Find VL lL lz and lR for the circuit shown below if RL is 20

Find V_L, l_L, l_z and l_R for the circuit shown below if R_L is 200 Ohm. Repeat for R_L 700Ohm. What is the value of R_L that will dissipate the maximum rated current for the Zener diode? Determine the minimum value of R_L which will ensure that the Zener diode is conducting and regulating. Show your work.

Solution

Ans)

Given Vz=10 V ,Pzmax=400 mW gives Izmax=Pzmax/Vz=400/10=40 mA

Izmax=40 mA

Part A)

We assume Zener is in constant voltage region initially

RL=200 ohms

VL=Vz=10V

IR=(20-VL)/Rs=(20-10)/220=45.45 mA

IR=45.45 mA

IL=VL/RL=10/200=50 mA

Iz=IR-IL=45.45-50=-4.55 mA (negative so our assumption is wrong zener is not working in constant voltage region)

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Part A

So it is open circuit then

VL=20*RL/(Rs+RL)=20*200/(200+220)=9.524 V

VL=9.524 V

IL=VL/RL=9.524/200=47.62 mA

IR=IL=47.62 mA

Iz=0 A (open circuit)

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Part B

RL=700 ohms

Assume VL=Vz=10 V initial assumption

IL=VL/RL=10/700=14.286 mA

IR=(20-Vz)/Rs=(20-10)/220=45.45 mA

Iz=IR-IL=45.45-14.286=31.16 mA (positive and our assumption was right)

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C) For maximum power dissipation Pzmax=400 mW

gives Izmax=40 mA

Vz=VL=10 V

IR=(20-Vz)/Rs=(20-10)/220=45.45 mA

IL=IR-Izmax=45.45-40=5.45 mA

RL=VL/IL=10/5.45m=1834.86 ohms

so RL for maximum power in Zener is RL=1834.86 ohms

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D) As minimu current required for Zener is not given assume to be Izmin=0 A

that gives IR=IL=45.45 mA as calculated above with VL=10 V

RL=VL/IL=10/45.45m=220 ohms

so minimum value of RL required is RL=220 ohms