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How would one make use of transistor logic gates ? Take for example this transistor AND gate for example .enter image description here

In all logic circuits, the output of one gate becomes the input of another . Here the output of the AND gate formed by Q1 and Q3 forms the input Q2 (which can be assumed as a NOT gate) . So , if and only if current flows through both bases Q1 and Q3 should current flow through the LED and to the base of Q2 ,thus enabling current to flow from the collector of Q2 to the emitter . This makes current flowing through "Out" essentially zero

This is how I understand AND gates . If A AND B are 1 then output is one . BUT If current flows through the base of Q3 (And not through the base of Q1) then that current will go through the emitter (with that 20k + 1k base emitter resistance{Measured with multimeter} resistance though) through the LED and to base of Q2 switching the transistor ON (The 20k resistance in conjunction with the voltage i'm applying at the base of the transistor (6v) is no where near the base emitter cut off current)

This is disastrous, the AND gate is not behaving like it should and is able to switch another base with current only applied at one base . How do you practically get around this ?

IMPORTANT : Ignore the first ground , it isn't supposed to be there

EDIT : The transistor i'm using is a 2N2222A not a 2N2222

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  • \$\begingroup\$ "In all logic circuits, the output of one gate becomes the input of another", This is not always true. Particularly if you are thinking on a transistor level. I believe there a quiet a few useful CMOS circuits that never have the output of one section feed to the input of another. Only ones I can think of right now are simple gates like AND, OR etc. \$\endgroup\$ – Lyndon White Aug 12 '15 at 10:00
  • \$\begingroup\$ That circuit would not work. It would output logic High all the time because the output is directly connected to the supply voltage. Also Q2 would blow up, because there is no current limiting resistor. Try to add a 10k resistor between supply and output to fix that. \$\endgroup\$ – Nils Pipenbrinck Aug 12 '15 at 10:00
  • \$\begingroup\$ R2 is also connected wrong. \$\endgroup\$ – Nils Pipenbrinck Aug 12 '15 at 10:01
  • \$\begingroup\$ @Oxinabox Alright , that was a generalization . Here i'm trying to make a half adder and one output does become the other input; the problem is , adding current to the base of the transistor from where the input is collected will still send current through the emitter and give 1 as input \$\endgroup\$ – Vrisk Aug 12 '15 at 10:04
  • \$\begingroup\$ @NilsPipenbrinck ,alright. a 330 Ohm resistor has been added to the collector of Q2 and R2 is directly connected to the emitter . What do you mean between supply and output , there is already a 10k resistor giving the path from base to emitter an equivalent resistance of 20k and this is faaaar below base emitter cutoff , adding another 10k wouldn't do much . \$\endgroup\$ – Vrisk Aug 12 '15 at 10:09
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Where did you get that diagram? (you copied an NMOS gate circuit??) It has some problems....

  • You mention that the first (leftmost) ground should not be there (because it would short Q2's input to ground, but what should be there is a resistor to ground that leaks the base cuirrent of Q3 to ground without switching Q2 on. With ~ 0.6mA base current I would take a 330 Ohm.

  • (not a fundamental problem, but now you could probably leave out R2, because Q2's base can take the maximum current the input stage can supply.)

  • The output has Q3 as pull-down, but it is shorted to power. You need a resistor between out and power. Its value is a balance between fan-out (the number of inputs that the output can drive) and power consumption. To simplify the disign, I would take 330 Ohm.

Now you have an RTL type gate, but with a somewhat uncommon input stage. The (old) designs I have seen use teh two input transistors in parallel, with or without an output trasistor.

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  • \$\begingroup\$ I'm guessing the 330 ohm resistor (i'll move the resistor from the emitter to the bottom) and the ground will leak enough to keep Q2 in cutoff . I tried to crunch the numbers and prove this will keep Q2 base emitter current below cut off , but I can't seem to work it out . Can you show me the numbers (with steps) to prove to me that Q2 base current will be below 9nA when only Q3 base is connected to positive 6v and greater than 9nA when Q1 and Q3 bases are connected . You see , I'm new to electronics and Kirchoffs laws are a wee bit confusing to me \$\endgroup\$ – Vrisk Aug 12 '15 at 10:28
  • \$\begingroup\$ Alright , I worked something out . 6/10k + V/330 = V/10k . solving for V gives me .2 and so the current flowing through the base of Q2 is .2/10k which is still no where near base emitter cut off current \$\endgroup\$ – Vrisk Aug 12 '15 at 13:50
  • \$\begingroup\$ Did you take into account that the BE junction is a diode, so it will drop a certain volatge, and won't conduct below ~ 0.6V? \$\endgroup\$ – Wouter van Ooijen Aug 12 '15 at 15:38
  • \$\begingroup\$ I'm not sure how this will help, using a voltage divider at emitter will permanently restrictions voltage to. 6 and will not change with current \$\endgroup\$ – Vrisk Aug 12 '15 at 17:14

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