Basic characteristic of the transistor
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The document summarizes some basic characteristics of transistors: 1. The forward current transfer ratio (CB) is represented by a horizontal line on an Ic-VCB graph, showing Ic=IE. The DC current gain (α) is defined as the ratio of collector current (Ic) to base current (IE). 2. The AC current gain (β) is the ratio of changes in collector and base currents. It is typically less than 1, but the transistor still provides voltage and power gain due to the high load resistance. 3. The relation between α and β is defined as β=α/(1-α) and α=β/(1-β).
Basic characteristic of the transistor
- 1. Basic Characteristic of The TransistorM. K. LaliwalaAssistant Professor
- 2. Basic Characteristic of The TransistorObjectives:Forward Current Transfer Ratio, CB
- 4. Relation Between α and β
- 5. Collector Cutoff Current,ICBO Forward Current Transfer Ratio, CB The graph of Ic-VCB for different values of IE is horizontallines. We also know, Ic=IB + IE, but in this case IB is very smalltherefor, , Ic= IE. D.C. Current Gain α or αdc or hfEα = αdC =hfE =Ic/IEWhere Ic is collector current due dc current IE
- 6. AC Current Gain Defined as ratio ΔIc/ ΔIE and denoted by αac or hfbAbility of the device to transfercurrent from low resistance inputco the high resistance output i.etransfer resistor=resistanceFrom the above graph, for VCB =10V, we have IE =4mA, Ic=3.92mAand when IE=6mA, Ic=5.88mAThus αac= hfb=ΔIc/ΔIE =(5.88-3.92)/(6-4) =0.98 <1Thus 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.
- 7. Relation Between α and βLet 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
- 8. Collector Cutoff Current,ICBOThe reverse current that is flowing through the base-collector junction, when the emitter is open is called collector cutoff current ICBOCollector Cutoff Current,ICBO In CB, we have Ic=IB+IE but IB is very small, therefore Ic=IE and α=Ic/IE, therefore Ic=αIEIncluding collector cutoff current ICBO we get, Ic= αIE+ ICBO but, ICBO=ICOTherforeIc= αIE+ ICO___________________(4) Equation (4) gives collector cutoff equation for CB amplifier
- 9. 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.