16 Aug
2004
16 Aug
'04
10:04 a.m.
On Mon, Aug 16, 2004 at 08:45:49PM +1000, Erik de Castro Lopo wrote:
On Mon, 16 Aug 2004 12:13:51 +0200
Alfons Adriaensen <fons.adriaensen(a)alcatel.be> wrote:
The current in a silicon diode is an exponential function of
the voltage. Consequently, the small signal impedance is
inversely proportional to the current. A diode does not
have a fixed voltage drop, it still depends on the current.
This is a bit too simple. The diode circuit does
not perform hard
clipping when driven at moderate levels.
Something like
out = tanh (in) (or tanhf() if available)
or
out = in / sqrt (1 + in * in)
Both of those would probably sound better than what I posted,
but the original circuit had a 12k source impedance. As soon
as the the input overcomes the forward voltage drop of the
diodes the impedance of the diodes drops to the low hundreds
of ohms. If you Spice the given circuit I think you'll find
that the
output looks a lot more like what I posted than either of
the above.
The transition from the linear to the clipped region would
indeed be a bit sharper than with either of the functions
I gave. But it's still a function with a continuous
derivative. This makes all the difference for the spectrum
of a 'clipped' signal.
--
FA