Help me walkthrough an oscillator cycle?

[varialbender]

I'm using the simple dual LFO from eaced.com, just running it off +/- 15V.

Can anyone help me walkthrough one cycle of the oscillator?
I really want to fully understand everything that's going on.
I'm hoping to add a tilt control, and hoping that a pot with center to the rate pot and ends to opposite diodes would work.
I could just add the few parts in the breadboard and find out, but I think it'd be helpful to understand why.
Thanks a lot.

[R.G.]

The drawing makes it a bit hard to understand, being physically oriented rather than electrically.

We can ignore the AD633 for the purposes of this discussion.

The left half of the TL072 we will call opamp A, and the right half opamp B.

The action of opamp A is an inverting integrator. The + input is grounded. The - input is fed current through the 10K and variable part of the 1M speed control. As that current comes into (for positive input voltages) or out of (for negative input voltages) the - input node, the action of the amplifier is to force its output in the opposite direction. The output pulls (pushes) current through the capacitor to equal the currrent from the input resistor and keep the voltage at the - input equal to the voltage at the + input. The only way you can cause a constant current to run into/out of a capacitor is by constantly changing the voltage, so the output of opamp A ramps up(down) at a constant rate, at least within the limits of the power supply.

The current into the - input of Opamp A is Vin/Rin, where Vin is the output of opamp B, and Rin is 10K plus the setting of the speed pot. Let's just call that Rin.

We know that for a capacitor, I = C dv/dt, and in this case I = Vin/Rin, so the output ramps with dv/dt = -Vin/(Rin*C). The minus sign just says that it goes the opposit way from Vin. If Vin is positive, the output ramps negative and vice versa.

Opamp B is set up as a schmitt trigger. If its output happens to be high (as high as the opamp output can go, usually about 2V less than V+), then the output itself puts a voltage on the + input through that 100K resistor. The - input is grounded, so for all input voltages on the 62K resistor that make the + input above ground, the output will remain positive. We can calculate that.

For Vout = 13V, if the + input were ... just barely... at ground, then there would be 13V/100K = 130uA going from the output to the + input. But the + input can't eat any of it, it's a high impedance. The current has to be balanced by a matching but negative 130uA being pulled out by the 62K resistor. So the 62K resistor has to have V = 130uA*62K = 8.06V across it, and that has to be negative to be sucking away the 130uA from the 100K. So if the voltage at the input side of the 62K resistor is more negative than 8.06V, the + input is more negative than the grounded - input, and the output goes from +13V to -13V.

But look at what happens. The instant the output B output pin drops from +13V, it can't supply as much current to the + input through the 100K resistor, so the output drops faster. And faster. And then the output is pulling the + input negative. Bang - the output slews to the negative limit as fast as it possibly can.

And now that it's negative, the output pulls the + input negative, so we have to pull the 62K resistor above ground by that same 8.06V to get the thing to switch again.

That's why it's called a trigger. It refuses to switch until you pull hard enough on the input, then it bangs to the other state and stays there until you forcibly pull it out of THAT state.

We have that schmitt trigger being fed by the inverting integrator's output, and the inverting integrator's input fed by the schmitt trigger's output. When power is turned on, the 5uF cap is initially uncharged and remains that way until it begins to be charged by the action of opamp A. There will inevitably be some tiny imbalance that will make the snap-action schmitt trigger of opamp b flip to either output high or output low. Let's guess that it happens to go high.

Opamp b's output at pin 7 is high. So that voltage (we guessed about +13V) is applied to its own + input to hold it there. And it's also applied to the Rin of the integrator. So the integrator sees an input of +13V across 10K plus the speed control. Current's coming into the - input. Opamp A starts pulling its output negative and pulling current out of the - input through the 5uF cap. The output of the integrator ramps negative. And that negative output is applied to the 62K input resistor of the schmitt trigger opamp B.

Nothing happens until opamp A's output crosses -8.06V, and then BANG Opamp B's output switches from +13V to -13V. That starts pulling current out of Opamp A's - input through Rin. Now opamp A has to ramp UP to balance that input current, and opamp A's output starts ramping up from -8.06V toward +8.06V where it trips opamp B over the other direction.

I've given you the description in a manner that lets you adjust the duty cycle. All you have to do is to use two diodes to separate the incoming + and - currents to the integrator to get a variable duty cycle.

There is an oscillator that does just this in schematic form at GEO, by the way.

[varialbender]

Thanks a lot RG!
I understand most of that, and will keep going over it until I either understand it all or know that I need to ask more questions.