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# Basic characteristic of the transistor

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### Basic characteristic of the transistor

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