TS-808 Circuit Question for the Tube Screamer experts

[gaussmarkov]

i've been doing some sleuthing to try to understand this ESR/dissipation stuff.  i think i have a handle on it, so i'll throw out a summary that makes sense to me and see what the feedback is.

ESR stands for Equivalent Series Resistance.  putting an ideal capacitor in series with an ideal resistor produces the equivalent circuit that represents the ESR of a capacitor.  so an ideal capacitor has an ESR equal to zero. 

most datasheets report the "tangent of dissipation factor" for capacitors instead of the ESR.  i suppose this is because ESR depends more on frequency than the dissipation factor does.  the relationship is ESR = tan(d) /(2 pi f C) where tan(d) is the tangent of the dissipation factor, f is the frequency, and C is the stated value of the capacitor.  for WIMA polypropylene box film capacitors (MKP2), tan(d) is less than 0.0005 so that at 1KHz the ESR for a 47nF capacitor is about 1.7 ohms, which is a small resistance.

if i am reading the AVX datasheet correctly, the BF/BQ caps have a tan(d) of 1% or less, making their ESR as much as 2 times higher. but this is still small.

Panasonic polyester caps don't come in values as low as 47nF, but they also have a tan(d) with this order of magnitude.

in the tubescreamer circuit, the 47nF cap that we have been talking about is in series with a 4.7K resistor.  so we can think of the ESR as just adding on to this resistance.  adding on a couple of ohms is well within the tolerance of a metal film resistor (1% or 47 ohms).  so ESR is inconsequential relative to the variation across builds due to variation in nominal resistor values.

as a series resistance, ESR does not affect the symmetry of the filter.  it attenuates.

in addition, ESR has nothing to do with leakage.  leakage would be modelled by putting a resistor parallel to an ideal capacitor.  at least for metal film capacitors like the AVX and WIMA ones, this factor does not appear in the datasheets AFAICT.  i wonder if this doesn't mean that it isn't worth worrying about?

so that's my understanding based on trying to piece together what i have found so far.  thoughts or corrections?

all the best, paul

[gaussmarkov]

for WIMA polypropylene box film capacitors (MKP2), tan(d) is less than 0.0005 so that at 1KHz the ESR for a 47nF capacitor is about 1.7 ohms, which is a small resistance.

if i am reading the AVX datasheet correctly, the BF/BQ caps have a tan(d) of 1% or less, making their ESR as much as 2 times higher. but this is still small.

oops.  20 times, not 2 times.  so a 47nF AVX box capacitor could have an ESR as large as 34 ohms.

[Gus]

most mylar 50V 1uf caps measure 1.5 ohms ESR on a sencore LC102 when I tested them.   

[gaussmarkov]

interesting.  your comment caused me to go pawing through the tech notes at Sencore.  they have one on ESR that contradicts what some of the manufacturers' literature seems to suggest. 

the document is Understanding ESR in Capacitors and it says towards the end that

A number of customers have been told that ESR must be tested at a particular frequency. This is not the case, because true ESR does not depend on frequency. ESR represents pure resistive losses, and resistance has the same impedance at any frequency. The Z METER only measures resistance.

The confusion appears to come from formulas which attempt to convert the "D" (dissipation factor) reading from an AC impedance bridge to an ESR value. ESR, however, is only one of the many capacitor imperfections that causes poor D readings. D also includes the effects of leakage resistance, equivalent series inductance, dielectric absorption, dielectric stress, and losses (such as water molecule resonance) in the dielectric. Most of these other losses are frequency selective, so any attempt to calculate ESR from D will make it seem that ESR varies with frequency.

the ESR definition on wikipedia is

ESR is properly the real resistive component of the complex impedance Z(ω) = R + j X(ω) of the device

so that is not frequency dependent.  but in contrast something called the Transwiki says

ESR is the sum of in-phase AC resistance. It includes resistance of the dielectric, plate material, electrolytic solution, and terminal leads at a particular frequency. ESR acts like a resistor in series with a capacitor (thus the name Equivalent Series Resistance). This resistor can cause circuits to fail that look just fine on paper and is often the failure mode of capacitors.

if the equivalent circuit under consideration is an RLC series (resistor, inductor, and capacitor in series) and ESR is the resistance of the resistor, then we have to go with the nonfrequency dependent definition as a theoretical matter.  but because datasheets tell us to translate from the dissipation factor, i guess manufacturers are using the second definition, or something similar.

[R.G.]

Here - let me confuse things with MY opinion!   

Capacitors have some real, measurable resistance that is unavoidable. This is the resistance of the leads going in to the plates, the resistance of the bonding to the plates, the spreading resistance of the plates themselves. Imagine yourself to be an electron on one of those light cycles like in Tron. You head down the pipeline tunnel of the lead wire, bump across the bonding connection, then the vast, slightly resistive plane of the plate spreads before you. Hating being next to the other electrons which have come in with you, you head for open spaces, trying to get to your own parcel of the plate as far from the other electrons as you can. In the process, you encounter real resistance, pure ohmic resistance to be slogged through.

That is the low frequency ESR.

Whenever you polarize a cap, you deform the insulation subatomically. There is both an immediate (dissipation factor) effect and a slow (dielectric absorption) effect. Both subtract energy from the electron motion in and out of the cap.

In power supplies at AC line frequency, the low frequency ESR is all that matters. You're not reversing the charge on the cap fast enough or far enough for the dissipation factor or dielectric absorption to eat anything noticeable.

As frequency goes up, that changes. The dissipation factor is the price you pay for the insulation withstanding the voltage, plus any funny effects it does like shape changing  (piezoelectrical response) that is does as a sideline. The subatomic movements and other funnies are paid for out of the energy that goes into the cap. It turns into heat eventually. Every place energy can go either looks like a resistor and burns up some as heat or looks like a storage pot and stores it into E-field or M-field, or some combination of all three. The DF is one way of looking at the heating of the dielectric by continually changing the E-field on it. The dissipation factor is one way of expressing this heating.

Another way of expressing the total losses, both through the real low frequency resistance and the dissipation and absorption of the dielectric is to measure the impedance at some frequency. The impedance you measure is going to have a real part and an imaginary part. At that frequency, the real part is going to be the sum of the energy losses. This can never be less than the real low frequency resistance of the wires and plates.

ESR is a negligible part of using caps unless:
1. You're pumping many amps into and out of it, as with low frequency power supply filters, high frequency switching power supplies, and pulse capacitors.
2. You're pumping a lot of power through the cap and expecting the cap to not eat much as heating.
3. You're really, really worried about the complex impedance of the cap, especially if you're doing a lot of power, as in RF work.

For audio work, you really have to strain to find a place other than power supplies where ESR is not neglegible compared to the other resistances in series with the cap.

Meanwhile back at the tube screamer question. There's a cap in series with a resistor. As long as the cap's ESR is well under the resistances in series with it and the currents are low, you can neglect it. For a cap in series with a 4.7K resistor, if the cap's ESR is under a few hundred ohms, neglect it.

As to the tonal differences of the resistor/cap going to Vr versus ground, they are identical to the degree that Vr has a low enough impedance to real ground. If Vr has a high enough impedance, to where it's not neglegible, it is just more terms in the impedance of the inverting input to ground.

Knowing what to neglect is the real issue hiding under here. That's one reason for pursuing ever-finer discrimination of what's really happening - you have a good feel for when you can just kick back and say "Shoot, this isn't that complicated."

[gaussmarkov]

super!  thanks, R.G.

[Gus]

 A little of topic. Lets look at a RAT

the 4.7uf  and 2.2uf.

  Maybe what people hear as changes in the RAT when using Al or Tant at the - to ground leg in the feedback section might be the ESR being a good % of the 47 ohm resistor value in the leg.

47 ohm leg has a 2.2uf
LC102
just measured two Al Pan hfs 2.2uf 50v    2.14uf, 2%DA, 2.7ohms ESR
pan BP 2.2uf 50V         2.2uf, 2%DA, 2.9 ohms ESR

Measured a pan 1uf tant    1uf, 2%DA, 2.3ohms ESR

Note the ESR also as posted before a 1uf film often measures 1.5ohms ESR

sets gain and RC for higher frequency boost

If using Al electros in that section of the RAT maybe adjust the resistor value

[ethanw]

Thanks for all of the replies guys!  Some of that went over my head, but I think that question was answered thoroughly!

Ethanw

[JDoyle]

Sorry, I was confusing ESR with leakage earlier; but my point is this:

You breadboard a TS-type stage up to mess with it.

You leave out the resistor in series with the gain pot, make the gain pot 100k, make the resistor in series with the cap 1k, and put in a .22 cap with, say, only 1 ohm of ESR to ground.

So your circuit has pretty much the same response as the original TS, but you can turn the gain down to 1 and have reduced the noise by reducing the resistance.

You have also just made the math below one hell of a lot easier for me.

In your experimental circuit, with Vbias at 4.5V and the gain pot at zero ohms, leaving the output directly connected to the input, the 4.5V bias on the negative input drops across the 1K resistor and the cap with an ESR of 1 ohm (in parallel with the Zin of the -input, but that is large enough to ignore). So the bias voltage drops across 1k + 1 ohm to ground and the cap drops 4.5mV of that 4.5V:

4.5 * 1/1000+1 = 0.0044955044955044955044955044955...

Which means that instead of being at 'virtual ground', the opposite end of the resistor is at +4.5mV, which then adds to the bias voltage on the negative input.

Which means that in order to keep the negative input the same voltage as the positive input, the output at rest is:

[(4.5V) - (4.5mV)] = 4.4955V

And to do this, the output sinks 4.5uA of current which has to traverse the feedback network from the output to the negative input. So if you increase the gain of your circuit by increasing that resistance, the output of the op amp then has to compensate by sinking more current, which itself has to pass through the feedback network dropping the voltage at the output further, and is also that much less current capacity the op amp has available when it comes time to actually amplify.

Said another way: the offset at the output is the offset on the negative input times the gain setting through the feedback network.

Or yet another way: the offset is equivalent to a constant input signal of the opposite polarity.

So by the time you turn the gain up to 100, the output, at rest, is:

Vout = Vbias + (-4.5mV * 100)
Vout = Vbias + (-0.0045 * 100)
Vout = Vbias + (-0.45V)
Vout = 4.05 [a -0.45V offset at the output]

Therefore, when you turn up the gain to max in this breadboarded circuit, even with absolutely no input signal present, the negative swing at the output will only have 0.15V of headroom before the diode clips and the positive swing will have a total 1.05V of headroom.

But if in the beginning you connected that cap to Vbias instead of ground, none of this occurs; the cap-and-resistor-combo has no voltage across it because both ends are at Vbias, so it doesn't drop any voltage and therefore doesn't not cause an offset, at rest or at any other time.

I am simplifying this a lot. The op amp has its own offsets, both output and input, and the diode has a leakage current, with the cap either adding or subtracting from both of them depending of the polarity of either, just to complicate things further.

But the focus of this thread is the effect of the cap and with everything else ignored or idealized...

Yes, in my opinion, there IS a difference between connecting the cap to ground and connecting it to Vbias. And further, while you may well be aiming for the asymmetrical clipping nature of that the offset induces, why let it hamstring you so that you can't get rid of it, even if you wanted to?

Regards,

Jay Doyle

[gaussmarkov]

In your experimental circuit, with Vbias at 4.5V and the gain pot at zero ohms, leaving the output directly connected to the input, the 4.5V bias on the negative input drops across the 1K resistor and the cap with an ESR of 1 ohm (in parallel with the Zin of the -input, but that is large enough to ignore). So the bias voltage drops across 1k + 1 ohm to ground and the cap drops 4.5mV of that 4.5V:

4.5 * 1/1000+1 = 0.0044955044955044955044955044955...

Which means that instead of being at 'virtual ground', the opposite end of the resistor is at +4.5mV, which then adds to the bias voltage on the negative input.

I don't think this logic is correct.  You are working with DC and your cap is blocking DC regardless of any ESR.  You need leakage to get a current to flow to ground through the cap and then the 1K resistor.  Without leakage, there is no current and no voltage drop.

[JDoyle]

I don't think this logic is correct.  You are working with DC and your cap is blocking DC regardless of any ESR.  You need leakage to get a current to flow to ground through the cap and then the 1K resistor.  Without leakage, there is no current and no voltage drop.

Good point, obviously I'm confused as well about which error is which...

Anyway, while I was obviously wrong, I stand by the difference with regards to leakage, and I'll stand by it even if we can find a actual measurement. The numbers I have seen have been in the 1 to 100 nA (.1 to .001uA) range.

Thing is, this is still enough to make a difference, and the reason is something I deliberately ignored because I'm not doing the math longhand online:

The current through a diode is:

Id = Is * [e^(Vd/nV_T) + 1]

Where:

'Id' is the current through the diode
'Is' is the reverse leakage current of the diode
'e' is the irrational number e, the base for natural logarithms
'Vd' is the voltage across the diode
'n' is the emission coefficient
and V_T = (^kT/_q); where 'k' is Boltzmann's constant, 'T' is the absolute temp in Kelvin and 'q' is the electron charge

If you feel like dealing with the math set the leakage current through the diodes at the 1nA we both seem to have found out there for cap leakage, and then take the gain to 100 (Vd is the resultant offset voltage) and note the difference.

So even at extremely small offset voltages like .45mV or even .045mV that would result from a leakage current of 1nA, any offset is going to make a difference in the way the diodes operate.

Whether it is a big deal to you or not, that is your own decision, but I'll stand by it either way: if there is some way for current to leak through that cap, the diodes are going to operate differently if the cap is connected to ground then it will if it is connected to Vbias...

[JDoyle]

I guess the biggest difference is the charge/discharge path of the cap is through the opamp when the cap is connected to ground. This takes away from the opamp's ability to do other things, like amplify.

[gaussmarkov]

Whether it is a big deal to you or not, that is your own decision, but I'll stand by it either way: if there is some way for current to leak through that cap, the diodes are going to operate differently if the cap is connected to ground then it will if it is connected to Vbias...

yes, i think you are right about that.  it's an interesting observation and it certainly did not occur to me until you pointed it out.

[R.G.]

Jay, I think you may be mixing up AC and DC effects. The ESR looks like it's in series with the cap. The 1K is definitely in series with the cap. Elements in series may be put in any order and not affect the voltages across or currents through the combination as long as you do not look at the inner nodes. In this circuit, the 1K, the cap and the 1 ohm are in series. We can reorder this to be 1K, 1 ohm, cap. So there is no difference between a 1001 ohm resistor and a 1K+1 ohm. The ESR does not cause offsets.

The parallel leakage resistance does. The parallel leakage resistance is the theoretical resistor in parallel with the cap that would leak the same current.

But leakage resistances are much bigger than 1M in general for electros. That runs the DC offset voltage down into the mud.