Question about coupling caps

[Rutger]

Hey,

I think it's a simple basic question, but I need to know it because it confuses me more and more  :loco
Sometimes you see two parallel resistors with a coupling cap in between, or two resistors that form a voltage divider but with a coupling cap in between.

Like the inputstage of this schematic for example:


Or the 1k/470n/22k after te 1e gainstage in this schematic:


I was learnt that a coupling cap decouples such resistors and that there is no interaction between them (for DC). So the parallel resistors aren't parallel anymore and the resistors don't form a voltage divider. But sometimes you read differently.

So what is true?

[polo16mi]

that i learned for coupling caps, is that they decouple the DC component for the signal, and depend of the value of the cap, they filter specific frecuency.

At first circuit that you show, 1M resistor present an 1M input impedance for Vin source. Then, 2.2uf and R2 in group, present a DC decouple and a high pass filter. (cut signals below 3.3Khz )

[J M Fahey]

In the first example, the 1875 power chip needs a DC path to ground for proper biasing; the 22K resistor used is a datasheet recommended value.
To avoid any possible DC coming from, say, a pedal connected to the input, they add a 2.2uF decoupling capacitor.
Which, by the way, cuts frequencies below around 4 Hz. 
The left side of the cap may float to any DC voltage , say a couple volts, and will pop loud when a guitar is connected there, so the 1M resistor is added to discharge said voltage to ground and avoid popping.
In the tube loop shown below, the 1K resistor in series with the 22K send pot is not really necessary.

[Rutger]

Thanks for the explanations, but unfortunately that's not really what I tried to point out. And probably that's because it's very obvious to you and it's my lack of knowledge (or english). So I'll try to make myself more clear.

I understand the functions of the resistors. But I wonder if, when you start calculating, you need to do the following:

example #1: when you want to know the input impedance of this circuit, for example to match it with the output impedance of a preamp,  is the inputimpedance 1M (R1) or 22k and 1M in parallel (R1llR2)?

example #2: when you want to know the behavior of the 22k pot, do you need to take the 1k resistor into account (1k+22k)?

[Enzo]

Caps block DC, so any remaining voltage division questions are stuck in the AC world.

In the second example, the 1k and the 22k are in series for any frequencies the 470nf cap passes.

[polo16mi]

1M in parallel with 22k is pretty near 22k, so AC talking, is same thing. Need take note of high pass filter.

Cant say nothing about 2dn example.

[Rutger]

Okay, so the decoupling is for DC only, and for AC it doesn't do anything: these resistors will still be parallel (#1) or in series (2#). Thanks for clearing that out

Actually 22k is quite a low input impedance in example #1, or is this usual in audio?

[J M Fahey]

Actually 22k is quite a low input impedance in example #1, or is this usual in audio?

It's fine, for a *Power* amp.
You are not meant to plug your guitar directly into it.
Preamps usually have 220K to 1M inputs, never less than 100 K at most.

[Roly]

example #1: when you want to know the input impedance of this circuit, for example to match it with the output impedance of a preamp,  is the input impedance 1M (R1) or 22k and 1M in parallel (R1llR2)?

The first thing to note is that the reactance "Xc", or effective resistance, of a capacitor depends on its value and the applied frequency;

Xc = 1 / (2 {pi} f C)

where;
Xc is the reactance in Ohms
f is the frequency in Hz
C is the capacity in Farads

At the frequencies of interest the reactance of the input coupling capacitor is low compared to the 1Meg and the 22k, so the effective input impedance is 1M||22k, which is roughly 22k for all frequencies of interest.

When we couple the output of a preamp to the input of a power amp our concern is voltage transfer, not power transfer, therefore an impedance match is not required; rather the load impedance, the power amp input, must be much greater than the source impedance, the output of the preamp.

In this case the input impedance of the main amp is about 22k and the source impedance of most preamps will be low to very low, from hundreds of ohms down to perhaps fractions of an ohm.

We can see that when the reactance of C1 is equal to the resistance R2 a 2:1 divider is formed giving half voltage or a 6dB drop.  By using the relationship above we can find the frequency that this 6dB point will occur at;

Xc = 1 / (2 {pi} f C)

make f the subject; multiply both sides by f, divide BS by Xc;

f = 1 / (2 {pi} Xc C)

f = 1 / (2 * pi * 22*10^3 * 2.2*10^-6) = 3.28832527Hz or 3.3Hz

We should not forget that the input impedance of the chip amp is also in parallel with the 22k resistor so the actual input impedance at this point may be lower again.

example #2: when you want to know the behavior of the 22k pot, do you need to take the 1k resistor into account (1k+22k)?

The object of the 1k appears to be to limit the available output drive should the Send Level be set at maximum.

The signal level available at the Send output will be 22/(22+1) of the signal level at the cathode.  22/23=0.95652174 or about 5% reduction which is quite a small loss for the benefit of providing some protection for external equipment connected to Send.  This stage may be capable of delivering signal levels in the tens of volts, but may be connected to solid-state devices that would be damaged by those voltages if unlimited current were allowed to flow.

Again we can find the low cutoff frequency, as above, from the 470nF and 22k.

f = 1 / (2 * pi * 22*10^3 * 470*10^-9) = 15.39216084Hz or 15.4Hz

HTH

[Rutger]

Thanks for al the answers and Roly for the clear explanation of all the theory, great help!

[Rutger]

Another, whole other question(s) about coupling caps!

#1
I see a lot of schematics with the following arrangement: gainstage-coupling cap-volumpot. The couplingcap and volumepot unsurprisingly form a highpass filter. But what suprises me is that there is a lot of basscut going on in these stages.

Like the last gainstage in the underneath schematic for example:



It seems that the value of the 68n couplingcap is calculated with the volumepot at max (100k), than the cutoff frequency is low (23Hz). But in real life you never ganna set the volumpot on 10, more like halfway. That is 10% of 100k which makes the cutoff frequency at 234Hz...??

Do I miss something or is this acceptable?

#2
I wonder, when there are other components between the coupling cap and volumepot, if they still form a highpassfilter?

Like for example the 22n cap/100k pot after the last gainstage of the following schematic:





I'm sorry for all the rog-schematics, I just happened to be on their website :cheesy:

[Roly]

In the first example just assume for a moment that the output wiper of the 100k volume control isn't connected to anything.  The CR circuit is 68nF and 100k no matter where the wiper is.

Now assume that the output goes to another stage which has an input resistance of 1 megohm.  With the wiper at minimum the CR is still 68nF and 100k, and with the wiper at maximum it's 68nF and 100k in parallel with 1 Meg, which is still pretty close to 100k.

Now if the load is only 100k then at maximum the CR would be 68nF and 100k||100k or 50k which would be double the cutoff frequency, but still quite low.

In either case with the volume pot set at 10% the R is still going to be very close to 100k.


The second example is rather more complex, but the answer is basically "yes" because the cutoff is still CR, but in this case it is the effective R of the following network (in this case the low pass formed by the two 10k's and two 2n2's, and the 100k volume pot).

Just by eyeballing this we can see the two series R's are each one-tenth the 100k volume pot, and similarly the two 2n2 in shunt are only one-tenth of the 22nF coupling cap, so the effect of the 10k's and 2n2's can be assumed to be fairly well removed from the CR formed by the 22nF coupling cap and 100k volume pot.

In the two LTSpice plots attached you can see that the low frequency rolloff in the second example is a bit higher than the first because the coupling cap is only one-third, but you can also see that the effect of the two 10k's and 2n2's is to provide some high frequency rolloff and doesn't really have any effect on the low frequency rolloff. {I've ignored the 1uF path in the second circuit for simplicity}

I suggest that you download Duncan's Tone Stack Calculator and fiddle about with the values in one of the simpler networks such as the Big Muff to get a feel for how changes in capacitance and resistance effect the frequency response.

HTH

[Rutger]

Thanks again Roly, it's all clear to me now! Great that you're willing to spend some extra time for folks like me  :dbtu:

I understand where I went wrong with question #1. I somehow assumed the wiper of the pot stated the resistance in the lowpass filter. But now I see that the resistance in the lowpassfilter will stay the same, no matter what the pot does.

I must have a Spice program myself, its seems so easy and the plots explain so much!

[Rutger]

Hooray, more questions!  :cheesy:

Looking at more designs, those coupling caps pop up or fanish at places where you don't expect it.

#1
In audio designs you often see coupling caps right after the input or just before the output of a design, to prevent the circuit form unwanted DC from outside. I don't see them often in guitaramp/stompbox designs. Why is that? Or is it just cost saving? Especially when there is a volumepot at the and of the circuit.

#2
Tonestacks and volumepots are mostly followed by a gainstage. Sometimes you see, but often you don't see a coupling cap between gainstage and pot. I expect you don't want DC flowing thru your pot (cracle-not-okay). So why do they take the risk? And often they are right, because it won't crackle...

Actually it would seem more accurate to me, to put a 1M resistor before the gainstage and a couplingcap between this resistor and the pot, if you know what I mean?

[Roly]

No problems - I've been doing this stuff for the best part of fifty years now and had trainees most of that time, and I consider it giving back to my late mentor Theo van Bemmel who saw it as his duty to pass on what he had learned.  In retirement I now spend about half of every day on forums and replying to e-mails explaining stuff, and I find it stops my brain turning to mush.

These questions are a bit sticker.  I often look at some circuit arrangement and wonder what the designer had in mind (and sometimes what they had been sticking up their noses).

Q1
If there were some consistent law that said "Thou Shalt Always Put a Blocking Cap on the Output", and if designers and manufacturers actually followed it, then I guess you would never see a cap on an input.

But it is, as they say, a Free Country, (or more accurately "there are no rules") so manufacturers do exactly what suits them.

When you decide to include an input blocking cap, for example, this also means you need to include a resistor to ground ahead of it to make sure there is no DC offset when you plug something in, and cause a loud "splat" (if not actual damage).  Since this is in parallel with whatever resistance is tying the actual electronics input (say an op-amp input) to ground it appears in parallel with it and therefore lowers your input impedance - but you typically want to keep your input impedance at one meg or greater, so the struggle with trade-offs begins.

There is not a lot in electronics that is a one-way deal, more of X is good and there is no downside - there is nearly always a downside.  As you bump up your input resistance so you start to get problems with voltage offsets due to tiny leakage out of the op-amp input; you start to get noise due to very high value resistors, so you have to make compromises.

Q2
Good practice is to keep DC off pots, but not everyone follows good practice.  It may be because they have an amp design that passes hardly any current through the pot, such as a FET or FET input op-amp, or perhaps it won't be a problem until the amp is out of warranty or they don't care because at its "price point" they don't mind if it's a bit crappy compared to their better offerings (the cynical approach).  Maybe it's purely cost, or maybe their "designer" simply doesn't know any better.

On the service bench you frequently encounter situations where somebody has said "That'll work/be enough" and it hasn't and isn't.

In no particular order examples that come to mind are - the power rating for resistors feeding zener droppers for preamp supplies; heatsinks - often; transistor voltage and current ratings, electrolytics near hot things, just about every PCB mounting external connector, etc and so on.

My personal credo is "If you can't make it indestructible, make it easy to repair".  Where possible try to allow for failure modes, for where things go wrong (such as accidentally plugging a guitar amp output into a mike channel input), and as far as possible prevent or limit damage.  Not everybody thinks like that.

So why do many stomps have no DC blocking on their inputs?  Perhaps the assumption that they will be alone, or at least first in the chain, or that any up-stream box will have an output blocking cap.  And mostly they are right.

You actually remind me of an old Marshall valve/tube PA that I used to look after.  The grids of the input triodes went directly to shorting microphone jacks.  That's it; no DC blocking, and the assumption that whatever was plugged in would provide the required DC path to earth for the valve.  I consider that a very brave assumption by Jim.

And this is part of the reason you normally see a resistor as the very last thing across the output after the DC blocking cap; mainly to charge the cap to de-thump a plug in, but to also DC tie the output line back to ground JIC - Just In Case.

I know these answers aren't very definite, but as I said, I sometimes look at circuits and equipment and wonder what on earth they were thinking.  :duh

HTH