Current amplifier 

Asked by Www.harshalhire51 | 20th Nov, 2019, 09:48: PM

Expert Answer:

Common-Emitter (CE) transistor amplifier is a current amplifier. CE-transistor amplifier circuit is shown in figure.
An ac input signal vi (to be amplified) is superimposed on the bias VBB (dc) as shown in Figure and the output is taken
between the collector and the ground.

The working of an amplifier can be easily understood, if we first assume that vi = 0.
 
Then applying Kirchhoff’s law to the output loop,we get,  Vcc = VCE + Ic RL   ..............................(1)
 
Likewise, the input loop gives  VBB = VBE + IB RB  ............................(2)
 
When vi is not zero, we get  VBE + vi = VBE + IB RB + ΔIB (RB + ri )
 
The change in VBE can be related to the input resistance ri of the transistor and the change in IB.
 
Hence  vi = ΔIB (RB + ri ) = r ΔIB
 
 
The change in IB causes a change in Ic.
 
We define a parameter βac  as,  βac = ( ΔIc / ΔIb ) = ic / ib
 
βac is also known as the ac current gain Ai.
  
The change in Ic due to a change in IB causes a change in VCE and the voltage drop across
the resistor RL because VCC is fixed.
 
These changes can be given by Eq. (1) as
 
ΔVCC = ΔVCE + RL ΔIC = 0   or ΔVCE = –RL ΔIC
 
The change in VCE is the output voltage vo.
 
vo = ΔVCE = –βac RL ΔIB
 
The voltage gain of the amplifier is, Av = vo / vi = ΔVCE / ( r ΔIB ) =  - βac ( RL / r )
 
The negative sign represents that output voltage is opposite with phase with the input voltage.

Answered by Thiyagarajan K | 21st Nov, 2019, 12:00: PM