# The 6.8V zener diode is specified to have Vz=6.8V at Iz=5mA, rz=20 ohm and Izk=0.2 mA. The supply voltage V¬+ is nominally 10V but can vary by + 1V. Find Vo resulting from connecting a load resistance RL that draws a current IL=1 mA. What is the minimum value of RL for which the diode still operates in the breakdown region? First, let’s sort out the basic questions.

First-: Vo resulting from connecting a load resistance RL that draws a current IL=1 mA.

To find Vo , we know that at no load, it must be equal to the Vz with the no load zener current.

But as said before Vz=Vzo+Izrz (this is called zener diode model also, so for this we use given values of Iz)

We have Vz=6.8V, Iz=5mA and rz=20 ohm. So we need to find Vzo

Solving, 6.8=Vzo+5*10^-3 *20

Vzo=6.7V.

At no load, zener current changes as-: (remove RL from second figure and apply KVL)

Iz=Is=(Vs-Vz)/(R+rz)

= (10-6.7)/(0.5+0.02)

= 6.35 mA

Hence-:

Vo=Vz=Vzo+Izrz

=6.7+6.35*10^-3*20

=6.82 V

Second question-:Find Vo(change) resulting from connecting a load resistance RL that draws a current IL=1 mA.

ΔVo=ΔIz*rz = ΔIz*20

We need ΔIz.

As soon as we connect RL, current will flow through load=1 mA so zener current will decrease by 1mA i.e

ΔIz=-1 mA

Hence

ΔVo=-20 mV

Third question-:. What is the minimum value of RL for which the diode still operates in the breakdown region?

Breakdown region means minimum current and minimum zener voltage region as well as rz=0.

We need to find RL=VLmin/IL

=Vzmin/IL

=Vzo/ILmin

Vzo=6.7V. but IL=Isbreakdown-Izmin

Isbreakdown=(Vs-Vzo)/(R+rz)

=(9-6.7)/(0.5*10^3)

=4.6 mA

So, load current=IL=4.6-0.2 = 4.4 mA

RL=6.7/4.4 = 1.522 k ohm.