Relationship Between Input Cap and Input Impedance?

[Ohioisonfire]

This is sort of my first post (save for a few embarrasing posts made years ago during a failed attempt at getting into DIY pedals), so I apologize in advance if I've done anything wrong.

I have spent a lot of time playing in LTspice lately simulating a few circuit ideas I have.  The primary goal is to get a very low input impedance (Z-in) out of an op-amp circuit to experiment with loading of the guitar (ie FF or RM).

I set up a generic inverting op-amp circuit with a large input cap value, with the intent of tweaking it later.  I then adjusted the two op-amp gain resistors (R3 & R5) to get the Z-in (~12K) and gain I was looking for.  Based on the simulations the gain and Z-in were great.  I tweaked in the input cap to give me the frequency response I was looking for, but then noticed I was no longer getting the same loading and associated high end loss.  I went back to double check the Z-in and noticed it was ~34K with the lowered input cap value.

I have done a lot of searching but not found the answer I am looking for, is there this strong of a relationship between input cap and Z-in?  or am I not understing how to use LTspice properly? I am still new to this world and trying to learn.  When I look at a circuit analysis (RM on Electrosmash for example) of low Z-in circuits I do not see them referencing the input cap in the calculations for Z-in.

Couple of notes:  I can get a simulated Z-in of ~12K if I use values of .013uF (C3) and 4.7K (R3), but this doesn't make sense to me.  I also have a similar circuit with a non-inverting op-amp and it does a very similar thing. 

Circuit "as designed":


Circuit tweaked to get "correct" simulated Z-in:

[antonis]

is there this strong of a relationship between input cap and Z-in? 

For frequencies at which cap's impedance (capacitive reactance, set by 1 / (2π*f*C) becomes comparable to resistive Zin, yes..
e.g. 5nF cap at 1kHz exhibits an impedance of about 31k8.. This impedance is considered in series with 12k till signal reaches virtual ground (inverting input)

P.S.
You'll always have to take into account signal sourse (Input) impedance, due to votage dividing effect..

[antonis]

On a second glance, you can face it as a C3/R3 High Pass Filter of 0.159 / (R3*C3) cut-off frequency..
(2650Hz and 2600Hz respectively)

So, for any given frequency, lowering 12k to 4k7 (2.55 times) needs to raise 5nF up to 13nF (2.55 times) to maintain same impedance..
(and vice versa, of course..)

[Rob Strand]

The way to estimate what is going on is,

XC = 1/(2*pi*f*Cin)

Zin = sqrt(Rin^2 + XC^2)       ; impedance of Resistor Rin in series with Capacitor Cin

At 1kHz and Cin = 5nF, XC = 31.8k
Zin = sqrt(12k^2 + 31.8k^2) = 34k

When the frequency is above the cut-off antonis mentions, f = 1/(2*pi*Rin * Cin) the input impedance is largely determined by  Rin (and nearly equal to Rin).

Below the cut-off the input impedance is determined by the capacitor.

The important thing is the impedance increase as the frequency decreases.  That's the nature of the circuit and you can't change it.     Looking at the impedance at 1kHz is to some extent arbitrary as at some lower frequency the impedance will be higher and at some higher frequency the impedance will equal Rin.

If the changes in input impedance need to be limited what you can do is add another resistor to ground before the input cap.  For a single supply you might find returning resistor to 4.5V is better in some cases.   What happens in this case is the added resistor puts a limit on how high the input impedance can go at low frequencies.

If the added resistor is 22k then input impedance never gets higher than 22k at low frequencies.    At high frequencies the input impedance is the added resistor in parallel with Rin=12k, Zin = 12k//22k = 7.8k.

If you use a lower value say 4.7k then the input impedance is 4.7k at low frequencies and  Zin=4.7k//12k = 3.4k at high frequencies.     The input impedance variation is  now less.

If the added resistor makes the impedance too low then you need to scale up all the resistors in the circuit and scale down all the capacitors.    The thing you don't want to do is set the added resistor to the impedance you want and make Rin enormous.  That will reduce the variation in input impedance but a high Rin will make the circuit noisy and susceptible to interference.   These things need to be done with some practical constraints.

[antonis]

Shouldn't be more convenient to represent capacitor impedance with XC (instead of XL) for not making us searching for some invisible circuit inductor..?? 

If the changes in input impedance need to be limited what you can do is add another resistor to ground before the input cap.  For a single supply you might find returning resistor to 4.5V is better in some cases.   What happens in this case is the added resistor puts a limit on how high the input impedance can go at low frequencies.

Or implement non-inverting configuration (signal at + input) and use that resistor for non-inverting input bias, placed after Cin, of course..
Obvious benefits should be low input impedance (almost constant for input HPF cut-off frequency set much  lower of the lowest frequency of interest) and no inversion of signal phase..

[PRR]

For an input (low-hiss) stage, inverting connection is almost always a poor choice.

Use standard non-inverting, fairly high bias resistor (1Meg) with like 0.01u input cap, then shunt a 12k across the jack. This can be varied to taste without changing gain (loading will chnage of course).

[Rob Strand]

Shouldn't be more convenient to represent capacitor impedance with XC (instead of XL) for not making us searching for some invisible circuit inductor..??

Thanks for picking that up.   I fixed it.  Not long ago I was doing a lot of calculations with XL and it somehow got stuck!  (For most stuff I actually use complex numbers: ZL=sL and ZC=1/(sC)).

[antonis]

For most stuff I actually use complex numbers: ZL=sL and ZC=1/(sC).

Do you perform calculations for transfer functions with zeros and poles by heart..?? 

[anotherjim]

One spec for opamps that gets little attention is the input resistance of the opamp itself. A NE5532 may sound different from a TL072 for this reason alone.

[Ohioisonfire]

Thanks for the input everybody!  Lot's of good points here that I have been playing with.

For an input (low-hiss) stage, inverting connection is almost always a poor choice.

Use standard non-inverting, fairly high bias resistor (1Meg) with like 0.01u input cap, then shunt a 12k across the jack. This can be varied to taste without changing gain (loading will chnage of course).

This was actually the "original" circuit I came up with.  Worked pretty well and simulated pretty well.  I was researching pros/cons of non-inverting vs inverting op-amp circuits and there were two things that stuck out to me and led me to investigate an inverting design:

• The inherent lower input impedance that would use less parts.
• The different clipping characteristics with diodes in the feedback loop (not shown in the schematic I shared)

I have modeled all of these other suggestions and have learned a lot from doing so.  Let me ask another question:

The Z-in of a rangemaster (used as an example here because it is a well known, very simple, low Z-in circuit) is said to be ~10-12K.  This is calculated based on the resistor divider and the Z-in of the transistor.  With a C-in of .005uF in front of that to create a HPF at 2600Hz, does the input cap change the Z-in of the whole circuit like it does with mine?

If the changes in input impedance need to be limited what you can do is add another resistor to ground before the input cap.  For a single supply you might find returning resistor to 4.5V is better in some cases.   What happens in this case is the added resistor puts a limit on how high the input impedance can go at low frequencies.

Some great insight in your reply Rob, thank you.  Very useful to improve my knowledge on the math behind these things.  I am curious about your statement regarding tying the resistor before the input cap to ground vs tying it to 4.5V.  In what cases would either option be better or worse?  If I tied it to 4.5v I would need a decoupling cap before it correct?

[antonis]

I am curious about your statement regarding tying the resistor before the input cap to ground vs tying it to 4.5V.  In what cases would either option be better or worse?  If I tied it to 4.5v I would need a decoupling cap before it correct?

Resistor before input cap forms a "direct" voltage divider with signal source impedance (as long as bias resistor is much bigger).. 

That's exactly what PRR said above..
.

[Rob Strand]

Do you perform calculations for transfer functions with zeros and poles by heart..?? 

To me it's pretty much the language of analog electronics.   You don't need to think about it after a while.
Some people have learnt to speak klingon!

Some great insight in your reply Rob, thank you.  Very useful to improve my knowledge on the math behind these things.  I am curious about your statement regarding tying the resistor before the input cap to ground vs tying it to 4.5V.  In what cases would either option be better or worse?  If I tied it to 4.5v I would need a decoupling cap before it correct?

If you just put a resistor in there then you need to consider the DC loading as well as the AC loading.  For example if your circuit is fed by an opamp biased at 4.5V then connecting the added resistor causes no additional (DC) current but if you connect it to ground then there a DC current will flow from the opamp IDC = 4.5V / Rgnd.

If the stage driving the resistor was a JFET biased to 6V then connecting to either 4.5V or ground will cause DC loading issues.  In this case it's best to put a big'ish cap between the JFET and your circuit to eliminate the DC loading all-together.

Both these examples are artificial to a large degree because the opamp case doesn't need the added impedance leveling resistor.  In the JFET case you might try to design the JFET stage to have a low output impedance so it doesn't care about the varying AC impedance and if that fails you might add a buffer to the output so it doesn't care about the loading.

The cases where you actually need to care about the loading enough to add an impedance leveling resistor aren't that common.  It would be more common to step around the problem so the design doesn't care the frequency dependent loading. 

It's OK to think about this stuff to improve your electronics ninja skills but it's more normal to step around things.

So I guess that leads to the question to why do you want to level the input impedance?

[antonis]

Rob, I think OP refered on placing  the input impedance setting resistor in place of bias one (+4.5V to + input) for a non-inverting configuration..

[Rob Strand]

Rob, I think OP refered on placing  the input impedance setting resistor in place of bias one (+4.5V to + input) for a non-inverting configuration..

I'm not *exactly* sure what he's doing.

FWIW, the input impedance calculation is sort of the same in both cases since both have a cap and resistor in series.

[antonis]

I'm not *exactly* sure what he's doing.

Me either..

I presume we have to wait for OP to clarify the signal input configuration (inverting or non-inverting) and the "impedance setting" resistor arrangement (series or shunt)..

[Ohioisonfire]

Rob & Antonis, thanks your continued engagement here.

I am curious about your statement regarding tying the resistor before the input cap to ground vs tying it to 4.5V.  In what cases would either option be better or worse?  If I tied it to 4.5v I would need a decoupling cap before it correct?

Resistor before input cap forms a "direct" voltage divider with signal source impedance (as long as bias resistor is much bigger).. 

That's exactly what PRR said above..
.

I understand the idea of placing the resistor prior to input cap to form a voltage divider.  That simulates really well and makes sense to me.  I was trying to ask Rob about the difference between tying it to ground, or tying to 4.5V bias.  Which he exaplined later, thank you.

So I guess that leads to the question to why do you want to level the input impedance?

Great question, I probably should have made that more clear in the original post.  I am trying to simulate the pickup loading a RM/FF/etc. would cause.  I want to experiment with that relationship using different circuits, diodes, etc.  Nothing very technical and probably not a good idea to start the circuit "wrong" to begin with, but just an idea I had.  Maybe there is a better/more elegant way to do it?

[antonis]

I am trying to simulate the pickup loading a RM/FF/etc.

Why don't you do it implementing "real" circuits..??
Range Master only needs 7 items and pickup only 3 (or 6 in case you include tone and volume controls..)

[Ohioisonfire]

Why don't you do it implementing "real" circuits..??

That is definitely the right question to ask, and probably (definitely) the right answer.  I am just experimenting and having a lot of fun doing it.  I am also doing a lot of studying/research in parallel on how to design things the right way.  This all started with a "what if" something along the lines of "how would that pickup loading work with different circuits/technologies that aren't FF/RM or GE transistor based?" 

Is there something to be found here?  Probably not, but I'm having fun learning about it.

[Rob Strand]

That is definitely the right question to ask, and probably (definitely) the right answer.  I am just experimenting and having a lot of fun doing it.  I am also doing a lot of studying/research in parallel on how to design things the right way.  This all started with a "what if" something along the lines of "how would that pickup loading work with different circuits/technologies that aren't FF/RM or GE transistor based?"

Is there something to be found here?  Probably not, but I'm having fun learning about it.

I think you are on the right track.   Loading is part of the sound on some pedals.

There's some things about spice worth noting.   This is a little technical.   The main point is not to detract you from what you are doing, which is fine, but to warn you sometimes there's more going on and spice has it's limits.   It's good to at least know about it.

One thing to watch out with spice is the AC analysis is for *small* signals.   [The underlying model is for small signal even when the signals aren't actually small!]

That means the effect of clipping diodes is essentially removed from the circuit.  When diodes clips the impedance across the drops and it can effect the circuit frequency response.   

The input impedance of the Rangemaster is the 68k in parallel with the 470k and the input impedance of the transistor.   The input impedance of the transistor depends on the biasing and the gain of the transistor, so you have play with the gain.  Spice will model all that with AC analysis no problem.

However if the signal is large enough to cause the transistor to saturate, in which case the collector voltage is pulled down to the emitter, the input impedance no longer works like a small signal circuit and AC analysis doesn't simulate what the circuit is actually doing.   You can try it.  Input 10mV, 1V, 100V and the AC analysis will show the same response just that the voltages are scaled up.   In reality the output will clip at some point (think also how a 9V circuit can general 10000V output, it can't!).  When the transistor clips the input of at the base of the transistor looks more like a clipping diode for positive input swings and is almost open-circuit for negative input swings.  The small-signal input impedance spice uses for AC analysis is no longer valid.

The way to look at these "large signal" effects is to use the transient analysis.     These produce waveforms in time at a signal frequency.   It's not as convenient as AC analysis which sweeps the frequency.   The waveforms are harder to interpret.   This is a whole topic in itself.

Dont' let all this distract you from using AC analysis.  I use it all the time for exactly the same things you are trying to look at.   The take home message is it has it's limits.   After have put some hours into spice at some point you might want explore transient analysis.

[ashcat_lt]

...then shunt a 12k across the jack. This can be varied...

Most guitars have this built in already.