Download presentation

Presentation is loading. Please wait.

Published byChloe Kane Modified over 4 years ago

1
CSE 177/ EEE 1771 Lecture-12 Bipolar Junction Transistors at DC

2
CSE 177/ EEE 1772 Ex-5.4:The transistor has β =100. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V BE = 0.7V =>V B – V E = 0.7 4 – V E = 0.7 V E = 3.3V So, I E = V E / R E = 3.3/3.3 = 1mA I C = αI E = 0.99×1=0.99mA {as, α= β/(β+1) = 0.99} V C = 10 – I C R C = 5.3V I B = I E –I C = 0.01mA V C >V B, so assumption is right. V E = 3.3V I E = 1mA I C = 0.99mA I B = 0.01mA V C = 5.3V

3
CSE 177/ EEE 1773 Ex-5.5:The transistor has β =50. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V BE = 0.7V =>V B – V E = 0.7 6 – V E = 0.7 V E = 5.3V So, I E = V E / R E = 5.3/3.3 = 1.6mA I C = αI E = 0.99×1.6=1.6mA {as, α= β/(β+1) = 0.98} V C = 10 – I C R C = 2.48V V C

4
CSE 177/ EEE 1774 Ex-5.7:The transistor has β =100. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V EB = 0.7V =>V B = 0 => V E = 0.7V So, I E = (10-V E ) / R E = 4.65mA I C = αI E = 0.99×4.65=4.6mA {as, α= β/(β+1) = 0.99} I B = I E – I C = 0.05mA V C = -10 + I C R C = -5.4V V C

5
CSE 177/ EEE 1775 Ex-5.8:The transistor has β =100. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V BE = 0.7V; here V E = 0V =>V B = 0.7 So, I B = (5- V B ) / R B = 0.043mA I C = βI B = 100×0.043=4.3mA V C = 10 – I C R C = 1.4V I E = I B +I C = 4.343mA V C >V B, so assumption is right. V E = 0V V B = 0.7V V C = 1.4V I E = 4.343mA I B = 0.043mA I C = 4.3mA

6
CSE 177/ EEE 1776 Ex-5.9:The transistor has β =100. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V EB = 0.7V =>V B 0 V E = 0.7 So, I E = (5-V E) / R E = 4.3mA I C = αI E = 0.99×4.3=4.257mA as, α= β/(β+1) = 0.99} V C = I C R C +(-5)= 37.57V V C >V B, so assumption is wrong. BJT is operating in saturation mode. V E = V B + V EB = V B + 0.7 V C = V E – V ECsat = V B + 0.5 Now, I E = (5-V E )/R E = 4.3-V B I B = V B / 10 = 0.1V B I C = (V C + 5)/10 = 0.1V B + 0.55 Now, I E = I B + I C => 4.3-V B = 0.1V B + 0.1V B + 0.55 Solving this, V B = 3.13V, V E = 3.83V, V C = 3.63V I E =1.17mA, I B =0.31mA and I C =0.86mA Now, β forced = I C / I B = 2.8 Β forced < β. So, BJT is in saturated mode. V B =3.13V, V C =3.63V V E =3.83V, I E =1.17mA I B =0.31mA, I C =0.86mA

7
CSE 177/ EEE 1777 Ex-5.10:The transistor has β =100. Find the node voltages and branch currents. Solution: Let, BJT is operating in active mode. V BB = (15×R B2 )/(R B1 +R B2 )= 5V R BB = R B1 II R B2 = 33.3kΩ now, applying KVL at loop, V BB = I B R BB +V BE +I E R E Substituting I B = I E /(β+1) and rearranging the equation gives I E = (V BB -V BE )/[R E +{R BB /(β+1)}] = 1.29mA I B = 0.0128mA I C = αI E = 1.28mA V B = V BB – I B R BB = 4.57V V C = 15-I C R C = 8.6V V E = I E R E = 3.87V Here, V C >V B. So, assumption is right..

Similar presentations

Presentation is loading. Please wait....

OK

Types of Number.

Types of Number.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on unstructured data analysis Ppt on ssa in india Ppt on medical tourism in dubai Ppt on content development examples Ppt on time division switching Ppt on powerline communications Ppt on regular expression generator Ppt on carbon and its compounds class Ppt on save heritage of india Ppt on information security management system