Hi again,
Frank, you havent yet drawn the circuit for us to see, so i can only
guess. If you draw the circuit we will solve this in 10 seconds. If
you dont draw the circuit it's going to take 10 posts instead.
Anyway, since you say that it is a negative going edge that triggers
the next stage then you do need the cap in series and the resistor
to pull up, but the diode can not be in series with the cap as you
seem to be suggesting it is now. Take out the diode and short it
(if it's in series with the cap) and everything works. A diode in this
application would be to protect the input of the next stage from
an over or undervoltage. In this case, when the previous stage goes
high the cap discharges completely (or nearly so), then when it goes
low the low appears at the input of the next stage for a short time
period depending on the size of the cap and the size of the resistor
and the input impedance of the next stage. All is well so far, but
now the cap is charged by the resistor in a way that makes the
right side positive (+5v approx), and now when the previous stage
goes back to positive the right side of the cap jumps up to +10v
(a voltage doubler circuit). This could blow out the next stage,
so a diode clamp is used to prevent the input of the next stage
from receiving +10v instead of the lower and safer +5v. To
get this protection the diode is connected with the anode to the
input of the next stage and the cathode to +5v.
Note that a diode in series with a cap does not work very well
anyway, because it depends on the leakage of the diode to operate,
if it operates at all.
The circuit i am recommending for your application would therefore
look like this:
Code: Select all
diode
+5 o--------------------+-------|<|------+
| |
| |
| |
R |
| |
| |
Sig o-----C--------------+----------------+-------o Sig Out
GND o---------------------------------------------o
Note you may wish to include a small 100 ohm resistor in series
with the cap to limit the discharge peak current through the diode.
This current might not hurt the diode, but it may cause the +5v
line to rise for a short time period which could bother other parts
of the circuit. With the extra 100 ohm resistor this doesnt happen.
Here are the waveforms and the circuits for negative edge triggering:
From these you can see that this should work very nicely.
If the input of the next stage has significant impedance, you may have
to increase the value of the capacitor.
LEDs vs Bulbs, LEDs are winning.