The graph of Ic-VCB for different values of IE is horizontal lines. We also know, Ic=IB + IE, but in this case IB is very small therefor, , Ic= IE. D.C. Current Gain α or αdc or hfE α = αdC =hfE =Ic/IE Where Ic is collector current due dc current IE
AC Current Gain Defined as ratio ΔIc/ ΔIE and denoted by αac or hfb Ability of the device to transfer current from low resistance input co the high resistance output i.etransfer resistor=resistance From the above graph, for VCB =10V, we have IE =4mA, Ic=3.92mA and when IE=6mA, Ic=5.88mA Thus αac= hfb=ΔIc/ΔIE =(5.88-3.92)/(6-4) =0.98 <1 Thus in CB ampifier current gain Ai is less than one, however, current in the output flows through large load resistance, it gives high voltage gain Av and therefore power gain Ap.
Relation Between α and β Let us consider IE=IB+Ic------------------------------------(1) α=Ic/IE ____________(2) β=Ic/IB____________(3) Substituting IB=IE-Ic from (1) to (3) we get, β=Ic/( IE-Ic) Dividing numerator and denominator by IE β=Ic/IE/( IE-Ic)/IE, Substituting α from (2) we get, β= α/1-α Similarly, we get for α=β/1-β
Collector Cutoff Current,ICBO The reverse current that is flowing through the base-collector junction, when the emitter is open is called collector cutoff current ICBO Collector Cutoff Current,ICBO In CB, we have Ic=IB+IE but IB is very small, therefore Ic=IE and α=Ic/IE, therefore Ic=αIE Including collector cutoff current ICBO we get, Ic= αIE+ ICBO but, ICBO=ICO TherforeIc= αIE+ ICO___________________(4) Equation (4) gives collector cutoff equation for CB amplifier
Similarly collector cutoff equation for CE amplifier is Ic=βIB+ICEO Here,ICEO=(β+1)ICO Therefore, Ic=βIB+(β+1)ICO_________(5) Comparing (4) and (5), we can say in CE amplifier leakage current is β times larger than CB. Since ICO is temperature dependent, when temperature changes large change in collector current occurs.