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enter image description here

I was referring to TTL inverter circuit in the book Digital fundamentals by Thomas Floyd and its working principles first proposition went as follows:

  1. When input is HIGH, Base emitter of Q1 is reverse biased and base collector is forward biased.
  2. This permits the current to flow though R1 and into Q2's base, hence turning on Q2.

My doubt is:

1.How can you say that base emitter of Q1 is reverse biased?

  1. We only know that Emitter of Q1 is high.
  2. We do not know what the voltage at the base of Q1 is because there is R1 between Vcc and the Base.

2.How can you say that the base collector is forward biased?

  1. We do not know the voltage at the collector of Q1, because we do know the state of Q2 or the rest of the circuit.

-Also how can Q1 conducts current through base collector, because if base emitter voltage is reverse biased the transistor is in open circuit or off state.

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  • \$\begingroup\$ Do you have a simulator to hand? \$\endgroup\$
    – Andy aka
    Commented Apr 23, 2020 at 14:33
  • \$\begingroup\$ No. I haven't started using simulation. \$\endgroup\$
    – Krishna M.V.
    Commented Apr 23, 2020 at 14:34
  • \$\begingroup\$ But I do have simulink and I don't know to use it properly \$\endgroup\$
    – Krishna M.V.
    Commented Apr 23, 2020 at 14:37
  • \$\begingroup\$ Can you see that the collector of Q1 is connected to R4 via the forward biased base emitter junction of Q2? \$\endgroup\$
    – Andy aka
    Commented Apr 23, 2020 at 14:38
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    \$\begingroup\$ And in parallel with R4 is Q3 b-e, forward biased. \$\endgroup\$
    – user16324
    Commented Apr 23, 2020 at 15:42

2 Answers 2

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Well, you do know Q1 base collector junction is forward biassed.

Looking at Q2 and Q3, you can put an upper limit on the collector voltage of Q1.

That allows you to determine an upper bound for Q1 base voltage.

enter image description here

Finally that lets you establish whether or not Q1 base-emitter is reverse biased for any value of the input (emitter) voltage.

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  • \$\begingroup\$ if you don't like the addition just roll-back \$\endgroup\$
    – Andy aka
    Commented Apr 23, 2020 at 15:46
  • \$\begingroup\$ @Andy it depends if we're giving him the whole story or just prompting him... \$\endgroup\$
    – user16324
    Commented Apr 23, 2020 at 15:49
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    \$\begingroup\$ Feel free to delete @brian it won't upset me (sob sob). \$\endgroup\$
    – Andy aka
    Commented Apr 23, 2020 at 15:52
  • \$\begingroup\$ Thank you that about does it \$\endgroup\$
    – Krishna M.V.
    Commented May 2, 2020 at 3:26
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because R1 takes some of the voltage and Vb is less than Ve of Q1 that is Vcc when it's high. Vb = Vcc - ( R1* Irev) Ve(High) = Vcc And about your second question you need a KVL through Vcc - BC of Q1 - BE of Q2 and BE of Q3.

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