Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Understanding Flip Flops by gavhays 11464 views
- Encoder and decoder by Then Murugeshwari 45111 views
- Treasures of e books by Mazhar Laliwala 2352 views
- Ict in t&l by Mazhar Laliwala 1302 views
- The 21st Century Learner by Kim Cofino 68729 views
- Middle School BE Courses: When They... by DCPS 6937 views

9,491 views

9,207 views

9,207 views

Published on

No Downloads

Total views

9,491

On SlideShare

0

From Embeds

0

Number of Embeds

895

Shares

0

Downloads

62

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Basic Characteristic of The Transistor<br />M. K. Laliwala<br />Assistant Professor<br />
- 2. Basic Characteristic of The TransistorObjectives:<br /><ul><li>Forward Current Transfer Ratio, CB
- 3. AC Current Gain
- 4. Relation Between α and β
- 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. 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. 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. 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. 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 />

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment