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 <V B, so assumption is wrong. BJT is operating in saturation mode. So, V CEsat = 0.2 =>V C – V E = 0.2 V C = 5.5V I C = (10-5.5)/4.7 = 0.96mA I B = I E – I C = 0.64mA Now, β forced = I C / I B = 1.5 Β forced < β. So, BJT is in saturated mode. V E = 5.3V I E = 1.6mA I C = 0.96mA I B = 0.64mA V C = 5.5V

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 <V B, so assumption is right. V E = 0.7V V B = 0V V C = -5.4V I E = 4.65mA I C = 4.6mA I B = 0.05mA

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..