# How to catch a high signal?

Discussion in 'Electronic Basics' started by Irina, May 25, 2007.

1. ### IrinaGuest

Hello all,

I have got a problem with a digital circuit. I got a signal from a
comperator, which changes from 0 to 1 and back. Now I want to catch
the first high signal and hold it until infinity. I try to use a RS-FF
with R = 0 and S = 0,1,0,1,... (my input-signal) , but I got, of
course, an undefind state until the first high comes. How can I get a
zero in the first moment and not an unknown state?

Thanks a lot!

Irina

2. ### Bob EldGuest

Use a "D" flop. Connect the "D" input high (5V) and connect your signal to
the clock input. Take the output on the "Q." When the first positive edge
of your signal appears, the "D" flop will change state to high out where it
will remain. The flop may be reset by using the reset input and should also
reset on power up.

3. ### BlakeGuest

It seems to me that an RS latch is just want you want for this application.

If I understand you correctly, by saying that R=0 and S=0 is an undefined
state, you are implying active-low logic. If that's the case, you want R = 1
for your circuit to operate. Your circuit can't latch if it's perpetually
reset.

But you do want a one-time reset when you set the latch into action. The
obvious solition is to use a momentary reset switch to reset it. Another
option is a high impedance pull-up resistor and a large value capacitor on
the reset pin. Maybe 10K and 1 uF. That way, on ititial power-up, the reset
pin will be held low for roughly one time-constant. You get a one-time reset
pulse, and then the latch is ready to catch the first (active low ?) pulse
on the set pin.

4. ### John FieldsGuest

---
View in Courier:

IN----------A
NOR Y--+
+--B | Vcc
| | |
| A--+ [0.1µF]
+--Y NOR |
| B-------+
| |
+-->OUT [100K]
|
GND

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