Calculating Resistor Value and Max Power Dissipation for 5.6V Zener Shunt Regulator

[Q] A zener diode exhibits a constant voltage of 5.6v for currents greater than five times the knee current. Izk = 1 mA . The zener is to be used in the design of shunt regulator fed from a 15 v supply the load current varies over the range of 0ma to 15ma. find the suitable value for the resistor 'R' . what is the max power dissipation of zener.

Hi Ansuman,

As I understand the statement,

@ resistance of the load with load current maximum which is 15mA

R(load) = Vz / (15mA) = (5.6V) / (15mA) = 373.33Ω

Solving for R(total)

with respect to I(zk) = 1mA

I(total) = I(zk) + I(load) = 1mA + 15mA = 16mA

R(total) = Vin / I(total) = 15V / 16mA = 937.5Ω

Solving for R

R(total) = R + R(load)

937.5Ω = R + 373.33Ω

R = 564.17Ω

assuming that the currents cannot flow to R(load)

therefore, 16mA will pass through the zener,

Pzmax = (5.6V)(16mA) = 89.6mW

It is not that clear, I think it is better if it has given figure with respect to the statement.

You can also check more about zener

link here-> http://www.allaboutcircuits.com/vol_3/chpt_3/11.html

Hope it helps.

God Bless.

Best,
Neil

thank you sir for your reply but the answers are different R= 470Ω and power dissipation = 112mW how to approach ???

Hi Ansuman

That's what it mean for this statement "a constant voltage of 5.6v for currents greater than five times the knee current" which it should be noted as Izmax or in figure, it is indicated accordingly.

then Izmax = 5(Izk) = 5 (1mA) = 5mA

same equation above with respect to Izmax

I(total) = Izmax + I(load) = 5 + 15mA = 20mA

R(total) = Vin / I(total) = 15V / 20mA = 750Ω

where,

R(total) = R + R(load)

Maximum Power Transfer Conditions

assuming that the current cannot flow to the zener
Vo = Vz = 5.6V

R(load) = Vo / I(total) = 5.6V / 20mA = 280Ω

R(total) = R + R(load)
750Ω = R + 280Ω
R = 470Ω

assuming that the current cannot flow to the R(load)

Pzmax = VzI(total) = (5.6V)(20mA) = 112mW

Hope it helps.

God Bless.

Best,
Neil

Thanks it works.