# Delay circuit

Discussion in 'General Electronics Chat' started by umphrey, Mar 8, 2013.

1. ### umphrey Thread Starter Member

Dec 1, 2012
39
1
Hello all,

I was wondering if you guys could help me out with something. I have a circuit carrying a logic signal, +5V or 0V, and I need to add a component such that when the voltage is transitioning from low to high it has very short prop delay (< 200 nanoseconds or so), but when it is transitioning from high to low it gets delayed, maybe by 5 microseconds or so. Any thoughts?

2. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
Assuming you are attempting to make the transitions more abrupt, look into Schmitt triggers. Namely, a Schmitt trigger buffer, or two inverters....

Edit: Sorry, I just reread the post... So a Schmitt trigger would work on making abrupt transitions. Would charging a discharging/capacitor work in your application?

3. ### umphrey Thread Starter Member

Dec 1, 2012
39
1
Well it's state dependent so hysteresis isn't really an issue, the signal is coming out of an AND gate so it's either high or low, but I need one transition to take longer than the other one.

Yes, but, only if I can make it so charge up time from 0V - 5V is =/= charge down time from 5V - 0V

4. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
Are you looking to delay the transition, which would essentially ramp up to a voltage, or are you just wanting the transition delayed from where it would normally be?

5. ### umphrey Thread Starter Member

Dec 1, 2012
39
1
It's just a digital signal, so if it's transitioning from high to low, it needs to get a few microseconds of prop delay, but if it's transitioning from low to high, delay needs to be minimal.

6. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
If that's the case, you could input a buffer, buffer charges a cpacitor, and teh capacitor goes to the input of a Schmitt triggered buffer, with the trip-up voltage set to close to Vdd and trip-down voltage to slightly less, given the impulse response of a capacitor, the transition point to a low output is shortly after the transition, using the step response, the capacitor will take some time to get up to the trip-up voltage of the Schmitt trigger.

umphrey likes this.
7. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
I am curious, why would you want to do this?

8. ### umphrey Thread Starter Member

Dec 1, 2012
39
1
If I get too much current, I have a transducer that sends a signal to break the circuit in 2 different places. But they need to open/close in a different order.