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transistor question????

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1- when i use a voltage divider bias or one resistor???????????



transistor.gif



what is the purpose of c3 ???????
 

You need a voltage gain of 100 which might not be possible if you bias the base with a negative feedback resistor from its collector. It depends on the signal source resistance. If the source resistance is high then the gain is low.

A transistor is never biased with just a single resistor from the supply voltage.
 
C3 is a bypass(shunt) capacitor. it is used to create an ac ground at the emitter. it is like short circuit during high frequency operations according to the relation.
Xc=1/2ΠfC

where
Xc=capacitive reactance

see more about bypass capacitors here
**broken link removed**
 
thank alll.........

please how i can calculate (input&output impedance ) for amplifier ?
 

ANS 1-
The use of single resistor and potential divider bias depend on your application.
fixed biasing is done with single resistor but having stability very low(s=1+\beta ).
and higher the value of S ,lower the stability.

but stability of divider ckt is too high,bcos S has lower value.
s=(1+\beta )Re+Rth/(Rth+(1+\beta )Re)

ANS 2
as posted by other friends.........
 
thanks allllllll...........

please how i can calculate (input&output impedance ) for amplifier ????
 

Input impedance is (R1 in parallel with R2 in parallel with the transistor input impedance). The transistor input impedance is (re * hfe). re is the dynamic emitter resistor which is 0.025/Ie at room temperature (Ie = emitter current in amps).

Output impedance is R4.

Keith.
 
keith1200rs said:
Input impedance is (R1 in parallel with R2 in parallel with the transistor input impedance). The transistor input impedance is (re * hfe). re is the dynamic emitter resistor which is 0.025/Ie at room temperature (Ie = emitter current in amps).

Output impedance is R4.

Keith.
Click to expand...


then>>>>>>.

input imedance =

r1 and r2 in parallel = 4.6 k

and

re=.025 / (9MA "approximately")=2.7 ohm

transistor input impedance is (re * hfe) =( 2.7 * 100hfe)=270ohm


total is 4.6k+2700ohm=7.3k input impedance

Is this true or false

thank all
 
Last edited:

Close. The 4.6k is in parallel with the transistor input impedance so 4.6k//2.7k = 1.7k.

Keith.
 
keith1200rs said:
Close. The 4.6k is in parallel with the transistor input impedance so 4.6k//2.7k = 1.7k.
Keith.
Click to expand...

Why 2.7k ? 2.7*1000 (that is: hfe=1000) ? What about 270 ohms for 9 milliamps?
 
Woops, yes you are right. I copied the 2700 ohms from the previous post without spotting it was wrong.

Keith
 
THANK ALL.....

INPUT IMPEDANCE = 4.6K / 270 OHM = 17K

I CORRECT IT IN MY previous post FROM 2700 TO 270 OHM
 

No, it is 4.6k in parallel with 270 ohms so 255 ohms. "parallel" is often written as "//".

Keith.
 
The purpose of C3 is to act as a by-pass capacitor for your amplifier, so effectively at high frequencies, the capacitor shunts the resistor and you have a common emmitter BJT amplifier. Also, in addition to using C3 as a bypass capacitor, if you choose the value properly, u might be able to use it as a bandwidth extension for your amplifier by choosing C3 such that it produces a zero to cancel out a pole at the collector :).
 
**broken link removed**

how i can calculate value of r1 and r2 !!!!!!!!1
 

الكثير من الشرف
;-)
 
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