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I was searching the internet for a circuit to generate a pulse when its input is held high, and I came across this circuit:

enter image description here

I tested it and it works like a charm.

What is the purpose of the last two transistors on the left? (Q3 and Q4)

I think they make a Darlington pair, but what is its purpose here?

I tried the circuit without it and it works without a problem.

I am new to this so please help me understand.

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  • \$\begingroup\$ Try with R6 > R7 and see what happens with transitions... \$\endgroup\$ Commented Apr 17, 2021 at 7:17

3 Answers 3

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Without load, the output is the inverted signal of Q2. The low value of R8 make sense if the Q3/Q4 is an open collector design, so can drain much more current than Q2 if you connect a load between DC supply and output terminal.

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  • \$\begingroup\$ So Q3 and Q4 are just to increase output current capability? \$\endgroup\$
    – Seif_1999
    Commented Apr 16, 2021 at 14:22
  • \$\begingroup\$ It looks like that was the intention, but Q3 emitter is not connected to Q4 base, so really no improvement in current. \$\endgroup\$
    – Bravale
    Commented Apr 16, 2021 at 14:51
  • \$\begingroup\$ Okay then, I will just eliminate them from the circuit. Thanks \$\endgroup\$
    – Seif_1999
    Commented Apr 16, 2021 at 14:54
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Looking at the schematic and your question, I would have to say the purpose would be to ensure absolute switching, and provide the pull up voltage for the next stage.

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I didn't want to give a long description of how a Schmitt trigger works or why the circuit needs one because there's lots online and I shouldn't have to. All the answer needs is the word "Schmitt." I went over because there's a minimum character limit.

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  • \$\begingroup\$ You should notice that R6 = R7 in the OP's diagram and pay attention to OP to correct it... \$\endgroup\$ Commented Apr 17, 2021 at 7:25

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