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To determine the value of the resistor \$\text{R}_\text{v}\$, we can use the formula that is given in the wiki page:

enter image description here

So, I did a little bit of maniplulation:

For the resistor \$\text{R}_{\text{v}}\$, we have:

$$\color{red}{\text{R}_{\text{v}}=\frac{\text{U}_{\text{R}_\text{v}}}{\text{I}_{\text{D}_\text{z}}+\frac{\text{I}_{\text{R}_\text{L}}}{1+\text{h}_\text{FE}}}\tag1}$$

Now, for \$\text{U}_{\text{R}_{\text{v}}}\$, we have:

$$\text{U}_{\text{in}}=\text{U}_{\text{R}_\text{v}}+\text{U}_{\text{D}_\text{z}}\space\Longleftrightarrow\space\text{U}_{\text{R}_\text{v}}=\text{U}_{\text{in}}-\text{U}_{\text{D}_\text{z}}\tag2$$

And for \$\text{I}_{\text{R}_\text{L}}\$, we have:

$$\text{U}_{\text{R}_\text{L}}=\text{U}_\text{out}=\text{U}_{\text{D}_\text{z}}-\text{U}_\text{BE}=\text{I}_{\text{R}_\text{L}}\cdot\text{R}_\text{L}\space\Longleftrightarrow\space\text{I}_{\text{R}_\text{L}}=\frac{\text{U}_{\text{R}_\text{L}}}{\text{R}_\text{L}}=\frac{\text{U}_\text{out}}{\text{R}_\text{L}}=\frac{\text{U}_{\text{D}_\text{z}}-\text{U}_\text{BE}}{\text{R}_\text{L}}\tag3$$

So, for \$(1)\$ we get (using \$(2)\$ and \$(3)\$):

$$\color{red}{\text{R}_{\text{v}}=\frac{\text{U}_{\text{in}}-\text{U}_{\text{D}_\text{z}}}{\text{I}_{\text{D}_\text{z}}+\frac{\text{U}_{\text{D}_\text{z}}-\text{U}_\text{BE}}{\text{R}_\text{L}}\cdot\frac{1}{1+\text{h}_\text{FE}}}\tag4}$$

Now, for \$\text{h}_\text{FE}\$ of the transistor we have:

$$ \begin{cases} \text{I}_\text{C}=\text{h}_\text{FE}\cdot\text{I}_\text{B}\\ \\ \text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\\ \\ \text{I}_\text{B}=\text{I}_\text{BS}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_\text{BE}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right) \end{cases}\space\space\therefore\space\space\space\text{h}_\text{FE}=\frac{\text{I}_\text{E}}{\text{I}_\text{BS}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_\text{BE}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right)}-1\tag5 $$

So, we get for \$(1)\$ and \$(4)\$ (using \$(5)\$):

$$\color{red}{\text{R}_{\text{v}}=\frac{\text{U}_{\text{in}}-\text{U}_{\text{D}_\text{z}}}{\text{I}_{\text{D}_\text{z}}+\frac{\text{U}_{\text{D}_\text{z}}-\text{U}_\text{BE}}{\text{R}_\text{L}}\cdot\frac{\text{I}_\text{BS}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_\text{BE}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right)}{\text{I}_\text{E}}}\tag6}$$

Now, for a diode the relation between voltage and current is given by:

$$\text{I}_{\text{D}_\text{z}}=\text{I}_\text{S}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_{\text{D}_\text{z}}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right)\tag7$$

So for \$(1)\$, \$(4)\$ and \$(6)\$ we get (using \$(7)\$):

$$\color{red}{\text{R}_{\text{v}}=\frac{\text{U}_{\text{in}}-\text{U}_{\text{D}_\text{z}}}{\text{I}_\text{S}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_{\text{D}_\text{z}}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right)+\frac{\text{U}_{\text{D}_\text{z}}-\text{U}_\text{BE}}{\text{R}_\text{L}}\cdot\frac{\text{I}_\text{BS}\cdot\left(\exp\left\{\frac{\epsilon\cdot\text{U}_\text{BE}}{\eta\cdot\text{k}\cdot\text{T}}\right\}-1\right)}{\text{I}_\text{E}}}\tag8}$$

Question: How can I find \$\text{I}_\text{E}\$? Can I make the approximation: \$\text{I}_\text{E}\approx\text{I}_\text{C}\$?

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  • \$\begingroup\$ If this is a real world design you are way over thinking it. You would only go into this sort of detail for a homework assignment and frankly I have not checked all of your calculations because given this design problem an engineer would not think like this. \$\endgroup\$ – RoyC Dec 25 '16 at 11:22
  • \$\begingroup\$ @RoyC It is a homeowork disign project \$\endgroup\$ – Kiopr Dec 25 '16 at 12:01
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    \$\begingroup\$ OMG. Are you serious ? This is not the way we design this type of a circuit. As for your question. In this circuit Ie = ILoad \$\endgroup\$ – G36 Dec 25 '16 at 13:30
  • \$\begingroup\$ If you want to stick to the passive sign convention, \$I_E \approx -I_C\$. More useful for design, \$I_E = -U_{out}/R_{load}\$. \$\endgroup\$ – The Photon Dec 25 '16 at 17:46
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Ie is the load current. The load current is Out/RL. Is that really your question?

As RoyC said it is not generally worth going to this level of calculation. You need to accommodate the tolerances of components, temperature variations and the input supply.

Some parameters such as hFE can vary over a wide range (3:1) so you have to be conservative and take worst case values to ensure it will work at end of the range. Then you have to make sure that other parameters such as heat dissipation in the components are acceptable at the other end.

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