First, let’s sort out the basic questions.

First-: V_{o }resulting from connecting a load resistance R_{L }that draws a current I_{L}=1 mA.

To find V_{o }, we know that at no load, it must be equal to the V_{z }with the no load zener current.

But as said before V_{z}=V_{zo}+I_{z}r_{z }(this is called zener diode model also, so for this we use given values of I_{z})

We have V_{z}=6.8V, I_{z}=5mA and r_{z}=20 ohm. So we need to find V_{zo}

Solving, 6.8=V_{zo}+5*10^-3 *20

V_{zo}=6.7V.

At no load, zener current changes as-: (remove R_{L }from second figure and apply KVL)

I_{z}=I_{s}=(V_{s}-V_{z})/(R+r_{z})

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

= 6.35 mA

Hence-:

V_{o}=V_{z}=V_{zo}+I_{z}r_{z}

=6.7+6.35*10^-3*20

=6.82 V

**Second question-:Find V _{o}(change) resulting from connecting a load resistance R_{L} that draws a current I_{L}=1 mA.**

ΔV_{o}=ΔI_{z}*r_{z} = ΔI_{z}*20

We need ΔI_{z}.

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

ΔI_{z}=-1 mA

Hence

ΔV_{o}=-20 mV

Third question-:**. What is the minimum value of R _{L} for which the diode still operates in the breakdown region?**

Breakdown region means minimum current and minimum zener voltage region as well as r_{z}=0.

We need to find R_{L}=V_{Lmin}/I_{L}

=V_{zmin}/I_{L}

=V_{zo/}I_{Lmin}

Vzo=6.7V. but I_{L}=I_{sbreakdown}-I_{zmin}

I_{sbreakdown}=(V_{s}-V_{zo})/(R+r_{z})

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

=4.6 mA

So, load current=I_{L}=4.6-0.2 = 4.4 mA

R_{L}=6.7/4.4 = 1.522 k ohm.

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