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  • Will CE amplifier with a DC input amplifies at the output?

    prashanth463

    Member

    Updated: Oct 26, 2024
    Views: 990
    it might sound to be a very basic question but i still keep confusing with this
    as far as i know the output of the ce amplifier when given with a dc input, the output is nothing but just the Vcc, how could this be considered as amplification?
    please correct me if i'm wrong
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Replies

  • Abhishek Rawal

    Member • Mar 1, 2013

    Take a look at this CE Amplifier image :

    [​IMG]

    At input you see Capacitor C1 - What capacitor does ? It blocks DC (Direct current) & allows only AC to pass.
    So, when you pass DC from input all DC current will be blocked.

    Not technically correct image but informative :
    [​IMG]

    R1,R2 & R5 are voltage dividing biasing resistors.
    +12V Vcc supply is DC supply which is given for that.Basically, +Vcc is not considered as "input" in amplifier, it helps in amplification of AC input.

    prashanth463
    the output is nothing but just the Vcc, how could this be considered as amplification?
    please correct me if i'm wrong
    I don't think we'll get any output as C3 will get short & all current will be grounded. (I am not cent percent sure about this, but i think this is how it works - learnt this subject 2.5 years ago in diploma as EDC-II , so correct me if I am wrong too )
    😀
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  • prashanth463

    Member • Mar 1, 2013

    #-Link-Snipped-# thanks for your reply
    as its a dc input we don't use the capacitors, which means that the output is +12V (according to one of my friend who has explained to me)
    looks like i'm again back at the initial stage of my doubt
    😀
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  • prashanth463

    Member • Mar 1, 2013

    #-Link-Snipped-# i think i have got an answer but i don't know whether it is correct or not
    if dc is given as an input, the whole circuit just acts as an NOT gate
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  • Jeffrey Arulraj

    Member • Mar 2, 2013

    True to an extent as the biasing of the ckt gets rapidly affected cos of the additional DC supply given at the base pushing it to the cut off or saturation region where the transistor is mostly in OFF state or in improper working state
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  • prashanth463

    Member • Mar 2, 2013

    Conqueror
    True to an extent as the biasing of the ckt gets rapidly affected cos of the additional DC supply given at the base pushing it to the cut off or saturation region where the transistor is mostly in OFF state or in improper working state
    true? you mean it acts as a NOT gate?
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  • Jeffrey Arulraj

    Member • Mar 2, 2013

    No I don't think so man it is not an inverter pal there is only a .7 voltage drop in the base and the output is 4.3V for 5 V
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  • Abhishek Rawal

    Member • Mar 2, 2013

    Conqueror
    No I don't think so man it is not an inverter pal there is only a .7 voltage drop in the base and the output is 4.3V for 5 V
    I don't think inverter & NOT gate have any relation between them.
    NOT gate has two states i.e 1(+Vcc) & 0(Gnd) while Inverter converts DC into AC.

    #-Link-Snipped-# I do think CE with dc i/p acts as NOT gate.
    I will confirm & inform you ASAP.
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  • Jeffrey Arulraj

    Member • Mar 2, 2013

    Yo #-Link-Snipped-# do tell why the hell I am not able to upload photo now

    I simulated this and some error is tossed if i try to upload the file

    The inverter is not which you get I simulated it and tested it in simulator
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  • prashanth463

    Member • Mar 3, 2013

    #-Link-Snipped-# waiting for your reply
    i'm a bit lazy to run over my last years text book, but according to what i have analyzed it looks like a NOT gate to me
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  • Jeffrey Arulraj

    Member • Mar 3, 2013

    It is not a NOT gate friend So you can relax for 5 V ip you are getting 4.3 V as op as similar to a diode in Forward bias the diode drop is only lost across the transistor
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  • prashanth463

    Member • Mar 3, 2013

    #-Link-Snipped-# thanks for the reply
    but i have a doubt, if you try to analyze the circuit, when the base is provided with a +dc input the transistor is forward biased and all the inputs will be grounded from emitter terminal hence the output of the ce amplifier will be '0' and when no dc is given at the base, none of the terminals will be grounded and the output will appear as +Vcc
    this is what i have analyzed, can you please correct me if i was wrong...
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  • soni.sapan

    Member • Mar 7, 2013

    I think output will be zero since all dc is blocked by capacitor at input stage & vcc is just for biasing BJT & to bring the Q point in active or cutt-off region(dont remember in which region BJT work as amplifier) & get the desire gain
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  • Jeffrey Arulraj

    Member • Mar 7, 2013

    soni.sapan
    I think output will be zero since all dc is blocked by capacitor at input stage & vcc is just for biasing BJT & to bring the Q point in active or cutt-off region(dont remember in which region BJT work as amplifier) & get the desire gain
    Check the whole thread buddy before posting

    I think the poster mentioned there is not going to be any capacitors to block DC
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  • Abhishek Rawal

    Member • Mar 7, 2013

    #-Link-Snipped-# Finally,got the answer.
    If dc i/p is applied by removing capacitors from all 3 ends in CE Amplifier,then Amplification will be obtained as per gain of the transistor.

    Eg : If i/p is 3V & Gain is 3 then 9V as o/p will be obtained.

    But if we don't ignore capacitors then all current will grounded & hence it will act like NOT gate.
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  • soni.sapan

    Member • Mar 7, 2013

    #-Link-Snipped-# did you check it on Multisim or any other software?
    I just tried it on multisim. When I used AC as an input, I got gain of 2.7. but when I used dc as input I got -ve pV & -ve microV in both case without capacitor & with capacitor😔
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  • soni.sapan

    Member • Mar 7, 2013

    & #-Link-Snipped-# I didnt get both of you😎
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  • prashanth463

    Member • Mar 7, 2013

    Troll_So_Hard
    #-Link-Snipped-# Finally,got the answer.
    If dc i/p is applied by removing capacitors from all 3 ends in CE Amplifier,then Amplification will be obtained as per gain of the transistor.

    Eg : If i/p is 3V & Gain is 3 then 9V as o/p will be obtained.

    But if we don't ignore capacitors then all current will grounded & hence it will act like NOT gate.
    thanks for the reply #-Link-Snipped-#
    #-Link-Snipped-# what aren't you getting the result as troll said? okay i will also try it on multisim and see what is the output
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