CS代考 ENGN 1218 Introduction to Electronics – cscodehelp代写

AUSTRALIAN NATIONAL UNIVERSITY Research School of Engineering
ENGN 1218 Introduction to Electronics
Week 10 Problems with Solutions: Zener Diode Circuits
Consider the circuit shown in Figure 1 with Vin = 14 V, VZ = 7.5 V, IZT = 20 mA, IZK = 0.25 mA, PZ(max) = 0.5 W,R1 =100ΩandRL =330Ω.
(a) Determine VL, IL, IZ, Iin.
IinR1IZ IL
(b) Repeat if Vin = 16 V.
Figure 1: The circuit for Question 1.
Consider the circuit shown in Figure 2 with VZ = 12 V, IZK = 0.25 mA, PZ(max) = 1 W, R1 = 0.1kΩ and RL = 0.5kΩ.
Figure 2: The circuit for Question 2.
Determine the minimum and maximum input voltages that can be regulated by the zener diode.
Repeat Q2 assuming IN4728, 1W Zener Diode (see attached data sheet).
Week 10 Problems
ANU ENGN 1218
Consider the circuit shown in Figure 3 with Vin = 30 V, VZ = 12 V, IZK = 0.25 mA, PZ(max) = 1 W and R1 = 0.2kΩ.
Figure 3: The circuit for Question 4.
Determine the minimum and maximum load resistances for which the zener diode will maintain regulation.
Repeat Q4 assuming R1 = 20Ω, Vin = 12 V and IN4733, 1W Zener Diode (see attached data sheet).
Week 10 Problems page 2
Complete Solution a
The given circuit is
AUSTRALIAN NATIONAL UNIVERSITY Research School of Engineering
ENGN 1218 Introduction to Electronics
Week 10 Problems with Solutions: Zener Diode Circuits
IinR1IZ IL
For Zener diode,
IZK = 0.25mA
IZM = PZ(max) =66.67mA VZ
Step 1: Consider the circuit external to the zener diode and find its Thevenin equivalent circuit The Thevenin voltage source VT H is equal to the open-circuit voltage of the original network.
Using node-voltage analysis technique, VTH−0+VTH−14 = 0
VTH = 10.744V
Week 10 Problems
ANU ENGN 1218
To find the Thevenin resistance RTH, we zero the sources in the original network (i.e. voltage sources are replaced by short circuit and current sources are replaced by open circuit) and then compute the resistance between the terminals.
RTH = R1RL =76.75Ω R1 + RL
Step 2: Determine IZ and VL using the Thevenin equivalent circuit
IZ = VTH−VZ =42.267mA RTH
We see that IZ > IZK and IZ < IZM so conditions of correct zener operation are satisfied. Hence, VL =7.5V Step 3: Determine Iin and IL using the original circuit Iin = Vin−VL =65mA IL = VL =22.727mA RL Check: Iin =IZ +IL. IZ = 62.3mA VZ = 7.5V Iin = =85mA IL = =22.7mA Week 10 Problems Complete Solution The given circuit is For Zener diode, IZK = 0.25mA IZM = PZ(max) =83.33mA VZ Step1: For Vin(min) WhenVin =Vin(min),Iin =Iin(min) andthisforcesIZ todecreasetoIZK. Hencewehave, IL = VZ=24mA Iin(min) = IL+IZK =24.25mA Vin(min) = VZ + Iin(min) R1 = 14.425 V Step2: For Vin(max) When Vin = Vin(max), Iin = Iin(max) and this forces IZ to increase to IZM. Hence we have, VZ=24mA RL IL+IZM=107.33mA VZ+Iin(max)R1=22.733V Hence input voltage variation from 14.425V − 22.733V can be regulated by the Zener diode. From data sheet, for IN4728 Zener diode VZ = IZK = IZM = Solution is Vin(min) Vin(max) PZ(max) =303.03mA VZ = 4.06V = 34.26 V Iin(max) = Vin(max) = Week 10 Problems Complete Solution The given circuit is For Zener diode, IZK = 0.25mA IZM = PZ(max) =83.33mA VZ Step1: For RL(min) When RL = RL(min), IL increases to IL(max) and this forces IZ to decrease to IZK. Hence we have, IZK = 0.25mA Iin = Vin−VZ =90mA R1 IL(max) = Iin−IZK=89.75mA RL(min) = VZ =133.7Ω IL(max) Step1: For RL(max) When RL = RL(max), IL decreases to IL(min) and this forces IZ to increase to IZM. Hence we have, IZM = 83.33mA Iin = Vin−VZ =90mA R1 IL(min) = Iin−IZM =6.67mA RL(max) = VZ =1.8kΩ Hence load resistance variation from 133.7Ω − 1800Ω can be regulated by the Zener diode. Q5 Solution From data sheet, for IN4733 Zener diode VZ = 5.1V IZK = 1mA We have R1 = 20Ω. Week 10 Problems IZM = PZ(max) = 196.08mA VZ ANU Solution is RL(min) RL(max) Week 10 Problems 14.82 Ω 34.24 Ω