TS-808 Circuit Question for the Tube Screamer experts

Started by ethanw, December 09, 2007, 12:50:35 PM

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ethanw

In the TS-808 and it's many derivatives, the often modded .047 cap and 4.7k resistor in the feedback loop are connected to Vr, the 4.5v part of the circuit. What does this accomplish? Why not just connect it to ground?

I read thru the Geo article and everything else I could find and didn't come up w/an answer. Thanks in advance!

Ethan

frankclarke

You would connect it to ground if you had a bipolar power supply that the opamp expects. The workaround is to have a voltage divider on the power supply and use 4.5V as opamp virtual ground.

ethanw

Thanks Frank. I'm working on a modded TS type circuit and acccidentally connected this leg of the circuit to ground. It works and sounds pretty good but seems a little dull sounding and maybe not as loud as it should be. Would a connection to ground affect the sound of the circuit? Sounds like the opamp wouldn't like it too much.

Ethan

d95err

Quote from: frankclarke on December 09, 2007, 12:56:29 PM
You would connect it to ground if you had a bipolar power supply that the opamp expects. The workaround is to have a voltage divider on the power supply and use 4.5V as opamp virtual ground.

Not correct in this case. The .047 cap blocks DC, so it doesn't matter if the 4.7k resistor is connected to ground or to the bias (4.5V) voltage. Both are the same from an AC perspective. I assume it just happened to be convenient to use the bias voltage instead of ground for layout reasons in the original Tube Screamer.


John Lyons

No doubt it "works" either way but is there a difference that someone can explain?
Put another way, will this make a difference soundwise?

Color me curious.

John

Basic Audio Pedals
www.basicaudio.net/

gaussmarkov

like d95err said,  it won't make any difference.  with a cap there, you are connected to a constant voltage so that AC signals behave the same whether it's ground, 4.5v, or even 9v.  remember that caps block dc.  as long as you do not exceed the voltage potential that a cap can handle (remember those voltage ratings?), it will wiggle with the AC signal no matter what DC voltage it is holding back.

John Lyons

Hmmm... ok.
The cap blocks DC and the RC combination off the FB loop is a low pass filter, but how does this work connected to either ground, Vb or 9v?
How does it compare to the vb connection to the 1M at the op amp input with no cap there? (showing my ignorance here :\

John

Basic Audio Pedals
www.basicaudio.net/

frankclarke

There is usually a 10uF cap from virtual ground to ground, so they end up as being much the same thing for AC.

gaussmarkov

a capacitor transmits AC.  the key thing that you need to remember is that theoretically, a cap will transmit AC the same way no matter what the DC potential is across it (within certain limits).  if there is no alternating current on either side of the capacitor, then no AC travels through it and everything is quiet.  if there is alternating current on both sides of the capacitor, the AC on both sides will travel through the capacitor in both directions and we get a mixture of the two signals.  a passive RC filter must have no AC on the outside lead of the capacitor but any constant DC level will do.  so any quiet power rail, 4.5V or 9V, can be the place that the low pass filter dumps the high frequency content that a low pass filter attenuates most.

as frank says, you need a big capacitor to keep the 4.5V supply line quiet.  that voltage is coming out of a voltage divider and every time you put a resistor or a capacitor in series with the DC supply you provide another opportunity for AC to reverberate.  so the 4.5V level can move around even if the the original 9V supply is rock solid.  a 10uF cap gives any AC that arrives on the 4.5V net a way to escape to ground.  here again, we could hook that cap up to 9V instead.  it is more probable that ground will be solid than 9V will be in most setups.

the 10uF capacitor isn't a perfect escape valve for AC.  some still remains.  the 10uF cap is acting as part of another low pass filter.

gaussmarkov

fwiw, i have an explanation of some of this on gaussmarkov.net.  if you want to jump into the middle of it, try Capacitors 3: In Series and In Parallel.  i use some math notation, so it won't help everyone. 

in words, what is written there is that like Ohm's law for resistors, V = I x R, where V is voltage, I is current and R is resistance, there is a law for capacitors where V is replaced by "rate of change in V" and R becomes 1/C where C is capacitance.  suppose we represent the rate of change in V by lower case v.  then  v = I / C  is the equation describing capacitor behaviour. 

if you can accept this "law" for capacitors, then you can note that it does not depend on the level of voltage.  it depends only on the rate of change of voltage.  and that confirms what we have been saying above:  AC goes through a capacitor the same way for various levels of an additional DC potential across the capacitor.

John Lyons

Ok, thanks for the detail gauss.
I think I'm getting it now.

John

Basic Audio Pedals
www.basicaudio.net/

ethanw


DougH

Some effect op amp circuits (Brown Source, for example) have Vr fairly underfiltered. Don't know if this is on purpose or not. But connecting "ground" points to Vr in this case will affect the sound of the circuit. IIRC, tube screamers have decent filtering for the Vr rail. But I suppose people who claim to hear what only dogs can hear may hear differences between using Vr or gnd. ( :icon_wink:)  YMMV...
"I can explain it to you, but I can't understand it for you."

JDoyle

'Ground' and 'Virtual Ground' are the same thing to our signal when looking at them through the cap, BUT, to a cap they are very different, they are a difference in voltage that can only be solved by charging or discharging that difference.

Basically, when connected to Vbias, the cap is 'passive' it just reacts to the signal by charging or discharging in accordance to the signals swing around Vbias. Vbias is it's zero point, so in absense of a signal both leads of the cap are equal in charge and it neither has to charge or discharge when at rest. It doesn't matter if it leaks current because with no voltage difference between it's leads there is no current (for any device, see Ohm's Law).

When the cap is connected to ground there is constantly a difference between the voltage at it's leads and except for the negative signal swing when it is draining, it is constantly in need of charging. And if it leaks (it does) it is CONSTANTLY charging and stealing current from the negative input (thus from the output of the op amp) reducing the gain available. This also means that the time it takes for it to charge to the crest of the positive swing is different from the time it takes to discharge (which it is constantly doing anyway) to the base of the negative swing, making the roll off point assymetrical for the different polarity swings. This is because for the positive swing the total current into the cap is what it would 'normally' be MINUS the constant leakage current; for the negative swing, the current being drained is the 'normal' value PLUS the leakage current. Thus, the assymetry.

Regards,

Jay Doyle

gaussmarkov

i'd like to hear more about this.  i was not aware that leakage was big enough to matter for 47nF caps.  obviously the simple capacitor theory that i was describing above does not include leakage.

i just went googling for info and here it is reported that a Panasonic polypropylene capacitor leaked 2.7mV per day starting at 9V.  that does not sound like an appreciable rate for a tubescreamer application.

i do not understand what the dissipation factor is but that seems to be what datasheets report.  for the Panasonics that smallbear used to carry (and probably still does) that number is around 1%.  again it sounds small, but i really don't know what it means.

so what size effects are you talking about?

JDoyle

A 6 mV change in the voltage across a diode causes the current through the diode to increase by a factor of 10.


gaussmarkov

Quote from: JDoyle on December 11, 2007, 03:35:40 PM
A 6 mV change in the voltage across a diode causes the current through the diode to increase by a factor of 10.

thanks.  i still don't see a significant effect.  diodes may be sensitive but there must be enough capacitor leakage for that to matter, right?

take the 2.7mV per day figure that i found and blow it up to 1000 mV per day.  that's 0.012 mV per second.  for an 800Hz signal, that's 0.0000145 mV per cycle.  that seems like an inconsequential effect.

JDoyle

Quote from: gaussmarkov on December 11, 2007, 04:48:13 PM
Quote from: JDoyle on December 11, 2007, 03:35:40 PM
A 6 mV change in the voltage across a diode causes the current through the diode to increase by a factor of 10.

thanks.  i still don't see a significant effect.  diodes may be sensitive but there must be enough capacitor leakage for that to matter, right?

take the 2.7mV per day figure that i found and blow it up to 1000 mV per day.  that's 0.012 mV per second.  for an 800Hz signal, that's 0.0000145 mV per cycle.  that seems like an inconsequential effect.

Yeah, I got that. But I need you to tell me how the length of time it takes for an open circuited, no contact to the outer world, cap to discharge has anything to do with a circuit that has a cap constantly charging and discharging and where millivolt differences on its leads are matched with exponential current changes. I'm just not believing that one has anything to do with the other. ESR and DA exist. 

Think of a sine wave going horizontally across the page and a line directly through that, the bottom of the page is 0V, the top is 9V, you just drew your signal on the cap across the page, and the center of your wave is Vbias. If the cap is connected to Vbias, the area between the interior of the sine wave and 0 (everything under the sine wave to 0 when it is going positive, and everything above the wave to 0 when it is going negative) is the time the cap spends charging and discharging. Now, if the cap is connected to ground, the entire bottom half of the page, everything below the sine wave whether it is swinging positive or negative, is filled in; because with the constant leakage, the cap is always charging.

My point is that any current that leaks out of the cap to ground has to be replaced by the op amp. And the leakage occurs at the negative input of the op amp so the op amp has to supply that current through the feedback nextwork making the sum needed the leakage current times the gain. And that current leaking creates a DC offset (the slower it leaks the closer to DC it gets), which allows the reverse diode leakage(x gain) and input current offset (1/2 x gain) to become much more of a problem. Plus, the more resistance you have in series with the cap, the higher the ratio of leakage current to signal current.

No, all of this is not extremely significant and tone shattering. But it there IS a difference between connecting the cap to ground and to Vbias and it DOES have an effect.

But hey, no need to take my word for it, try simulating it in PSPICE, you will see less total gain with the cap to ground versus the cap to Vbias. Unless I screwed something up when I did it...

Regards,

Jay Doyle

gaussmarkov

#19
Quote from: JDoyle on December 11, 2007, 07:43:41 PM
But I need you to tell me how the length of time it takes for an open circuited, no contact to the outer world, cap to discharge has anything to do with a circuit that has a cap constantly charging and discharging and where millivolt differences on its leads are matched with exponential current changes. I'm just not believing that one has anything to do with the other. ESR and DA exist.

hmm.  maybe that was the experiment.  it's hard to tell.  i assumed, perhaps incorrectly, that leakage would be measured with a ground connection.

Quote from: JDoyle on December 11, 2007, 07:43:41 PM
My point is that any current that leaks out of the cap to ground has to be replaced by the op amp. And the leakage occurs at the negative input of the op amp so the op amp has to supply that current through the feedback nextwork making the sum needed the leakage current times the gain. And that current leaking creates a DC offset (the slower it leaks the closer to DC it gets), which allows the reverse diode leakage(x gain) and input current offset (1/2 x gain) to become much more of a problem. Plus, the more resistance you have in series with the cap, the higher the ratio of leakage current to signal current.

No, all of this is not extremely significant and tone shattering. But it there IS a difference between connecting the cap to ground and to Vbias and it DOES have an effect.

i did not ask about the effect of leakage.  leakage surely has an effect.  i asked whether the effect was big enough to matter in this case.  maybe it isn't?

Quote from: JDoyle on December 11, 2007, 07:43:41 PM
But hey, no need to take my word for it, try simulating it in PSPICE, you will see less total gain with the cap to ground versus the cap to Vbias. Unless I screwed something up when I did it...

what spice model are you using for the capacitor?  the default in LTSPICE is an ideal capacitor without leakage and, of course, there is no difference.

cheers, paul