Filters and Impedance

[LaloFP]

Hi guys!

Because an effect that adds tons of bass, I'm thinking of adding a togglable HighPass Filter at the output of the pedal (after an inverting opamp that controls the general volume with a pot in the feedback loop).

(assume that the excess of bass cant be fixed before... for the sake of my doubt that points to something I can learn! )

If I simply replace the decoupling cap and resistor at the end with lower ones (say 1uF and 2k), I worry 3 things:
1 - that is going to make the input impedance of the next effect smaller, like 2k, right? That will drive the circuit, right? Having before that an ideally infinite opamp impedance, that isnt a problem... right  ?
2 - can the corner frequency of the filter be affected by the next effect in the chain?
3 - I never like to put such a small resistor to ground. It feels current "wasteful"

I think I know the basics of this subject but I doubt about the implications and "rules of thumb"

Thanksss !

[ElectricDruid]

If the last stage is an inverting op-amp, why not add the highpass filter to the input of that? You probably already have a cap in that position (ahead of the input resistor) so you could simply make it switchable. The highpass frequency depends on the input cap and the input resistor, so it won't be affected by your variable-gain volume control in the feedback loop.

[antonis]

an effect that adds tons of bass,

A bass booster..?? A treble cutter..??   

[FiveseveN]

1. No, the input impedance of the next effect is a property of that circuit, what you're changing is the output impedance of this effect, but

Having before that an ideally infinite opamp impedance, that isnt a problem... right  ?

An ideal op amp has infinite input impedance and zero output impedance, the other way around. 2K in parallel with ~0 isn't going to make a difference on that front, but it may load the op amp unnecessarily. 2K should be fine really but why not ~10K and a smaller cap?

2. Yes, anything in parallel with the R of your RC filter is going to change its value. So you need to pick that value such that it's much smaller that the load. For the "standard" 1M effect input Z anything under 100K will work, again 10K is even better.

Like ElectricDruid noted, the proper way to do it is to not assume it can't be fixed before the output stage and then you'll have less to worry about.

[LaloFP]

Hi! Sorry for the late response, Im moving to another place and everything is a mess 

Yeah, I know that the proper solution FOR the circuit could be in other places. But, honestly, Im more concern about fixing my ignorance in that point, rather than to fix the circuit 

Even, at the input I have an non-inverting opamp buffer with the usual decoupling cap and resistor to VRef. I can diminish the resistor to VRef to alter the filter and compensate there, not changing the input impedance of the pedal (but it is the same "problem"... right?)

BUT, I care more about learning and this idea that I have bothers me a lot: "3 - I never like to put such a small resistor to ground. It feels current 'wasteful' " . That bothers me because its something I FEEL, rather than something I know/understand, and I want to secure my knowledge there.
I don't usually see 2k to GND in pedals and that reinforces my believe.
I know that if the Opamp has zero output impedance, 2k doesn't change anything in the ideal universe. But I don't how safe it is in a real TL072 and the limits in practice

2K should be fine really but why not ~10K and a smaller cap?

I always assumed that what matters is the relationship and cutoff freq, no the actual values, because swapping the values (if constant "ratio") dumps the same frequencies to gnd in the same amount.
Is that wrong?

an effect that adds tons of bass,

A bass booster..?? A treble cutter..??

its a tremolo! But Its a harmonic one, and If I make the HPass part become a ALLpass filter, I end up with only the High frequencies being modulated. BUT the High freqs peaks at the same level that the constant bass and then goes lower (its centered lower than the [constant] bass), giving the bass more perceived volume in time.

But I don't want that aspect to distract us from the matters I point, because my ignorance is over those! haha

[FiveseveN]

I don't how safe it is in a real TL072 and the limits in practice

See Figure 6: Maximum peak-to-peak output voltage versus load resistance in the datasheet.

I always assumed that what matters is the relationship and cutoff freq, no the actual values

As per the above, there are practical limitations. Like I said, 2K is fine, 10K or more is marginally better.

[ElectricDruid]

Even, at the input I have an non-inverting opamp buffer with the usual decoupling cap and resistor to VRef. I can diminish the resistor to VRef to alter the filter and compensate there, not changing the input impedance of the pedal (but it is the same "problem"... right?)

Sorry Lalo, I don't understand you.

In the usual non-inverting op-amp buffer it's the resistor to Vref that *sets* the input impedance (because it's usually far smaller than the op-amp's own input impedance). In which case, changing that resistor to change the high pass cutoff *does* change the input impedance of the pedal. The usual way to make tweaks there is to change the input *capacitor* instead, since that changes the cutoff but *doesn't* affect the impedance.

BUT, I care more about learning and this idea that I have bothers me a lot: "3 - I never like to put such a small resistor to ground. It feels current 'wasteful' " . That bothers me because its something I FEEL, rather than something I know/understand, and I want to secure my knowledge there.
I don't usually see 2k to GND in pedals and that reinforces my believe.
I know that if the Opamp has zero output impedance, 2k doesn't change anything in the ideal universe. But I don't how safe it is in a real TL072 and the limits in practice

Like you, it's not something I'd do. Values of a few K downwards do start to use "significant" current, at least for a pedal. Of course, we're only talking millliamps, but back when I built pedals with batteries, I used to think 5mA was a significant draw for an entire pedal. It's easy to do the sums, basic Ohm's law stuff: V=IR, so I=V/R. Stick in the biggest voltage the resistor is likely to see and work out the maximum current.
In other situations, people make different choices. In analog mixing desks, for example, the biggest problem is noise, and the power supply can be built to be as large as you need, so resistor values tend to be smaller than pedal builders would typically use (to reduce noise) and they don't worry about the increased current.

2K should be fine really but why not ~10K and a smaller cap?

I always assumed that what matters is the relationship and cutoff freq, no the actual values, because swapping the values (if constant "ratio") dumps the same frequencies to gnd in the same amount.
Is that wrong?

No, it's not wrong, it's the "R x C" combination that sets the cutoff. But that's exactly FiveseveN's point - that means there's never any *need* to use a really small resistor value. You can easily scale the resistor up by a factor of (say) 5 or 10, and then scale the capacitor value down by the same factor to keep the same cutoff.

HTH,
Tom

[antonis]

that means there's never any *need* to use a really small resistor value.

Unless you have an op-amp of really large bias current..

[LaloFP]

I don't how safe it is in a real TL072 and the limits in practice

See Figure 6: Maximum peak-to-peak output voltage versus load resistance in the datasheet.

Awesome, so with a 9v pedal I could use a minimum of 250ohms aprox. (stressing the power supply, but inside of the TL072 specs, at its limit)
Thank youuu FiveseveN! 

The usual way to make tweaks there is to change the input *capacitor* instead, since that changes the cutoff but *doesn't* affect the impedance.

THAT was my misunderstanding! Because two filters with the same cutoff but different values (say 1nF-10k and 10nF-1k) has the same dbReduction, so same Voltaje reduction across the spectrum, I confused that with same current behaviour. But I needed to do the math (online calculator) and watch more videos about the internals of the caps to see that thats not the case. The impedance changes. In my mind I conceptualized the capacitor more as a freq-variable resistor than what it really is. That's what I was wrong about, thank youuuu 

BUT, I care more about learning and this idea that I have bothers me a lot: "3 - I never like to put such a small resistor to ground. It feels current 'wasteful' " . That bothers me because its something I FEEL, rather than something I know/understand, and I want to secure my knowledge there.
I don't usually see 2k to GND in pedals and that reinforces my believe.
I know that if the Opamp has zero output impedance, 2k doesn't change anything in the ideal universe. But I don't how safe it is in a real TL072 and the limits in practice

Like you, it's not something I'd do. Values of a few K downwards do start to use "significant" current, at least for a pedal. Of course, we're only talking millliamps, but back when I built pedals with batteries, I used to think 5mA was a significant draw for an entire pedal. It's easy to do the sums, basic Ohm's law stuff: V=IR, so I=V/R. Stick in the biggest voltage the resistor is likely to see and work out the maximum current.
In other situations, people make different choices. In analog mixing desks, for example, the biggest problem is noise, and the power supply can be built to be as large as you need, so resistor values tend to be smaller than pedal builders would typically use (to reduce noise) and they don't worry about the increased current.

Awesome, thats the logic behind my "fear"  Thank you tom!!

Im so relieved now 

[antonis]

The usual way to make tweaks there is to change the input *capacitor* instead, since that changes the cutoff but *doesn't* affect the impedance.

THAT was my misunderstanding! Because two filters with the same cutoff but different values (say 1nF-10k and 10nF-1k) has the same dbReduction, so same Voltaje reduction across the spectrum, I confused that with same current behaviour. But I needed to do the math (online calculator) and watch more videos about the internals of the caps to see that thats not the case. The impedance changes. In my mind I conceptualized the capacitor more as a freq-variable resistor than what it really is. That's what I was wrong about, thank youuuu 

I think you've misunderstood what Tom said..

A series capacitor with a shunt resistor don't alter impedance 'cause impedance is set by the resistor..

[LaloFP]

The usual way to make tweaks there is to change the input *capacitor* instead, since that changes the cutoff but *doesn't* affect the impedance.

THAT was my misunderstanding! Because two filters with the same cutoff but different values (say 1nF-10k and 10nF-1k) has the same dbReduction, so same Voltaje reduction across the spectrum, I confused that with same current behaviour. But I needed to do the math (online calculator) and watch more videos about the internals of the caps to see that thats not the case. The impedance changes. In my mind I conceptualized the capacitor more as a freq-variable resistor than what it really is. That's what I was wrong about, thank youuuu 

I think you've misunderstood what Tom said..

A series capacitor with a shunt resistor don't alter impedance 'cause impedance is set by the resistor..

Yesyes, I understood that (I think)! 
If Im getting it right: the capacitor does not set the impedance because it doesnt alter the current, just the voltages. Is that right? The Resistor is altering the current.

[antonis]

the capacitor does not set the impedance because it doesnt alter the current, just the voltages. Is that right? The Resistor is altering the current.

No..

[ElectricDruid]

A series capacitor with a shunt resistor don't alter impedance 'cause impedance is set by the resistor..

Yesyes, I understood that (I think)!
If Im getting it right: the capacitor does not set the impedance because it doesnt alter the current, just the voltages. Is that right? The Resistor is altering the current.

I think you *did* understand it, but your explanation of *why* it's true is way off.

If I were you, I'd forget about the voltage and current while thinking about the impedance. For input impedance, the way I think about it is to imagine it as the load presented to the previous the circuit. This can be imagined as a resistor between the signal and ground. Now, in this case we've got two values that fit that pattern. One is the input impedance of the actual op-amp which we can't see because it's internal to the chip, but luckily for us we can ignore it anyway because it's usually enormous. That imaginary internal resistor is in parallel with our actual shunt resistor outside the chip. Say our resistor is 1M and the internal impedance is 100M. The parallel combination is 990K, which is close enough to 1M that we don't care. The internal impedance is likely even more than this, which gets us even closer.

Once you've got that impedance value value worked out, you can work out the cutoff by doing the usual 1 / (2 x Pi  x R x C) or better still getting your favourite online calculator to do it for you.

As far as capacitors go, there are only three points you need to worry about: DC, Infinity, and the rest.
At DC, a capacitor is an open circuit. Imagine you pulled it out of the circuit and threw it away. Now what does the circuit look like?
At an infinite frequency, the capacitor is a short circuit. Imagine you replaced it with a bit of wire. Now what does the circuit look like?
The rest, all the stuff in-between DC and infinity. Well, the circuit goes from doing what happened at DC to doing what happened at infinity. The exact frequency that this changeover occurs is what working out the RC combinations is for!

HTH,
Tom

[LaloFP]

Is that right?

No..

If I were you, I'd forget about the voltage and current while thinking about the impedance. For input impedance, the way I think about it is to imagine it as the load presented to the previous the circuit. This can be imagined as a resistor between the signal and ground. Now, in this case we've got two values that fit that pattern. One is the input impedance of the actual op-amp which we can't see because it's internal to the chip, but luckily for us we can ignore it anyway because it's usually enormous. That imaginary internal resistor is in parallel with our actual shunt resistor outside the chip. Say our resistor is 1M and the internal impedance is 100M. The parallel combination is 990K, which is close enough to 1M that we don't care. The internal impedance is likely even more than this, which gets us even closer.

Once you've got that impedance value value worked out, you can work out the cutoff by doing the usual 1 / (2 x Pi  x R x C) or better still getting your favourite online calculator to do it for you.

I get that part, the "only resistors" scenario.

As far as capacitors go, there are only three points you need to worry about: DC, Infinity, and the rest.
At DC, a capacitor is an open circuit. Imagine you pulled it out of the circuit and threw it away. Now what does the circuit look like?
At an infinite frequency, the capacitor is a short circuit. Imagine you replaced it with a bit of wire. Now what does the circuit look like?
The rest, all the stuff in-between DC and infinity. Well, the circuit goes from doing what happened at DC to doing what happened at infinity. The exact frequency that this changeover occurs is what working out the RC combinations is for!

The input impedance (of the next circuit, the one starting with the filter):
At DC: is infinite
At infinite freq: The impedance is the previous-only-resistors impedance (the resistor to gnd value)
So in the middle we never "lose" impedance, it can only be greater than the resistor value

... right? 
This remembers me when once I finished an exam in probability and statistics and told a friend that I was getting a 9 (out of 10) surely and with confidence. I got a 2 and I instantly thought "okay, I didnt understand s**it) hahah

[ElectricDruid]

I get that part, the "only resistors" scenario.

Ok, cool. That's a good start.

As far as capacitors go, there are only three points you need to worry about: DC, Infinity, and the rest.
At DC, a capacitor is an open circuit. Imagine you pulled it out of the circuit and threw it away. Now what does the circuit look like?
At an infinite frequency, the capacitor is a short circuit. Imagine you replaced it with a bit of wire. Now what does the circuit look like?
The rest, all the stuff in-between DC and infinity. Well, the circuit goes from doing what happened at DC to doing what happened at infinity. The exact frequency that this changeover occurs is what working out the RC combinations is for!

The input impedance (of the next circuit, the one starting with the filter):
At DC: is infinite
At infinite freq: The impedance is the previous-only-resistors impedance (the resistor to gnd value)
So in the middle we never "lose" impedance, it can only be greater than the resistor value

I suppose that might be true, but it's beside the point because there's a much bigger effect going on. Say the output impedance of the previous stage is 1K and the input impedance of the next is 1M. We can look at that as a voltage divider, and 1K on the top with 1M on the bottom is 1000/1001, so we'll get 0.999 of our signal. If the cap increases the input impedance by even several magnitudes, the amount of signal we get only goes up by a tiny fraction. So it's not important.
You're right though, in as much as the capacitor's "effective resistance" value varies across the frequency range from "total resistance" at DC up to "negligible resistance" at infinity.


So what's the much bigger effect I mentioned? There's *another* voltage divider at play here.
The signal comes in at one end of the cap, and goes out to the op-amp at the other. From that point we have a resistor to ground. Now, those two components form a voltage divider. Except that in this voltage divider, one of the components isn't a fixed resistor, but instead is a capacitor. So what happens?
Well, at DC we known the cap has a huge resistance, so the top of the voltage divider is far larger than the bottom and no signal gets through. At low frequencies this continues to be the case, although gradually less and less so.
At high frequencies we know the cap has a very low resistance, so the bottom of the divider (our 1M resistor) is much larger and virtually all the signal gets through.
So low frequencies are blocked, and high frequencies are passed - it's a highpass filter!
The frequency where it swaps over from blocking to passing is the cutoff frequency, and it obviously depends on the values of the C and R in the voltage divider.
The classic lowpass RC filter works the same way, as a voltage divider, but with the two parts the other way up.

HTH,
Tom

[antonis]

Maybe it should be more comprehensible by facing it up from voltage dividing effect viewpoint..



P.S.
Set Rs = 10k, R = 1M and C = 4.7nF and see what happens for 100Hz, 1kHz and 10kHz..

[LaloFP]

Good morning guys!

Yeah yeah, I understood the filter part, the implications in voltage and the independence of that from the values of R and C (with constant "ratio"). That was what led me to wrongly assume that the same would apply to impedance (Wrong concept: if the ratio of the R and C is constant, the impedance is constant. Correct concept: the impedance depends on the R to gnd)

Thank you again my friends, this post clarified a lot for me