How to introduce delay to a signal

I need to introduce a delay to the output of an astable, it needs to be in the order of 10 micro seconds.

I have read that optical cable is good for delays, which makes sense for small delays, but I have no access to optical cable for the project I am working on. The ideal solution would be some components / IC's. I am working with a square wave, so the signal can be assumed to be digital

• Perhaps D flipflop clocked suitably (you need to take into account propagation delays, too)
– Deep
Jan 11, 2018 at 14:16
• I assume this is for the rising and falling edge. Both have to be delayed by ~10uSec but the duration of the signal is guaranteed longer then that. Any accuracy required? Jan 11, 2018 at 14:22
• What is the minimum time that a zero or one will be present on the astable output? Jan 11, 2018 at 14:22
• I think you could use a circuit similar to the one in this link: angelfire.com/al2/Comunicaciones/Laboratorio/Image722.gif This circuit generates a square signal, so if you feed it with square signal the output will be shifted square signal. Anothe option would be: researchgate.net/figure/… playing with RC in both cases you change the discharge time which changes the delay time. Jan 11, 2018 at 14:29
• Even if you had an optical cable, you'd need 3 km of it to delay 10 us. Since electrical signals also travel with the speed of light, any electrical cable would give the same delay but more attenuation. With optical cables attenuation is less of an issue. Many PAL televisions from the 1980s had a delay line of 64 us. These worked in a piezo mechanical way I believe. I do expect that these delay lines will be hard to find these days. Jan 11, 2018 at 14:39

simulate this circuit – Schematic created using CircuitLab

There is a tolerance on Vt+ and Vt- that shifts withtemperature that will make the delays asymmetric.

Also if the waveform is not repetitive, it will take 20% longer for the 1st edge.

This is my approach if the delay tolerance is adequate.

Since the Schmitt trigger thresholds are 1/3 to 2/3 each delay is 2/3 of V+ which is very close to linear approximation of the RC exponential decay.

• If the astable output is at a frequency much less than 1/10us then this works. If the frequency is higher then it doesn't work. Jan 11, 2018 at 15:21
• Andy , you are correct. "MHz clock" would imply a digital delay needed while the old "astable square wave" vernacular implies to me one is using a lower frequency and will work with this analog delay. Yes, we should be more explicit. Jan 11, 2018 at 15:30
• Instead of Schmitt, you can feed the signal to various logical gate inputs (double not worked for me) Mar 16, 2019 at 3:58
• @SunnyskyguyEE75 I replaced the incorrect part 7414 by 74HC14, and with the indicated $5V 30kHz$ I recovered the same results posted in this answer. Thanks again. Mar 19, 2019 at 20:57
• any time........ It's not really an astable, just an RC delay with 1/3 hysteresis Mar 19, 2019 at 21:07

(Similar to @oldfart 's suggestion...)

Look at the 74HCT595 (5 V) or 74LV595 (3.3 V) 8-bit shift register. This gives a serial input and serial output with an 8-CLK delay in between. You can select your clock frequency to get the delay you want, where the total delay is 8 / fclk with 1/fclk of jitter.

If you want to increase the precision and reduce the jitter, you could cascade several 74x595 in series. For example, using three of these gives you 24 fclk-periods of delay for less than a pound.

My first approach would be a set of shift registers plus an oscillator. You can get 8 registers in a package. The input goes in on one side the delayed signal comes out at the other side. The uncertainty is about 1 clock cycle thus for 800KHz that would be 1.25us. (Sample point at the input as your signal is asynchronous to the shift clock).
You can change the delay by adding more registers or change the clock frequency. With the latter you also influence your uncertainty.

simulate this circuit – Schematic created using CircuitLab

Post edit:
Sorry: corrected my numbers!