Re: idea about differential circuit
- From: "Joel Kolstad" <JKolstad71HatesSpam@xxxxxxxxx>
- Date: Thu, 6 Sep 2007 18:25:10 -0700
<mookiewookie@xxxxxxxxxxx> wrote in message
news:1189126404.917758.178350@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Is there a way to build a two-way differential circuit i.e. a circuit
that compares a varying voltage to a reference, and if that voltage is
rising, it gives a certain value (say 1), while if its decreasing, it
gives 0. I was thinking of using a quad comparator but is there a
simpler solution? TIA
How about: Take a capacitor, connector one side to your signal (assumed to be
low impedance), the other side to a small resistor. Size the resistor and
capacitor such that the reactance is much greater than the resistance (in
general, this means a rather small R). The idea is that the circuit looks
like just a capacitor to the input, so I is still very near equal to C*dv/dt,
but you generate an output voltage across the resistor (Vres=C*dv/dt*R) which
is the derivative of the input signal. The sign of that voltage tells you is
the voltage is rising our falling -- use a single comparator to "square it up"
if you need to.
You'll probably also need a bit of filtering/hysteresis in the system so that
noise doesn't cause the output to oscillate wildly when the input signal is
stationary.
.
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