SV: Re: [sdiy] BBD clock circuit

[jhaible at debitel.net]

> JH, how does the 1/X function work?
> Is it doing log arithmetic?

Yes. Starting from the back end, Q1A performs an expo function.
(collector current vs. BE voltage). It's important that you need
a _negative_ BE voltage here. So in all the log arithmetic below, negative 
voltages are counted as "+", and positive voltages are counted as "-".

Now starting at the front end. Consider pin5 of U1B at a fixed voltage.
Let's asume 0V for now.

Then R3 sinks a current i3 to a virtual GND node (pin6), and the "diode"
Q2A performs a log function. (Don't bother about constants here - they
will all cancel in the end.) Important: As R3 is sinking, the log'ed
voltage goes up (opamp output), and we count this as "-".

The voltage from the opamp output is now downshifted by two diode voltage
drops (Q2B and Q1B). We count this as a "+" contribution. Important: the
current thru these diodes is different: It's the current i1 set by R1.

Now the expo transistor sees a voltage at its base which is basically
(without constants) -log(i3) + 2*log(i1). 

After performing the expo function, the log's dissapear, subtraction becomes
a division, and a multiplication with 2 becomes squaring.
So the collector current of Q1A is 

i_out = i1**2 / i3.

i3 was out input current, amd it's magically found in the denominator.

But the _real_ magic is that the same circuit performs an expo function
from the other input of the opamp (pin5) at the same time. (Which is
easy to show.)

I hope this was of some help, even though omitting all the constants
may look criminal. They really all just cancel in the overall function.

I think whoever invented this circuit is a true genius. Dynacord,
for comparison, made the 1/x function in the TAM-19 with a 3080 in
an opamp feedback path to divide, and a thousand of trimpots.

JH.

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