In a transistor if #I_c/I_e=alpha# and #I_c/I_b=beta#, if #alpha# varies between #(20)/(21)# and #(100)/(101)#, then the value of #beta# lied between?

Question

4376 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T08:14:03

#20 and 100#

Representative circuit in Common emitter amplifier configuration is given below
dnatechindia.com

Given is the ratio of #I_c/I_b=beta#
And the ratio of #I_c/I_e= alpha#

From above expressions we have
#I_c=beta I_b=alphaI_e# .....(1)

In this configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor. As such we have

#I_e = I_c + I_b# .....(2)

To have a combined the expression for #α and β# we divide both sides of (2) by #I_b# and use (1) to get

#I_e/I_b = (I_c + I_b)/I_b#
#=>beta/alpha=beta+1#
#=>beta=alphabeta+alpha#

#=>beta=alpha/(1-alpha)#

For #alpha=20/21#

#beta=(20/21)/(1-20/21)=20#

Again for #alpha=100/101#
#beta=(100/101)/(1-100/101)=100#
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Note: value of #alpha# will always be#<1#.
and typically, #beta# values lie between #20 and 200# for most general purpose transistors.