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I'm trying to self study electronics. I started reading a book on digital electronics and I got stuck on the first chapter. It's about the logic gate circuit. There is a two input AND gate circuit using three transistors:

enter image description here

In my physics book, the circuit for an AND gate shown is pretty clearly. It used two transistors and I get that theory, but I don't get this.

In this image, why is it that when either A or B or both are at 0 V, X=0 V? I know that in these cases, at least one of the transistors is off, so current from 5V passes through R and to the base of transistor T3, making it on. The current should also flow through the resistor across R, so why does X have an output 0?

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    \$\begingroup\$ X will have an output of Vce of T3, which is low when T3 is conducting (think, tha you can roughly approximate it as short to ground). Not zero though. \$\endgroup\$
    – Eugene Sh.
    Apr 26, 2021 at 14:53
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    \$\begingroup\$ The three resistors and two transistors on the left create a NAND gate. The two resistors and transistor on the right are an inverter that convert it to AND. \$\endgroup\$
    – Dave Tweed
    Apr 26, 2021 at 15:26
  • \$\begingroup\$ Like you said, transistor T3 is on. Being on just means it conducts current and has (almost) 0V over it so X will be 0V. \$\endgroup\$
    – Justme
    Apr 26, 2021 at 16:53

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  1. T3 goes conducts and goes to saturation when either A and B or both are A and B are O. Since path from 5v via T1 , T2 is not complete and in one way open circuit. Since in saturation has Vce less that .3V, you get X=0

  2. When A and B are 1 , path from 5v via T1 and T2 is complete and both the transistor will be in saturation. Hence T3 will not turned on, as Vbe of T3 will be less than 0.7. hence you get X as high

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