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why do we want to amplify.the circuit works on open swith .so bas voltage is set by Rb resistance.if i use some sensor and i gave that output to input of base..i agree your point of amplification..but why its here used,we can also make it with single transistor.right
its quite simple.as q1 is npn transistor it need is positive supply to make the circuit work while switch is in close the negative power is supplied hence it wont work i mean alarm in off. once switch opens..positive power is applied to switch through resistor .hence circuit works..if u have any...
now i am clear with what you say.as there are positive and negative logic system setting on and off.if i connect one input has +5v and other is -5 .what will u consider..a postive logic or a negative logic system
i am new to electronics
in above not gate circuit diagram how i am getting hight output for no input and no output for high input
doubt(even if its hight input..i have collector voltage am i right)
in the above circuit.what are all the voltage drop that i getting is given to led..am i right or wrong.if wrong is there any action realted to voltage drop of resistor to led clarify me plzz
i seen in it in a page it is mentioned we are using positive logic systems that is above +2v for on and below 0.8 to for off.but in below statement its said to ground for 0 .if i do so it looks like - voltage ..help me plzz
"we can define the ideal Digital Logic Gate as one that has a "LOW"...
i seen in it in a page it is mentioned we are using positive logic systems that is above +2v for on and below 0.8 to for off.but in below statement its said to ground for 0 .if i do so it looks like - voltage ..help me plzz
"we can define the ideal Digital Logic Gate as one that has a "LOW"...
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