Something that does not make sense (to me) about the TS-808 schematic

[Hardtailed]

Let's start with said schematic (from Geo)


I'm sorry, maybe I'm just stupid , but I just can't understand something. I have some but limited knowledge about electronics, mostly I've never dealt with audio processing and with semi-conductors before, but I'm getting closer to my goal of understanding everything! I don't want to clone a TS-808 but I'm working on a design of my own which will gather ideas from the TS and the Guv'nor (my current OD pedal, black serie of course).

Ok, to my question, in the clipping amp section, we see that a 4.5V is applied just before the opamp input, this I suppose is to bias the signal to half of the available power (I guess we could call that "class A" biasing). Then we have the feedback loop, I understand most of it, how the parallel condensator limits treble content, how the condensator in serie to ground limits low frequency content.

What I can't understand is, if it's biased to 4.5V, how can the diodes clip both side of the signal? Technically, what comes out of the opamp is pretty much all positive with the middle point being around 4.5V right? So one diode would be always off, and the other always on.

Please help me figure out that one, I will love you forever afterwards!

Thank you!

[cd]

You need to read two things:

Technology of the Tubescreamer: http://geofex.com/Article_Folders/TStech/tsxfram.htm

Effects FAQ (specifically, the clipping section): http://geofex.com/effxfaq/distn101.htm

The first picture in Distortion 101 is what you want.  In the Tubescreamer, the part marked "zero volts" is 4.5V, and the signal goes above/below 4.5V (biased to half the power supply - NOT NECESSARILY CLASS A!!!)  The clipping diodes limit the amount the signal can go above/below 4.5V.

[Transmogrifox]

Another way to look at it is this:

The feedback loop in an op amp causes the opamp to ensure that the inverting terminal is always the same as the noninverting input.  In other words, the inverting input voltage tracks the noninverting input almost perfectly in a good op amp.

Since the inverting input is coupled to ground through a capacitor, there is not DC current flowing through that branch, so the DC voltage on the output is the same as the inverting input, and the same as the noninverting input.

For example, if you changed that 4.5Volt DC bias to 2 volts, the output would be 2 volts DC, and the inverting input would be 2 volts.

IN any case, the voltage drop accross the diodes in steady state mode is 0 Volts ( 4.5V-4.5V = 0)

When you apply an AC input to the opamp, things change.  Now because you're charging and discharging that capacitor to ground, you now have current flowing from the noninverting terminal to ground.  Because the noninverting terminal must stay at 4.5Volts plus the AC input, all that current to ground through the series R and C must come through the feedback resistor (500k pot & 51k Resistor).  Therefore, the output voltage must be great enough that the output minus the input divided by the feedback resistance equals the current flowing from the inverting terminal to ground.  If you took the capacitor out, that would be true for DC inputs as well, and the op amp would be in the condition that you described, where one diode is on all the time.

Now with the diodes, as you apply AC input, the equivalent impedance with the diode decreases exponentially with the voltage change across it.  The diode "on" voltage is defined near the "knee" of the exponential curve.  After that, the change in the output of the op amp gets to be less and less sensitive to the input and so the signal is nonlinearly clipped.  This is effectively a logarithmic amp.

The output looks like : Vo = Vin  + (n* k*T*ln(Vin/Is*R) + 1

where
n is an integer
k= boltzman's constant
Is= a constant dependent on the diode's physical properties
T = termperature (usually assumed ~25 C)


The above is approximately right on high gain settings where the current through the pot and 51k resistor are very small compared to the current through the diode.  It begins to fail somewhat when you put a parallel resistance in with the diodes that is closer (within an order of magnitude) in value to the 4.7k resistance to ground.

As we now consider the capacitance to ground, we see that we are boosting higher frequencies more than the lows, and are lessening intermodulation distortion cross-terms.

I think RG's articles about the tubescreamer (as linked above) are an excellent resource for understanding this circuit.  If I was effective at confusing you, I hope RG's articles will set you straight.

[Hardtailed]

Thank you for the information.

I did read Geo's pages (Distortion 101 and Technology of the Tubescreamer, which is where I got the schematic from).

However, I still don't see how the diodes would not "see" the voltage of the DC part of the signal. I do understand how the capacitor blocks it from going to the ground, but the potential difference would still exist between the two (which is what is charging the capacitor when the circuit is first engaged). Guess I still have a few things to learn...
I do have "The art of electronics" on order.

Anyway, thank you

[Khas Evets]

Shouldn't the 4.7K and .0047uF go to 4.5V rather than ground? I've seen it both ways on different schematics. I know the SD-1 is to 4.5V. For the purposes of this discussion, it doesn't really matter since the cap blocks DC, just curious.

[Transmogrifox]

Thank you for the information.

I did read Geo's pages (Distortion 101 and Technology of the Tubescreamer, which is where I got the schematic from).

However, I still don't see how the diodes would not "see" the voltage of the DC part of the signal. I do understand how the capacitor blocks it from going to the ground, but the potential difference would still exist between the two (which is what is charging the capacitor when the circuit is first engaged). Guess I still have a few things to learn...
I do have "The art of electronics" on order.

Anyway, thank you

The diodes do "see" the DC part of the input.  They see the same DC voltage on both sides, so neither is forward biased until there is a changing voltage on the input that would cause the capacitor to charge or discharge more.  If there was a changing voltage on the input (guitar signal), then it gets to where the capacitor is charged and discarged through the diodes.

The 4.5V bias is called a "virtual ground".  It just DC shifts the input signal's zero crossings to a safe margin above and below the negative and positive supply rails. If you referenced the input to ground, the op amp would be a half wave rectifier, since it would be saturated to ground for all the negative peaks of the signal, and the positive peaks would likely bias it into a linear region.

When I was just learning electronics, I made this mistake.  I breadboarded a 741 op amp set up for very high gain with a 9Volt battery for supply.  I referenced the noninverting input to ground and I couldn't figure out why it had such a gated fuzzy bad sound and didn't have as much gain as I thought it would.  As I kept with it, I quickly learned that what I was doing would have worked fine if I had either used a second 9Volt battery to create a lower negative supply, or I could have split the 9V in half and use the center (4.5V) as ground.   That way you have a power supply of +4.5V, Ground, -4.5V...or you can just do like the tubescreamer and your battery negative is still ground, you just use the half 9V supply as a center reference for all AC signals.

I hope my rambling was useful to you.  It does take some time to fully absorb the "why". It's good you ask the questions even if it feels like we all think you're stupid and just don't get it because you've become the savior to all the people who don't have the nerve to ask.  It also means that the question bothers you enough that you'll put in the effort it takes to understand the answer.  Keep learning and one day, at 1:00 AM, you'll wake up and it will dawn on you and you won't be able to sleep because your mind is figuring out too many things to rest.  I think some more thinking upon RG's techonology of the tubescreamer will help bring the picture together.  If you have a breadboard, wire up an op amp and start experimenting with it.  Use DC blocking capacitors for some experiments, and omit them for others and see what you can learn.  Keep a few spare  resistors and op amps around because you'll probably see the magic smoke come out.  I have let quite a bit of magic smoke out by experimenting.  I think it's fun.