pedal/amp they're plugged into. The voltage seen at the input of said pedal/amp is proportional to the voltage produced by the pickups multiplied by the input impedance and divided by the sum of input + output impedances.

Therefore as input impedance decreases in relation to the preceding output impedance, the voltage transfers decreases as well. Note that pickups contain real and reactive (inductance/capacitance) impedances which produces a curve of impedance by frequency. That is, the output impedance changes with frequency.

This is why treble loss is a common side effect of low input impedance. Because at high frequencies, the output impedance of a pickup rises due to the inductance.

 

[YourGuitarist]

Low input impedance is IMO best thought of in terms of voltage transfer. The output impedance of your pickups (or preceding pedal) form a voltage divider with the input impedance of the pedal/amp they're plugged into. The voltage seen at the input of said pedal/amp is proportional to the voltage produced by the pickups multiplied by the input impedance and divided by the sum of input + output impedances.

Therefore as input impedance decreases in relation to the preceding output impedance, the voltage transfers decreases as well. Note that pickups contain real and reactive (inductance/capacitance) impedances which produces a curve of impedance by frequency. That is, the output impedance changes with frequency.

This is why treble loss is a common side effect of low input impedance. Because at high frequencies, the output impedance of a pickup rises due to the inductance.

That being said input impedance can also be of consequence when following an active stage (ie pedal outputs).

In regard to your discussion of current, active elements (such as BJTs, opamps) have limited current capabilities which when paired with too low a load impedance, will clip due to the load drawing more current that the BJT etc can provide.

 

[Feral Feline]

I would just stick a buffer on the front.

Nonetheless, I've been wanting to better understand this topic that I've been passive-aggressively avoiding for so long, so ...
I think figuring out the fix (well, somebody explaining it to me in an Electrosmash-way) will do more to edjumicate me, than me tactlessly tacking a buffer up front.

Instead of sticking a Band-Aid buffer on the front end, what pray-tell can be done to fix the circuit?



Everything I've read thus far is oversimplified, simply "stick a resistor on the front end of the circuit" yet without addressing the associated reduced gain and possible frequency shift what with any surrounding capacitors... I just realised maybe they don't mean in series, but to ground or Vref or...?

• How to determine how big the resistor needs to be while allowing the most gain (if in series)?
• How much thermal-noise is introduced into the signal path (not a problem for a distortion pedal, I suppose, but a compressor or EQ)?
• What about the circuits that slap a variable resistor on the input, ie a pot as a pseudo gain control such as the Mr Squishy, how much variance to the circuit's input-impedance is the pot introducing?
• What are the questions I should be asking that I don't even know enough yet to ask?
• etc...

...The voltage seen at the input of said pedal/amp is proportional to the voltage produced by the pickups multiplied by the input impedance and divided by the sum of input + output impedances.
...

Those final two words in the quoted sentence, "output impedances" — is that referring to the output-impedance of the pickup/preceeding pedal, or the output impedance of the circuit suffering low input-impedance?



Build a pedal, they said; it'll be fun, they said.
Nobody mentioned how much dad-gummed learned maths would have to be mangled in the making...

 

[YourGuitarist]

Just the output impedance of the preceding stage.

 

[YourGuitarist]

Two details I see get glossed over are:

• Corner frequencies are NOT ALWAYS as simple as 1/2pi*R*C.
  • This formula applies only for 1st order filters composed of only a single resistor/capacitor.
• DC voltage sources act as AC GND paths. That is, a reference voltage such as the DC midpoint is still a pathway to GND. It is NOT GND though. Looking at the pedal below as an example.

Notice how Vref connects to R11/R12 & C11/C12. It can be seen that R12/C12 form a path to ground in series with R2. R12 is purely resistive while C12 forms a frequency dependent impedance that is often treated as a short to GND at audio frequencies. That is because capacitors provide an impedance that decreases with rising frequency.

Formally, this is Zc=1/(j*w*C). Where Zc is impedance of the capacitor in ohms, C is the value in Farads and "j*w" is a complex expression for frequency in radians. J is an imaginary number = to the square root of '-1'. The math involving j is really beyond the scope of this forum.

Also note for inductors, ZL=j*w*L; this better explains my point earlier about treble loss and pickup inductance.

A more rigorous understanding identifies Vcc as an AC GND as well, especially due to C11 forming a low AC impedance connection between the two. Therefore, R11/R12/C12 are effectively in parallel with each other and series with R2. Technically, the Vcc has an output impedance of its own, that to my understanding, is small relative to the circuit (think 10s of ohms compared to 1000s), which is why we assume it is GND for the signal.

Now returning to the more traditional input impedance section, we have R1/R2/R4/C1/C4.

Series impedances sum and parallel impedances (denoted by '||') can be treated as a single impedance = Z1*Z2/(Z1+Z2). Where Z1 & 2 are each path.

From here, the total input impedance can be found (neglecting the impact of the power supply):

R1 || (C1+ (R2 || R4 + (C4 || Zopamp) ) ).

One characteristic of parallel impedance is that it is ALWAYS lower than the impedance of the individual pathways. This is why despite the opamp having an input impedance of on the order of Tera ohms, the pedal's AC impedance will still be closer to ~500k.

If you want more detail, just ask, though I can get somewhat lost in this stuff.

 

[Feral Feline]

Thank you YourGuitarist. I've read through your post very carefully but am still somewhat lost in this stuff — any more detail would be more confusing, probably. I'll have to re-read it a few times and hope more of it sinks in each time. I'm basically mathematically-illiterate, so...

I'd have to look up the impedance of the op-amp in its datasheet to work out the example given, the LM308 of the Muroidea, to determine the Muroidea's input impedance — except the datasheet doesn't mention impedance of the op-amp at all. All I can do is plug in the values of the components:

1M || (22n + (1M || 1k + (1n || ?Zopamp? ) ) ).


It's late, I shouldn't be attempting to understand this when tired; It'll be difficult enough to ken when I have all my faculties together...

 

[Stickman393]

Super helpful information here. Greatly appreciated!

 

[Stickman393]

Two details I see get glossed over are:

• Corner frequencies are NOT ALWAYS as simple as 1/2pi*R*C.
  • This formula applies only for 1st order filters composed of only a single resistor/capacitor.
• DC voltage sources act as AC GND paths. That is, a reference voltage such as the DC midpoint is still a pathway to GND. It is NOT GND though. Looking at the pedal below as an example.

View attachment 81420

Notice how Vref connects to R11/R12 & C11/C12. It can be seen that R12/C12 form a path to ground in series with R2. R12 is purely resistive while C12 forms a frequency dependent impedance that is often treated as a short to GND at audio frequencies. That is because capacitors provide an impedance that decreases with rising frequency.

Gonna follow along here. Vref connects to R2, R12, and C12. Both R2 @ 1 Megaohm and R12 @ 100k ohms are resistive. C12 has an impedance that becomes smaller as frequency increases, and can be treated as a short between VREF and Ground at audio frequencies. This essentially creates a short to ground at audio frequencies.

That "short" is in series with R2, but parallel with R12. As far as audio frequencies are concerned, R12's impact is negligable..

Formally, this is Zc=1/(j*w*C). Where Zc is impedance of the capacitor in ohms, C is the value in Farads and "j*w" is a complex expression for frequency in radians. J is an imaginary number = to the square root of '-1'. The math involving j is really beyond the scope of this forum.

Oh my. I've gone cross-eyed.

Also note for inductors, ZL=j*w*L; this better explains my point earlier about treble loss and pickup inductance.

A more rigorous understanding identifies Vcc as an AC GND as well, especially due to C11 forming a low AC impedance connection between the two. Therefore, R11/R12/C12 are effectively in parallel with each other and series with R2. Technically, the Vcc has an output impedance of its own, that to my understanding, is small relative to the circuit (think 10s of ohms compared to 1000s), which is why we assume it is GND for the signal.

The previous example can be applied to the rest of the power supply insert as well. C11 connects Vcc to Ground at audio frequencies, which means that as far as audio frequencies are concerned Vcc and Ground are at the same AC potential. We can therefore think of both as being "ground" as far as audio signals are concerned.

Now returning to the more traditional input impedance section, we have R1/R2/R4/C1/C4.

Series impedances sum and parallel impedances (denoted by '||') can be treated as a single impedance = Z1*Z2/(Z1+Z2). Where Z1 & 2 are each path.

From here, the total input impedance can be found (neglecting the impact of the power supply):

R1 || (C1+ (R2 || R4 + (C4 || Zopamp) ) ).

The equation that determines the input impedance is found via the following:
1Meg || (0+ (1Meg || 1K + (0 || holyfuckton)))= Zin
1Meg || 1Meg || 1k || bigfuckinnumber= Zin
1/1meg + 1/1meg + 1/1K + 1/justgiveup = 1/Zin
0.000001 + 0.000001 + 0.001 + peepants = 1/Zin
0.001002 + doesn't really make a difference =1/Zin
998ohms=Zin @ audio frequencies

Is that correct? Because signal has a parallel path to ground through R4 & C4, and the resistance of that pathway is closer to 1k ohms. The input impedance is lower than one would presume though just looking at R1 and R2.

One characteristic of parallel impedance is that it is ALWAYS lower than the impedance of the individual pathways. This is why despite the opamp having an input impedance of on the order of Tera ohms, the pedal's AC impedance will still be closer to ~500k.

If you want more detail, just ask, though I can get somewhat lost in this stuff.

I'd love your feedback to confirm if I absorbed all that correctly, or if I need to re-think anything. That is extremely helpful in shaping the way I think about input impedance.

Much of my knowledge is borrowed from my trade as a commercial HVAC mechanic: previously I had been thinking about this in terms of VA vs Watts/ the power factor of transformers. This adds a new dimension that I'd love to chew on.

 

[Feral Feline]

...The equation that determines the input impedance is found via the following:
1Meg || (0+ (1Meg || 1K + (0 || holyfuckton)))= Zin
1Meg || 1Meg || 1k || bigfuckinnumber= Zin
1/1meg + 1/1meg + 1/1K + 1/justgiveup = 1/Zin
0.000001 + 0.000001 + 0.001 + peepants = 1/Zin
0.001002 + doesn't really make a difference =1/Zin
998ohms=Zin @ audio frequencies
...

Man, Stickman, you're like a Bounty, slurpin' up all that impedance info...

While I can only fantasize about soaking it all in...

 

[Stickman393]

In pee dance? I 'ardly know her! I'm just an HVAC guy. Beer can cold, good nuff. What what.

I probably should have mentioned that the way I went from parallels to a mathematical equation is the law of parallel resistances: 1/Rt=1/R1+1/R2+1/R3...etc.

My knowledge base contains lots of little pockets that I haven't quite peered into yet. Like...filters. 1/(2pi*R*C). Thats a new one to me.

But...I can see that the combination of R4 and C4 creates a simple low pass filter. Lets see here:

1/{2*3.14*1000(ohms)*0.000000001(farads)}=low pass frequency
1/0.00000628 = Low pass frequency
159,235-ish HZ.

All frequencies below that point will pass, all frequencies above that point will be shunted to ground if I understand correctly.

Does this mean that, above that frequency impedance is essentially 0, but below that point the impedance curve will be closer to that 998 ohm range?

 

[YourGuitarist]

Here's what I simulated. Converting from 113.77dB, I get ~488k peak impedance. That sounds pretty reasonable to me. Note that I added a dummy resistor R7 to measure the current into the entire input circuit (it small enough to be negligible). Measuring current over one component would give inaccurate results.

I am trying to calculate the impedance by hand and plot with Matlab but my algebra skills are s***. So attempt 3 it is.




Here's my approach so far...




I can see from my Matlab plot that there is an error. This is because intuitively, the impedance of the of the capacitors approaches zero with increasing frequency. The circuit will eventually behave as though the capacitors are 'shorted'. Thus, the HF impedance is dominated by the values of R1/R2/R4 but really just R1/R2 bc R4 is so much smaller. Imagine that the capacitors are removed, then R1/R2 are directly in parallel and thus can be simplified to a single resistance that is half their value.

The op-amp input impedance is so high that it virtually forms an open circuit at its input. Since it is in parallel to ground with all of the resistors and capacitors preceding it, and because a parallel impedance is ALWAYS LESS THAN the lowest branch, the impedance of the opamp can be ignored.

This is a good exercise for me, reveals where I need practice.

I would invite @Chuck D. Bones or someone else with some math/Matlab chops to critique my work.

 

[YourGuitarist]

Gonna follow along here. Vref connects to R2, R12, and C12. Both R2 @ 1 Megaohm and R12 @ 100k ohms are resistive. C12 has an impedance that becomes smaller as frequency increases, and can be treated as a short between VREF and Ground at audio frequencies. This essentially creates a short to ground at audio frequencies.

That "short" is in series with R2, but parallel with R12. As far as audio frequencies are concerned, R12's impact is negligable..



Oh my. I've gone cross-eyed.



The previous example can be applied to the rest of the power supply insert as well. C11 connects Vcc to Ground at audio frequencies, which means that as far as audio frequencies are concerned Vcc and Ground are at the same AC potential. We can therefore think of both as being "ground" as far as audio signals are concerned.



The equation that determines the input impedance is found via the following:
1Meg || (0+ (1Meg || 1K + (0 || holyfuckton)))= Zin
1Meg || 1Meg || 1k || bigfuckinnumber= Zin
1/1meg + 1/1meg + 1/1K + 1/justgiveup = 1/Zin
0.000001 + 0.000001 + 0.001 + peepants = 1/Zin
0.001002 + doesn't really make a difference =1/Zin
998ohms=Zin @ audio frequencies

Is that correct? Because signal has a parallel path to ground through R4 & C4, and the resistance of that pathway is closer to 1k ohms. The input impedance is lower than one would presume though just looking at R1 and R2.



I'd love your feedback to confirm if I absorbed all that correctly, or if I need to re-think anything. That is extremely helpful in shaping the way I think about input impedance.

Much of my knowledge is borrowed from my trade as a commercial HVAC mechanic: previously I had been thinking about this in terms of VA vs Watts/ the power factor of transformers. This adds a new dimension that I'd love to chew on.

I apologize but I'm not great at following other's work. One thing I see that I didn't explain correctly is that the resistance of a capacitor used IN EQUATION is the expression (1/Cs). S is equal to J*W where J is square root of -1 and W is the radian frequency.

 

[m4268588]

There is no need for R7. Click on the top part of the plot window to V(v1)/I(V1). Next, click the left end to change the selection from Decibel to Linear.

The input impedance is as follows:
1/(1/R1+1/(1/C1+1/(1/R2+1/(R4+1/C4))));
To make it polynomial
factor(1/(1/R1+1/(1/C1+1/(1/R2+1/(R4+1/C4)))));
Enter this at the Maxima prompt. If you don't have Maxima on your workbench yet, you can use it online. (http://www.dma.ufv.br/maxima/index.php?c=0&n=1)
I don't know how to use MATLAB, but I think there are similar functions.