Emitter bypass cap roll off calculation...

[Phorhas]

Hello,

How do I calculate the hi-pass roll-off frequency due to an emitter bypass cap?

[Jason Stout]

Sorry I can't answer, but I can tell you that I *think* it is usually (but by no means does it have to be) overwhelmed by coupling RC and/or tone controls and output RC.

[Phorhas]

from my experience the cap value has quite an impact on the freq. response

[Transmogrifox]

***EDIT: The following is irrelevant to the question asked, but still good power supply info ***


For a bypass cap, the actualy 3-dB roll-off is not as useful of number as the rejection at a given frequency.  For most audio circuits, you have your 9V adaptor generating a rectified 60 Hz signal, and so the frequencies you will hear power supply noise will be in multiples of 60 Hz...so try to design for a certain rejection at 60Hz and all the harmonics will get even more rejection with a simple RC bypass filter.

You notice many pedal designs put a 100 ohm resistor in series with the power supply at the input.  This does 2 things:

1.  Increases the RC time constant (duh)
2.  Increases total rejection due to the capacitor's ESR (Equivalent series resistance) which is often about 10 ohms.  So with a 100 ohm resistor, the maximum rejection you can get on a signal is 10/(10+100) = 1/11 rejection ~20dB-->not very good for a noisy power supply.

Ok, so it's not that bad because the power supply itself can have a an output resistance of about 50 ohms, so this can provide maybe 24 dB rejection, still not phenominal.

So here are some rules of thumb for decent rejection on a pedal bypass:

60 dB rejection on 60 Hz is ideal.
40 dB on 60 Hz is still good.

-3dB roll-off point is 1/(2*pi*R*C) and on the simple 1rst order RC filter, decreases by 20 dB/decade...or a more simple way to deal with it is that the magnitude of Vout/Vin (filtered supply divided by unfiltered supply) is:

Vo/Vi =                  1/ squareroot[ (2*pi*R*C*f)^2  + 1)]

where f is the frequency dependence.  You can see that if f=0 (DC) then this expression simply reduces to 1.  If it is very large, it approaches 0.

So if you assume a 50 ohm supply output impedance, ignore ESR (for now, tell you how to deal with that later), and want 60 dB rejection on 60 Hz.  What size capacitor do you want?

Well, dB (volts) = 20*log10(V)

so Vo/Vi in dB = 20 * log10(Vo/Vi)  
or...

= Vo/Vi in dB = 20*log10{ 1/ squareroot[ (2*pi*R*C*f)^2  + 1)] }

and you want Vo/Vi in dB = -60 dB

So here's the equation you need to solve:

-60 = 20*log10{ 1/ squareroot[ (2*pi*R*C*f)^2  + 1)] }

--->  10^(-60/20) = { 1/ squareroot[ (2*pi*R*C*f)^2  + 1)] }

----> [1/{10^(-60/20)}]^2 = 2*pi*R*C*(60Hz) +1

so  C =  { [1/{10^(-60/20)}]^2 - 1 } / (2*pi*(50ohms)*(60Hz))

C = 5300 uF

Which is way more capacitance than what is reasonable in a guitar pedal.  There are two solutions:  Add some series resistance, which comes at a voltage headroom cost, or try for 30 or 40 dB rejection instead...realizing that this will have even more rejection at 120Hz 180Hz etc.

This is why you see some pedals using a transistor and zener to regulate the supply to about 8V instead of trying to simply filter the 9V.  If you're using a battery or well regulated pedal supply, you have little to worry about and you may not even need to filter at all, but 20 dB rejection is plenty for these cases.


NOw for dealing with ESR:  Use a few capacitors in parallel to get your equivalent needed capacitance for the rejection you desire.  Using the example above, to get your 5300uF capacitance, instead of going and buying an oil canister, use 10 x 530uF capacitors in parallel.  Then the ESR is reduced to 1 ohm, which gives you a maximum of 1/50 rejection ~-34dB, or with the 100 ohm series resistor ~ -40dB.

Now you can see why the power supply calculations aren't truly that useful, because the ideal is not at all economical.  Just use a 100 ohm resistor and a 470 uF cap or come up a notch and use 3 x 220uF caps in parallel and you may get about -30dB rejection at 60 Hz.

[Transmogrifox]

Oh shoot!  I totally responded to the wrong question.  I can't believe I misread the original question so terribly.  You weren't talking about power supply.


Yes this roll-off frequency is 1/2*pi*R*C.

The tricky part is determining R.  Usually the equivalent output resistance of the emitter is much smaller than the external bias resistor.  Rule of thumb, you can guess about 25 ohms for about 1mA bias on a BJT, or 100 ohms at 100-500uA, and maybe 100 to 150 on a FET.  

The BJT output resistance (Re) is Beta/((Beta+1)*gm)  and gm~(Ic/Vt)
where Vt ~ 25 mV at room temp, Ic is the collector bias current.
so if collector current is, say, 500uA and you're using a 3904 (Beta~170), then Re~50 ohms.

So you can see for large Beta, this reduces to Re~1/gm= Vt/Ic.

So just take .025/Ic and this will give you a good estimate of the equivalent Re.

[WGTP]

I don't understand all that, but I will say that it explains another reason why plugging different transistors into your Fuzz Face makes them sound different.

In tweaking different circuits I think it is very important to socket these so you can adjust to taste.  Also, with 2 or more stages, like the ROG emulators or the Vulcan, be sure and try different values in the different stages or none at all to lower the gain.

If you start with a 47uf cap and hear nothing different as you go down in value until you get to a 10uf cap, the 10uf cap is probably starting to roll off the bottom octive of your guitar between 80Hz and 160Hz.  

Am I correct in this?  

[Mark F]

I think it is very important to socket these so you can adjust to taste.  Also, with 2 or more stages, like the ROG emulators or the Vulcan, be sure and try different values in the different stages or none at all to lower the gain.
Am I correct in this?  

I Believe the bypass cap  in BJT circuits does more than set the rolloff. I think it also stabilizes the circuit ;ie prevents oscillation. So, I don't know if I would totally eliminate it. I don't know if this holds true for JFETS

[niftydog]

I'm assuming you're talking about a common emitter amp?

There are three factors that might determine the Æ' rolloff. The input cap, the output cap and the emitter bypass cap. The -3dB point is determined by the "worst" of these formulas.

For Æ'low due to Cin, you need to know the signal source impedance and the input imedance of the amplifier itself. Not a trivial excercise, so I won't go down that path, unless you really want me to!

For Æ'low due to Cout;

Æ'low >= 1 / [2pi.Co(RC +RL)]

Co = output cap, RC = collector resistor, RL = load resistor.

For Æ'low due to CE;

Æ'low >= 1 / [2pi.CE(re +RE1)]

CE = emitter cap, RE1 = non-bypassed emitter resistance.

re is the dynamic resistance of the emitter and can be approximated by;

re = 30mV / IE  (±40%)

Have fun!   :wink:

[brett]

In a Fuzz face, Q1 has an Re of about 100 ohms (=0.025/Ic).  That value is handy if you are considering adding an emitter resistor to "tone it down" (e.g. easyface) and lower the input impedance.  ie. adding a 100 ohm resistor roughly "halves" the gain and input impedance, which makes common Si transistors like the 2N3904 work much better.

Q2 has a MUCH lower Re (=0.025/Ic) becoz Ic is so much higher (2 to 3mA).  You can assume about 10ohms.  

for 22uF, fc = 1/(2.pi.R.C) or about 680Hz
For single-pole filters like this, you don't notice much effect until somewhere down around fc/2 (around "middle A" in this case)

I *think* these calcs are right. :wink:

[WGTP]

I admire your formulas and calculation and appreciate the explainations.

The reason I presented it in based more on sonic results, is that thru experimentation you can sort of figure out when the -3 is high enough to effect your tone, I'm guessing around 100Hz.  You all have presented explainations as to why it might be a 22uf cap for a BJT and a 4.7uf cap for Jfets.  

On my Vulcan I use a pretty small  1 or 2.2uf cap in the first stage which acts sort of like a treble booster into the 2nd and 3rd stage.  I then don't use the parallel cap/resistor between stages.  The 2nd and 3rd stage have either 10uf or 22uf caps.  To tone it down I have left out the last stage by-pass cap, it still seems to work.  Of course there is no bass roll of then.

As mentioned, a single stage has 3 single pole filters, so the roll off at some point becomes 18db/oct.  That times 3 stages results in a very steep bass cut.  So some of the caps need to be large enough that the cut doesn't occur too high up in the bass response of your effect.  

To me, it appears some designers make the caps large enough in most points in the ciruit that only 2 or 3 are audibly effecting the tone.  This makes it much easier to tweak and the rate and amount of roll off greatly effect the overall sound of the different distortion devices.  It has been argued that this is more important than the method used to actually clip the signal.

It seems most times we want to roll off the bass to get a smoother distortion.  The exceptions being Fuzz Face type circuits where 47uf caps are used to maximize gain, in addition to the low input impedance which rolls off the highs.

Make since/am I correct?  

[Kent S.]

Just placing a bump here to track this thread.

[Transmogrifox]

nifty dog did an excellent job of covering the formulas, and even took more effects into consideration than I did.

I was just thinking that all of that info that nifty dog and myself posted makes the head spin.  Here's the summary that has relevant information that will get you into the ballpark:

****************************

f3dB ~ 1/ (2*pi*R*C)

*****************************

and R is most significantly defined by the emitter output resistance, so ignore the rest of the effects and just calculate R as

*****************************

R ~ Re ~ Vt/Ic =  (.025)/  (Collector Bias Current) at room temperature.

*****************************

[WGTP]

So, could you give me a ballpark figure for a 2N5089 in a Vulcan?  

[aron]

I thought this was in the FAQ and it is, but the question is different.

[Transmogrifox]

For the  Vulcan, we may safely assume the Ic is on the order of 450 uA.  At room temperature, Vt = .025.

Since a 2N5089 is a very high gain transistor,
Re is then Vt/Ic (approximately) = .025/(450e-6) = 55 ohms.


The cut-off frequency is 1/2*pi*R*C

and since 1k (emitter bias resistor) is much larger than 55 ohms, and Ro (resistance from emitter to collector) is on the order of 100k usually, which is negligible compared to 55 ohms, so Re can be used safely for R.

So, Joe uses 22uF caps in the Vulcan.

fo = 1/2*pi*55*(22e-6) =  130 Hz  !!!

So why does this not get as mushy when a TS-9 cuts off at 700 Hz???

The answer is that there are 3 cascaded stages.  By the end of the last stage, frequencies at 130 Hz are actually down by 9dB.

The 3dB cut-off point by the end of the last stage is actually at about 400 Hz if I remember my pole approximations correctly.

However, there's another pole with that .0022uF cap and 220k resistor at about 350Hz (at max gain, it changes slightly with the gain pot).  

Either way, the approximate poles are a little lower than the average distortion pedal.

I would say that if you need to tighten it up at all, play with the bypass cap on the second stage first, then tinker with the others.  For noise purposes, you would ideally do it on the last stage first, but a significant amount of clipping occurs on the second stage, thus we're more concerned with sound than following the ideal "funnel" structure for reducing total noise power.

[WGTP]

Excellent example.  Thanks.  

[Kent S.]

This is why you see some pedals using a transistor and zener to regulate the supply to about 8V instead of trying to simply filter the 9V.  If you're using a battery or well regulated pedal supply, you have little to worry about and you may not even need to filter at all, but 20 dB rejection is plenty for these cases.

Now this answers the question I had in another post about why BOSS tends to do this in some of their latter pedals. Icouldn't figure why they were using a JFet and Resistor to drop the voltage down from 9V to 8V;the switching and Led circuits ran off 9V, but the audio portion of the circuit ran off 8V 9bias at +4V ... instead of +4.5V ....

NOw for dealing with ESR:  Use a few capacitors in parallel to get your equivalent needed capacitance for the rejection you desire.

As a follow up question, what's the best formula and method for calculating a cap(s) ESR , other than reading it on a suitable meter? Thanks.

[WGTP]

Again I appreciate the explainations, but I have some more questions.

With the input/output caps, don't you have even more poles and roll off?

Do cap's in parallel with resistors have 6db/oct. roll off rates?  

[Transmogrifox]

For Kent S.:

There is no good reason to calculate ESR.  Most engineers get this information from the datasheets accompanying the particular caps they buy.  The calculating of ESR is more of a Materials Science problem and I don't know of a text that does that specifically so I would have to drag out my old Materials Science book and derive it---and the method for calculating it would change between tantalum, electrolytic, ceramic, etc., capacitors as the metal leads aren't the only source of resistance.  There's generally a certain degree of loss in the dielectric material (which is a different material in the above caps)

You're pretty safe to assume an ESR of about 10 ohms for large electrolytics.  The main thing to realize is that there IS such a thing and know how to deal with it.

Once you assume an ESR of one capacitor of a kind, then you can just treat the ESR like regular resistors for high frequencies.  So if you stick caps in series together, the ESR is double (or the sum of the two ESRs if different).  For parallel resistors, if the resistors are all the same value (ie 10 ohms), then you can get the total resistance by simply dividing by the number of parallel resistors.  For example, if you put 10 resistors of 10 ohms in parallel, the equivalent resistance is 10/10 which is 1 ohm.   If you have 15 ohm resistors and put 3 in parallel, the equivalent is 15/3 = 5 ohms.  

So putting the caps in parallel reduces the total ESR so you get better rejection on the power supply.
Now for WG:

The input/output caps do introduce further poles (or zeros) in the circuit, however, Joe Davisson has those set so low that they don't have a very significant effect on the overall roll-off frequency for distortion purposes.  You could probably replace the .1uF at the input with a 470 uF and not hear a difference.

And secondly, yes, the cap in parallel with a resistor has a 6 dB/octave slope.  If you looked at a plot of the frequency response, you would see the gain on low frequencies is about what it would be if the capacitor wasn't there, and the gain on the high frequencies would be what you would expect if somebody shorted a wire across the capacitor, and there is a section in between where it rises from the lower level to the higher level at a rate of 6dB/octave so that it would look like a rounded stair-step up in the frequency domain.

[WGTP]

Again many thanks.  Verifies several theories I had.  Very important info to know when "voicing" distortions.  The by-pass caps can be used to tweak the low and the parallel r/c in the signal path can be used to tweak the highs.  Actually they probably are pretty interactive so it's both.   Good places for the cap blend circuits or some swithces.