1. Home
  2. ⏚ Electrical Engineering
  3. Need to generate a TTL signal from a footswitch

Need to generate a TTL signal from a footswitch

I have an optical shutter controller which drives a shutter using either a TTL-level pulse or a footswitch. Unfortunately, there is only one input on the controller. I'd like to construct a circuit with a junction box that will allow me to use both inputs without having to manually disconnect/reconnect each time. I plan to build a logic Or- gate to handle the two inputs. The TTL line is already compatible. I need to build a little circuit which will convert the pressing of the footswitch into a TTL-level signal. The leads of the inactivated footswitch are grounded (that is, when I measure the resistance between the two leads it shows 0 ohms). When the footswitch is activated (pressed) the resistance between the leads is infinite (that is, open). So, there are no active elements in the footswitch, it looks like 0 or infinite resistance depending on whether or not its pressed. Can anyone help me with a circuit that will generate TTL-low (0 Volts) when the switch is inactivated (leads grounded), and TTL- high (5 V) when the switch is activated (leads open)?

Thanks very much!

Reply to
mike
Loading thread data ...

Connect theswitch between signal ground and TTL input. Connect one suitable resistor (for example 1 kohm to 5 kohms) from +5V to that input pin. Now when there is switch installed and not pressed (resistance 0 ohms), you get logic 0. When switch is pressed (open circuit), the input is logic 1.

This pull-up resistor does not cause problems if you drive the input with soem TTL signal source (TTL output can handle this extra pull-up load).

Reply to
Tomi Holger Engdahl

As another poster pointed out, add a resistor to +5V and it's TTL-compatible. However:

1) A long lead may act as an antenna or pick up static electricity, so perhaps a buffer circuit may be necessary; 2) Many switches make intermittent connections at the moment when the switch makes or breaks. TTL may be fast enough to see this as many very rapid openings/closings of the switch. This may or may not be a problem for your circuit. If it is, a one-shot circuit can clean up the signal. It's possible to combine that with the buffer.
Reply to
Michael Moroney

Thanks very much for your replies. I had planned on using a battery for power and I thought this circuit (pull-up resistor) would drain the battery especially because the shutter spends most of the time closed (so switch is shorting). Is that true, or am I worrying for nothing?

Mike

ama.net/

Reply to
mike

Indeed!

From the time the first switch/button was connected to the first logic circuit, the problem of "de-bouncing" had to be addressed.

As to the circuit that "powers" the foot switch: I would prefer a voltage divider that makes the "high" level on the order of 3 volts and a series resistor than makes the "low" level on the order of .3 volts. If the driven circuit has a bit of hysterisis in its input characteristics, a small cap. across the switch may be sufficient for "debounce."

Reply to
John Gilmer

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.