common emitter

Thread Starter

muni

Joined Jul 29, 2008
45
sir, i've completed my AMIE degree through INSTITUTION OF ENGINEERS. Now i'm preparing for competitive exams.i request your help and guidance on my way.

i request to clear my doubt with reasonable explanation.


a common emitter transistor amplifier has a collector current of 1.0 mA when its base current is 25 micro ampere at the room temperature. it's input resistance is approximately equal to------------?



the answer given inthe book is 1000 ohms
 

S_lannan

Joined Jun 20, 2007
246
give that Ic = 1ma and Ib = 25ua
beta is roughly ic/ib so that makes beta 40

input resistance is = 're * beta

're is 25mv/IE = 25

soooo input resistance = 1000

checks out to me...
 

Thread Starter

muni

Joined Jul 29, 2008
45
give that Ic = 1ma and Ib = 25ua
beta is roughly ic/ib so that makes beta 40

input resistance is = 're * beta

're is 25mv/IE = 25

soooo input resistance = 1000

checks out to me...
sir, thank you. the book has also given me same solution.but in the book after calculating beta value=40, it is given that

for CE, Rm = (Vt(millivolts)(b+1))/1.025 milli amps

here at this point i was unable to undersatnd.

thank you for your solution sir.

i,m preparing for INDIAN ENGINEERING SERVICES. I've so many such unanswered problems.

sir i'm very new to this society. i don't know how to attach a file along with this quiery . i request you to suggest the way how i can add my files.

my e mail address is psmuni@gmail.com
 

studiot

Joined Nov 9, 2007
4,998
Your book solution was just being more exact (pedantic) than S Lannan.

The equations are the same.
The 1.025 arises because Ie =Ic+Ib , Vt = 25 millivolts is an assumed standard value. the factor Beta+1 is more exact but most normally use just beta since beta >>1.

For most purposes 'rule of thumb methods' are good enough.
 
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