Understanding high- and lowpass filter

[Gorn]

Hi guys!

This is my first post here, so i hope my questions are not to supid...

I started building pedals from preetched PCBs a while back and I'm trying to understand the circuits a little bit and mod them to my taste. Because I like to play some Doom and Stoner as well as 60/70s rock, I thought adding a pregain bass control and a postgain treble knob would be useful and interesting as a first mod. 

What I understand is this:

Highpass: Cuts lows, the capacitors comes first in the signalpath, after it a resistor linked to the ground.
Lowpass: Cuts highs, the resistor comes first in the signalpath, after it a capacitors linked to the ground.
Cut frequency can be calculated: f=1/2πRC

So if one takes a look at the Muff schematic, the input and output filters look exacly like this.
I wanted to try the Fuzz Factory and the Legend of Fuzz, but i cant identify the filters.

So my question is: How do I identify the capacitors that are used as a Hi/Low Pass if they are not in arangend in the simple form I described above?
I think the resistance of a capacitor depends on the frequency, the higher the less, and the capacity of it. Does that mean that there is no highpass needed, if you use one or more capacitors with a low capacity in the signalpath?

For the FF I read that the Input cap is not the one that cuts the bass as it is quite big (10uF) but the one between Q1 and Q2, but it still dosent look like a Highpass to me.

I think I'm missing something quite simple, but I cant figure it out by myself...

Any help is welcome, it is hard to search for a solve for a beginner, because often times you dont now what the problem is or how it is called, so pointing me toward resources about electronics of audiocircuits is also be appreciated

Thanks guys
Gorn

[Gorn]

I wasnt sure if posting a schematic or a link of the Fuzz Factory or legend of Fuzz is against the rules, so i didnt, if its ok I still could do it in this post

Gorn

[garcho]

Filters are complicated! You won't find many "a-ha" answers. Reactive capacitance is the phenomenon that turns capacitors into frequency dependent resistors for AC (not for DC, which is moot: no frequency!). What makes filters complicated is the interaction they have with the rest of the circuit. Read up on tone stacks and active op amp filters and patterns will start to sink in.

The FF cap between Q1 and Q2 has a path to ground via a 47k resistor and a 10k pot, hence the HPF.

[garcho]

I almost forgot...

Welcome to the forum!!

[antonis]

Hi and Welcome..

It isn't so complicated as long as you are able to identify equivalent impedances.. 
https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html



At point B there is a HPF made up of 10μF & 220k..
(clearly distinct, I presume..)

At point A, there is a HPF made up of 10μF input cap and 2N3904 Base input impedance in conjunction with bias configuration..
The later is the parallel equivalent of all the possible ways signal "sees" to GND (or to a point of lower voltage level) ..
From Base to GND via Emitter: equals to transistor h_FE times 0.026/Collector current (grounded Emitter)
From Base to Collector (via 220k resistor): 220k/(2N3904 stage Gain + 1)
Both the above impedances are considered in parallel so their equivalent shunt impedance is the parallel combination..

P.S.
There is also a HPF made up of 10μF Drive wiper cap (AC128 Emitter resistor bypasws cap) but let it be for the moment..

[Gorn]

Hey,
Thanks a lot for the answers!
I will look up tone stacks, active op amp filters and equivalent impedances.

Does the resistor in case of the HPF not need to go directly to GND, only to lower voltage level like you stated Antonis? Does the calculation of the cutoff frequency change because of that?

-Gorn

[antonis]

Does the resistor in case of the HPF not need to go directly to GND, only to lower voltage level like you stated Antonis? Does the calculation of the cutoff frequency change because of that?

It does (vicariously)..

What counts is the apparent value of the resistor which simply is the voltage difference (drop) across it divided by the current flowing through it.. (Ohm's Law..)

Of course, there should be some way from the "lower" voltage point to GND (arbitrary but conventionally set as 0V reference point) so you simply have to take the whole resistive string into account, in case of possitive voltage..
In case of negative voltage (in relation with signal instant possitive voltage polarity), apparent resistor value is considered its "nominal" value divided by voltages ratio absolute value plus unity..

e.g. 1k resistor on your hand is considered as 1k when placed between +1V and GND but is considered 100R between +1V and -9V..
(Ohm's beneficiaries would sue me for his Law cruel violation..)

P.S.
Sorry, I'm pretty sure I confused you more but neither my teaching nor my English skills are adequate.. 

[Gorn]

I have to admit i did not understand everything you two wrote, but I also didnt expect that because I am very new to this and I am not an native steaker as well 
So I  go step by step and take my time and if i get stuck I know there are a lot of nice people here that try to help. The mods just have to wait  bit

Thanks again!
-Gorn

[Digital Larry]

It isn't so complicated as long as you are able to identify equivalent impedances.. 

Yeah, that's the actual challenge... considering EVERYTHING that's connected to a given node of the circuit.  Looking into the base of a transistor, you have to consider what's on the other side (on the emitter, mostly) and what is the gain of the transistor, which you can read off a spec sheet but is usually something like 150 - 500. 

I do recall a certain incident when studying for my EE.  The prof put a simple RC circuit on the board and said "what is the time constant"?  Everyone reached for their calculators.  He screamed:  "STOP!  You are engineers, not scientists!"  The lesson being, get a first order answer as quickly as possible.  It may be all that you need.  Refine if necessary, but the 8-digits-of-precision answer your calculator tells you may not be close to reality no matter how good it looks.

[garcho]

In case of negative voltage (in relation with signal instant positive voltage polarity), apparent resistor value is considered its "nominal" value divided by voltages ratio absolute value plus unity.

e.g. 1k resistor on your hand is considered as 1k when placed between +1V and GND but is considered 100R between +1V and -9V.

I'm struggling to understand this. To what does the resistor appear? Is this for an AC signal "riding" at 0VDC? I also don't understand what "plus unity" means in this context. I'm confused!
A lot of audio circuitry is at ±15VDC, or ±12. Do the resistors in those circuits "appear" as their nominal value because the ratio of the bipolar power supply's voltages is 1? If the AC signal in your example above was centered around -4VDC, would the resistor appear as its nominal value? What would "plus unity" mean in the context of symmetrical power supplies like ±15? Sorry to hijack the thread.

[antonis]

 

We're talking about the way signal facing some kind of obstacle, Gary..

If you take in mind Miller effect, where cap value is seen as gain + 1 times bigger, you'll get the idea for C-B feedback resistor seen as gain + 1 times smaller..
(after all, gain is the ratio of Vout/Vin..)

[Rob Strand]

Antonis is on the right track but if you haven't seen the Miller effect before it might not be obvious what is going on.

The Miller effect is when you place a resistor across the input and output of an amplifier and the input impedance becomes different to the value of the resistor.

You need to think of input current.    If we have circuit and we want to know the input impedance we can in theory apply a voltage to the input then measure the input current.   The input impedance is then,

    Zin =  V_applied / I_measured.

When you have just a resistor to ground we know from Ohms law we would get I_measured = V_applied / R,  so Zin = R.

Imagine now taking the ground wire on the resistor and connecting it to the output of an amplifier with a gain of -100.  The resistor is connected between the input an output of the amplifier.   This is more or less the situation with the 220k on the first BJT.

If we applied a 1V signal to the input the output of the amplifier would be -100V.    What changes is the current through the resistor.
From the resistor's perspective it has 1V at one end but -100V at the other.  The voltage across the resistor is 101V.    The current is then I = 101 / R.   101 times greater than the current through just a resistor to ground.   If we were to do our input impedance test we would see a current I_measured = 101 / R.    When we work out the input impedance we would get,

     Zin   =  V_applied / I_measured 
              = 1 / I_measured
              = 1 / (101 / R)
              = R / 101

So the input impedance 1/101 times less than R to ground.

The general result is,

      Zin  = R / (1 - A)

where A is the amplifier gain.

The input becomes less when the amplifier gain is inverting because that makes the voltage across the resistor higher.  If we have a positive gain the input impedance can become less.  Think of a buffer if we connect a resistor from input to output the voltage across the resistor is zero.   No current will flow so the input impedance is infinite (very high).   In the general formula that's like putting A = 1.

With the transistor amplifier it's a little more complicated because you have the Miller effect on the 220k resistor but input impedance of the transistor is still going to ground.   So you need to workout the Miller effect on the 220k resistor then put that in parallel with the input impedance of the transistor.   The parallel combination is the effective input resistance that the cap sees when forming the hpf.

[antonis]

As of now, OP must be able to locate stages in-between HPF made up of 100nF nad 47k feedback resistor and calculate its corner frequency..