Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Given Information:
Voltage across zener = Vz = 5.6 V
Knee current = Izk = 1mA
Supply voltage = V = 15 V
Load current = IL = 15mA
Required Information:
Resistor = R = ?
Power dissipation of zener = P = ?
Answer:
R = 470 Ω
P = 0.112 Watts
Explanation:
R = supply voltage - zener voltage/Total current
Where Total current = zener current + Load current
zener current = 5*knee current = 5*1 = 5 mA
Total current = 5 mA + 15 mA = 20 mA
R = (15 - 5.6)/0.02
R = 470 Ω
There maximum power will be dissipated in zener when load current is zero so all the current will flow through the zener
Iz = (15 - 5.6)/470
Iz = 0.02 A
P = IzVz
P = 0.02*5.6
P = 0.112 Watts
Therefore, the maximum power dissipation of zener is 0.112 Watts.
