Issue with using a TTL signal to Short two lines

[SpencerPallas]

Hello,

I have a piece of equipment that sends out a trigger for synchronization which is a 3.3V TTL signal lasting ~200us. I need to use this to synchronize with another piece of equipment which requires a short to trigger it. The two lines are a +3V line and the signal in line.

I have tried using a resistor to NPN transistor circuit to accomplish this but it hasn't worked. I've also tried adding 5V relay and an outside +5V power source to see if that would stay closed longer but to no avail.

Any words of wisdom would be greatly appreciated!

Cheers

[pebe]

How long is the trigger pulse that your other circuit needs, and what current does it draw when shorted?

[SpencerPallas]

I'm getting around 18 uA when it is shorted. As for the time the triggered device needs to be triggered, I don’t know.

[pebe]

Sorry I missed your reply. The transistor idea you tried should work OK but the triggered device may require a longer pulse.

Try adding a 1µF cap between C and E of the transistor to lengthen the time that the collector is low. To ensure the transistor switches on hard enough to discharge the cap, feed the base of the transistor from your TTL signal source via a 2k2 resistor.

[SpencerPallas]

It worked!

Thank you very much but could you explain how? I tried using a cap earlier to keep the transistor switched open for longer by attaching it to the base thinking that it would be charged by the signal. How does having the cap between the emitter and the collector keep it open?

Cheers!

[pebe]

As the piece of equipment you are using requires a short to ground to trigger it, then it has to have some form of pull-up on its input. So a 1µF capacitor fitted there would normally be charged by current through the pull-up. The idea was to use your transistor’s collector current to short out the capacitor quickly to produce the negative pulse.

In the circuit, the 2.2K base resistor fed from 3.3V would give about 1.2mA of base current during the 200µS pulse. Most small signal transistors have a gain of at least 50, so that would result in a collector current of 60mA to discharge the cap.

A current of 1A into or out of a 1F cap will change its voltage by 1V in 1sec.
So from the formula T = C x V / I, the time to discharge the 3.3V on the cap at a rate of 60mA is 1e-6 x 3.3 / .06 = 55µS, so the 200µS pulse of current into the collector is ample time to discharge it.

At the end of the pulse the 1µF cap would start to charge via the pull-up and that would determine the length of the pulse delivered to the other piece of equipment.

I hope that explanation helps.