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

Replies

• 

  Abhishek Rawal

  Take a look at this CE Amplifier image :
  
  
  
  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 :
  
  
  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 )
  😀
• 

  prashanth463

  #-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
  😀
• 

  prashanth463

  #-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
• 

  Jeffrey Arulraj

  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
• 

  prashanth463

  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?
• 

  Jeffrey Arulraj

  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
• 

  Abhishek Rawal

  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.
• 

  Jeffrey Arulraj

  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
• 

  prashanth463

  #-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
• 

  Jeffrey Arulraj

  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
• 

  prashanth463

  #-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...
• 

  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
• 

  Jeffrey Arulraj

  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
• 

  Abhishek Rawal

  #-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.
• 

  soni.sapan

  #-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😔
• 

  soni.sapan

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

  prashanth463

  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