I want to find the minimum resistance \$R_L\$ so as to maintain \$V_z\$(voltage of the zener corresponding to the minimum current \$I_{z_0}\$) across the same resistor \$R_L\$.
simulate this circuit – Schematic created using CircuitLab
I have two ways of looking at this, one of them is wrong and I need your help to figure it out!
First method:
I take off the diode as the load and find thevenin voltage :
$$V_{th}=\frac{R_L}{R_l+R_s}V_s$$
Since
\$V_{th}=V_{R_L}\$
I'll equate it with \$V_z\$ which gives :
\$\frac{R_L}{R_L+R_S}V_S=V_z\$
thus
\$R_L=\frac{R_S.V_z}{V_S-V_z}\$
Second method:
If \$R_L\$ is minimal then \$I_L\$ passing through it would be maximal and \$I_z\$ minimal (\$= I_{z_0}\$) thus we'd have :
$$I_S = I_L+I_{z_0} \Leftrightarrow \frac{V_S-V_z}{R_S}=\frac{V_z}{R_L}+I_{z_0}\Leftrightarrow R_L=\frac{R_S.V_Z}{V_S-V_Z-R_S.I_{z_0}}$$
(I'm skeptical concerning the implication \$I_L\$ maximal \$\Rightarrow\$ \$I_z\$ minimal, or in other words \$I_z = I_{z_0}\$)
Best Answer
A typical 300mW Zener uses a test current Izt=5mA with an incremental resistance Zzt and a threshold current, at 1/10 Izt or It @ 0.5mA which results in a knee resistance of Zzk approx. 10x Zzt.
The same is true for LEDs.
Make a Thevenin equivalent circuit for 5mA at a low power.
This gives the low rated resistance, Zzt and is important to include in the equation to solve for R1.
simulate this circuit – Schematic created using CircuitLab
Now you can easily solve for R1