How do I control input and output impedance?

[nashville]

Hello! I want to put together breadboards of the Fender, Vox, and Marshall tone circuits as they're shown on this tone stack calculator:

https://www.guitarscience.net/tsc/fender.htm#RIN=38k&R1=100k&RT=250k&RB=250k&RM=10k&RL=1M&C1=250p&C2=100n&C3=47n&RB_pot=LogA&RM_pot=Linear&RT_pot=LogA

I have all of the parts, but I don't know how to control the input and output impedance.

Fender: 38k input, 1M output
Vox: 717 input, 600k output
Marshall: 1300 input, 517k output

I tried the Fender circuit with my HX Stomp set to line out (10k output) before it, and a Boss pedal (1M input) after it, and the circuit functioned like the TMB knobs on my Twin, so I know getting input and output impedance in the ballpark is necessary. I just don't know how. Help me, please. Thanks!

[antonis]

Can't figure out the way for Zin & Zout calculations made..

R_IN should be the output impedance of preceding stage and R_L should be the input impedance of succeeding one..
IMHO, the "correct" way to make comparisons between various tonestack configurations is to implement the same (constant) R_IN & R_L for all of them..

Of course, there are ways to calculate any Tonestack In & Out impedances, by their own (Zin at R1/C1 joint and Zout at RT wiper) but it should need anvanced math knowledge due to each particular frequency band filter loading the others..

[Rob Strand]

You will find how the 38.5k ohm and 777 ohm  (slightly different to 717) values are calculated here,

https://aikenamps.com/index.php/designing-common-cathode-triode-amplifiers

Thanks to Mr Aiken.  That stuff has been on the web a long time.

A lot more good reference info here,
https://aikenamps.com/index.php/white-papers

[antonis]

You will find how the 38.5k ohm and 777 ohm  (slightly different to 717) values are calculated here,
https://aikenamps.com/index.php/designing-common-cathode-triode-amplifiers

Sorry but who said that OP's Tonestacks are driven from a CC Triode amp stage..?? 

[Rob Strand]

Sorry but who said that OP's Tonestacks are driven from a CC Triode amp stage..?? 

There's some secret handshakes going on here  .

The values of the input and output impedances in the Tonestack calculator come from the original circuits.   So you have to look-up the original circuit and then analyse the circuits to work out what part values and Rin and and Rout to use.

Then there's the question of which version of the circuit we are actually talking about!   The resistor and cap values help narrow it down but at the end of the day you have to go hunting for the corresponding schematic.

Another way to say it is not all Fender/Vox/Marshalls have the same circuit or same tonestack values.   If you look at enough schematics you will see all sorts of circuit and part value variations.  In some cases even different Rin and Rout!   If you look close you will also see the pot tapers are different on different models - so tone control 5 isn't always the same response!

Then you have people modding tone stacks using Marshall part values but the circuit they are modding is a Fender circuit with a completely different "Rin" to the Marshall circuit.   The response will be quite different.

20 years ago I spent ages looking into this stuff.     There's headaches in the details ...

[antonis]

There's some secret handshakes going on here  .
The values of the input and output impedances in the Tonestack calculator come from the original circuits.

My bad..

I thought that R_INs was set by OP for some Tonestack driving stages of his own awareness..

[Rob Strand]

I thought that RINs was set by OP for some Tonestack driving stages of his own awareness..

(Normally) You can set the values in the Tonestack simulator software but they start off with a default set of values that matches a particular model amp.

[Vivek]

My spice analysis of active baxandall type tone controls showed that the input impedance of that circuit varied by position of tone controls and input frequency, as well as choice of component values.

The effective input impedance would typically be from 500 ohms to 25k ohms.

That lead me to believe that the baxandall should be fed by an earlier stage with very low output impedance, maybe below 100 ohms.


I did not study effect of different output impedances of earlier stage on the characteristics of the baxandall.

[antonis]

the baxandall should be fed by an earlier stage with very low output impedance, maybe below 100 ohms.

Not necessarily but it should be good to know driver stage impedance to predict Tonestack particular attitude..

[nashville]

My #1 question right now is what do I put on the breadboard to guarantee the tone stack sees "X" input impedance and "Y" output impedance?

[anotherjim]

That's what I thought you wanted. If you use an opamp before and after the tone stack...
The opamp has a low output impedance that can be considered near-zero, then you simply insert a series resistance of the required Rs value into the front end of the tone stack.
After the tone stack, a non-inverting opamp +in pin can be considered an infinitely high load impedance, so you place a resistance there to 0v or Vref of the required RL.

[antonis]

After the tone stack, a non-inverting opamp +in pin can be considered an infinitely high load impedance, so you place a resistance there to 0v or Vref of the required RL.

Actually, RL should be considered the parallel equivalent of bias and shunt resistors..
Or, simply implement desirable RL value as non-inverting input bias resistor..

Or, are we talking baout the same case..?? 

[anotherjim]

Ermmm, what I wrote is that...

...non-inverting opamp +in pin can be considered an infinitely high load impedance, so you place a resistance there to 0v or Vref of the required RL.

[zesak]

You could use LTSpice, and it will tell you what the input and output impedance of the chosen tone stack is.

[PRR]

> I don't know how to control the input and output impedance.

Clarify your question. You don't have Fender tube stages on hand so you want to fake it with a signal generator and load resistor?

The Fender 12AX7+100K plate node tends to look-like 39k +/-10%.

A signal generator tends to be under 600 Ohms, often under 100 Ohms, so pencil 150 Ohms (H-P 200AB is 130 Ohms).

39,000 - 150 = 38,850 Ohms in series with the signal generator.

Nearest standard value is 39k.

The cathode follower tends to be under 1k. Taking 750r as a fair average of numbers people toss in:
750 - 150 = 470 Ohm nearest standard value.

[Rob Strand]

There's

My spice analysis of active baxandall type tone controls showed that the input impedance of that circuit varied by position of tone controls and input frequency, as well as choice of component values.

The effective input impedance would typically be from 500 ohms to 25k ohms.

That lead me to believe that the baxandall should be fed by an earlier stage with very low output impedance, maybe below 100 ohms.

I did not study effect of different output impedances of earlier stage on the characteristics of the baxandall.

Those are all valid points.

Many active Baxandall circuit use an input cap to roll-off the low end.  The input impedance depends on the bass control setting (more boost => lower impedance).  So you can't choose a cap for a single low-frequency cut off.

u could use LTSpice, and it will tell you what the input and output impedance of the chosen tone stack is.

There's two separate issue here:
- In a real circuit we have the source impedance of the circuit driving the tone-stack and the load impedance placed on the tone stack.  These affect the response and in order to achieve the same response as the targeted amplifier the tone-stack simulators add extra parts Rin and Rout so the simulated response matches the amplifier.
- The input impedance and output impedance of the tone-stack circuit itself.
   Those aren't directly useful but they will give an an idea how much the response will be affected *when* we
   add Rin and Rout.

In a real amplifier Rin and Rout parameters are already present.  They aren't part of the tone stack but they are required to emulate the effect of the surrounding circuit in a real amplifier.   The idea of the tone-stack simulator is to emulate the *actual* response of amplifier.   

If we wanted to build a circuit to accurately emulate the response of an amplifier then then we need to add Rin and Rout as real parts, unless the circuit already does this.   For example a JFET circuit driving the tone stack might have an output impedance of 10k, we could then for example add 27k in series with the output to emulate the 37k Rin.  On the other hand if we have an opamp driving the tone stack then it has a low output impedance and we would need to add 37k in series with the opamp output so the tone stack works normally.

A completely different strategy would be to not add any series resistances and try to tweak *remaining parts* to get a response close to the original amplifier.  That's a lot harder to do and I suspect things don't quite work out.

[nashville]

Clarify your question.

Desired signal chain:

1. Electric guitar > Fender/Vox/Marshall tone circuit > rackmount solid state power amp > guitar speaker cabinet

also

2. Laptop (for sine sweeping and getting a frequency plot) > interface > Fender/Vox/Marshall tone circuit > rackmount solid state power amp > guitar speaker cabinet

Problem:

When I only put the 3 caps, 3 pots, and 1 resistor in the tone circuit with no attempt to control the input or output impedance, the circuit doesn't work as expected. When I put a buffered guitar pedal before the circuit and a buffered guitar pedal after the circuit (10k, 1M) the tone circuit works much closer to what I expected. I understand that Fender/Vox/Marshall amps have different input and output impedances to their tone circuits and I've experienced what it's like to get those numbers wrong so I'd rather try to get those numbers right.

Hypothesis based on help from people in this thread:

If I put a Boss pedal (10k out, 1M in) before the tone circuit, here is how I could modify the circuit to get the same input impedances as the circuit would have in an amp:

Fender 38k input Z: add 28k series resistance after the Boss pedal
Vox 717 input Z: add 770 resistor to ground after the Boss pedal (10k and 770 in parallel = 707 total?)
Marshall 1300 input Z: add 1.5k resistor to ground after the Boss pedal (10k and 1.5k in parallel = 1300 total?)

If I put a Boss pedal (10k out, 1M in) after the tone circuit, here is how I could modify the circuit to get the same output impedances as the circuit would have in an amp:

Fender 1M output Z: no changes necessary, Boss pedal is already 1M
Vox 600k output Z: add 1.5M resistor to ground before the Boss pedal (1M and 1.5M in parallel = 600k total?)
Marshall 517k output Z: add 1.1M resistor to ground before the Boss pedal (1M and 1.1M in parallel = 517k total?)

___________

Would my problem be solved as simply as this, or are there factors beyond my perception that also need to be considered?

[antonis]

As PRR almost permanently says, the aim is to realize that everything is a "voltage dividing" matter.. 

P.S. So ridiculously simple but so true and correct..!!
It took me enough time (counted in years) to embend it..

[Rob Strand]

As PRR almost permanently says, the aim is to realize that everything is a "voltage dividing" matter.. 

P.S. So ridiculously simple but so true and correct..!!
It took me enough time (counted in years) to embend it..

It's true but the parts in the voltage divider aren't so obvious.    Interpreting what the equivalent divider is isn't always simple.

Take a bridge-T circuit,


Take the lower T-part then convert the T into a delta.
https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/

Following the naming on that web-site of A = in , C = out, B = ground.

If you do the T to delta conversion you end-up with a voltage divider.   You still need to put the bridge R2 in parallel with ZAC to form the voltage divider with ZCB.    The final voltage divider is ZCB / (ZCB + R2//ZAC ).   If you plug in all the values you will end-up with the transfer function in the pic.

If you go through the details:

The T (or Y) parts are:

    ZA = 1/sC,   ZB = R1, ZC = 1/sC

then the delta part values are,

    ZCB = (ZA ZB + ZA ZC + ZB ZC) / ZA    =  2R1 + 1/sC
    ZAC = (ZA ZB + ZA ZC + ZB ZC) / ZB    =  2/sC   + 1/(s^2 C^2 R1)

   [equations not checked]

The ZCB part can be interpreted in terms of normal parts:  a resistor of value 2R1 in series with a cap C.

However the impedance ZAC isn't so easy to interpret in terms of common parts.

I'm not going find what makes up ZAC.  It's likely to end-up with inductors even though we only started with R's and C's.   (For those interested look-up "network synthesis".)
[There's also a cheat's answer that it's a cap in series with an FDNR:
https://en.wikipedia.org/wiki/Frequency_dependent_negative_resistor]

Don't get caught-up on the math.  The main message here is things can be interpreted as voltage dividers but it's not always straight forward.    The parts in the voltage dividers aren't just resistors and caps they are complex circuits.  We can end-up with inductors even when there were no inductors in the original ckt.

[antonis]

The main message here is things can be interpreted as voltage dividers but it's not always straight forward.    The parts in the voltage dividers aren't just resistors and caps they are complex circuits.  We can end-up with inductors even when there were no inductors in the original ckt.

So, do you suggest to implement less scientific practices like estimating impedance by blending salamander blood via our lucky rabbit's foot under full moonlight (3^rd movement)..??