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555 trigger level converter for Flextronics

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kinjalgp

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I don't think the circuit will work. The base of transistor will be always 0.7V less than its emitter. So it won't be biased at all. This thing can work for PNP transistor but not NPN.

Input should be applied to base and the diode's anode will go to base and cathode to emitter.
 

Phasor

Member
The base of transistor will be always 0.7V less than its emitter
But the diode is reverse biased, so the 0.7V doesn't apply :?. It looks to me, like it should work.
 

Roff

Well-Known Member
bogdanfirst said:
just curious of something........why did you add the capacitor?
it could work without it too.....
Yeah, I didn't have the capacitor at first, but I decided it would be a more general solution for a non-inverting translation of a low voltage signal to the trigger input of a 555. With the cap, the DC level of the input signal is irrelevant, and besides, you need a differentiator anyway if the input signal duration is greater than the pulse width of the one-shot you are triggering. The cap kills 2 birds with one stone.

Kinjalgp, it works. Breadboard it if you don't believe it. The capacitor charges one way through the diode, and the other way through the emitter.

Ron
 

kinjalgp

Active Member
This assumes that your 0 to -2v source can drive the load, which is basically 4.7k
Oh.....Sorry I didn't read that. I thought the clock transition is from +2V to 0V. But actually it is from 0V to -2V. In that case it will work.

You have not shown -2V in the schematic. :)
 

Roff

Well-Known Member
kinjalgp said:
This assumes that your 0 to -2v source can drive the load, which is basically 4.7k
Oh.....Sorry I didn't read that. I thought the clock transition is from +2V to 0V. But actually it is from 0V to -2V. In that case it will work.

You have not shown -2V in the schematic. :)
As I said in my previous post, the DC level doesn't matter. The transition only has to be a -2 volt transition. It could be from 50 volts to 48 volts, or -67.3 volts to -69.3 volts, and it will still work. Of course, you have to wait for initial conditions to settle, as in all AC coupled circuits.

Ron
 
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