[30] IV. Assume following data from the Zener diode data sheet: Zener knee curre

Question

[30] IV. Assume following data from the Zener diode data sheet: Zener knee current- output voltage 7.5V. For the circuit below answer the following questions: -0.25mA, Zener Max Current 100mA, and max de power dissipation - 1 W, and regulated a) b) c) What range of input voltage can be regulated using this information? Considering the input Avg. voltage found in (a), what will be min. value of R2? If for a change of S0mv Zener voltage the amount of changes in Zener current is 50 mA what is the Zener impedance for this Zener diode?

Explanation / Answer

Given data,

Zener knee current(Iz(min))=0.25mA,

Zener maximum current(Iz(max))=100mA,

Zener maximum Power dissipiation(Pz)=1W,

Regulated Voltage(Vz)=7.5V,    R1=1k(ohm)

Vs=V1+Vz=( Iz(min)*R1)+Vz=(0.25*10^-3*1*10^3)+7.5=7.75V Minimum Input Voltage

Vs=V1+Vz=( Iz(max)*R1)+Vz=(100*10^-3*1*10^3)+7.5 = 107,5V Maximum Input Voltage

Input voltage range is (7.75V to 107.5V).

2. Input Average voltage is

Vs=( Vs(min)+ Vs(min))/2=(7.75+107.5)/2

Assume no load condition & find the current(Is),

V1=Vs-Vz=57.625-7.5=50.125V

Hence, Is=V1/R1=50.125/(1*10^3)=50.125mA, Current

Zener current at this loadis, Iz=Iz(max)-Is=100-50.125=49.875mA

Now, R2=Vz/Iz=(7.5/49.875)*1000=150.375(ohm)

3. Zener Impedance is,

Zz=Vz/Iz, here Vz & Iz is the rate of change of Zener Voltage & Current

Zz=(50*1000)/(50*1000)=1(ohm)

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