Stereo Tremelo Panner Idea

[zpyder]

Look at most any discreet octaver, and look for a transistor with equal value resistors on the emitter and collector, driving the rectifier.  Like the second tranny in this:
http://www.fortunecity.com/tinpan/humperdinck/223/diffoct.gif
Ben

This link doesn't appear to work... I get the fortunecity logo

zpyder

[puretube]

copy&paste...

[zpyder]

no dice - is it working for you??

zpyder

[Transmogrifox]

www.geocities.com/transmogrifox/osc.pdf

That is the configuration I was recommending for the LED pair.  2.7k resistors are what you'd want to use in a 9V supply. If that isn't bright enough, get more efficient, brighter LEDs, because that schmitt trigger isn't rated to give you much more current than that (2.2k is probably ok).  Note the extra buffer after the oscillator probably isn't necessary, but since the 40106 has so many devices in the chip, you'd just as well use them.   I can't see any reason why the LED illumination is assymetrical unless something in the circuit is assymetrical.  This could be due to mismatched (badly--like different colors or totally different types), or mismatched resistors.  If you were operating the schmidt trigger on 5V, but the LED's on 9V, that would completely explain it.  The CD40106 can operate on a Vdd of 10V according to the datasheet, so it doesn't need to be 5V supplied.

Now to the strangeness of your frequency calculations:  You were using the wrong calculation.  The 1/(2*pi*R*C) formula only applies to finding the frequency where the magnitude is -3dB on a first order filter, and it also applies to certain oscillators and other kinds of 2nd and multiple order filters that are configured in such a way that resonant frequencies can be boiled down to this.  All the same, it is not a magic formula that explains every oscillator, and particularly Schmitt trigger types.

The datasheet says the hysteresis voltage is typically about 3.1 Volts when Vdd is 10Volts when the temperature is 25 Celsius.  These kinds of oscillators will drift in frequency due to temperature, variation of hysterisis to supply voltage ratio, and tolerances on components (caps and resistors). We'll just call this voltage 3V on a 9V supply just for ease of calculations.

Here's what happens:
You turn on power, the input is probably near ground potential (assuming the capacitor has no charge stored on it).  The output swings high and starts to charge the capacitor.  The Schmitt trigger input voltage slowly begins to climb as the capacitor charges, then at about 6 Volts, the output swings low, thus discharging the capacitor slowly, until the input voltage swings down to about 3 Volts.  Then the output swings high again until the cap charges back up to 6 Volts again...and on and on it goes forever.  If you buffer the voltage off the input of this oscillator, you will have a triangle wave shape (approximately).

So...here's how the formula works.
A capacitor's voltage/current relationship is this:

I(t) = C *(dV(t)/dt)

If you haven't had calculus, basically the above states, in words:
"The current as a function of time equals the rate of change of the voltage applied times the capacitance"

So, the faster the voltage changes, the more current will be flowing through the capacitor.  If you have something like a resistor to limit the current, then the voltage can only change so fast, and that's what allows you to tune the frequency of this thing. 

The voltage on the output of the schmitt trigger changes almost instantaneously.  If it had the power to drive a capacitor directly, then the amount of current would be almost infinite in the capacitor.

You put that resistor in there, then the current basically gets as large as it can due to the voltage drop across the resistor, and charges the voltage on the capacitor as fast as the current can trickle in.

So...if you do the differential equation and all that and find the solution, the answer boils down to this:

Vf(t) = Vi + Va*exp(-t/RC)

where Vf(t) is the final voltage on the capacitor
Va, is the voltage applied at the other end of the resistor
Vi is the charge initially stored on the capacitor.  We'll assume this is 0

Now to find the frequency of oscillation, we only need to concern ourselves about the 3 V hysteresis.  Let's just find out how much time it takes for a capacitor to charge from 0 Volts to 3 Volts (which is the same as finding out how long it charges from 3V to 6V).

This will give us the time of half the period of oscillation frequency.  This is because it will take the same amount of time during the other half cycle to discharge from 6V to 3V.  Then it repeats the same thing over again.

So here's what we're doing:
charge time = discharge time
and one charge plus one discharge equals one period of oscillation.

Let's put that formula in a usable form with a little algebra.

Assuming Vi = 0 We have,
Vf(t) = Va*exp(-t/RC)

Now, we know what Vf(t) is.  We want to find time.  We also know Va.  Because we're calling 3 Volts to be 0 Volts, then we need to adjust Va to be 9V-3V = 6V.

Now let's algebraically manipulate this thing:

-t/RC = ln( Vf/Va)
t = -RC*ln(Vf/Va)

Vf = 3V
Va = 6V

t = -RC*ln(3/6)
= -0.69*(-RC)
=0.69RC

This is 1/2 period, so one full perion is
T=2*0.69*RC = 1.39*RC

The frequency in hertz is 1/T, so
f = 1/(1.39*RC)

Let's see how well that matches your observation:
you used,
R = 1 Meg
C = 4.7 uF

f = 1/(1.39*1*4.7) = 0.15 Hz, which is about 6.5 Seconds for a cycle.

I think that sounds closer to what you observed than to what you calculated.

Just note, that the frequency will be approximately the same for any supply voltage (Vdd).  The frequency is dependent on the ratio of the hysteresis voltage to the supply voltage, which is relatively linear...at least enough that it won't make a huge difference in the final answer.