# Voltage regulator transistor circuit analysis (question about the emitter current)

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

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: $\text{I}_\text{E}\approx\text{I}_\text{C}$?

• 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. – RoyC Dec 25 '16 at 11:22
• @RoyC It is a homeowork disign project – Kiopr Dec 25 '16 at 12:01
• 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 – G36 Dec 25 '16 at 13:30
• If you want to stick to the passive sign convention, $I_E \approx -I_C$. More useful for design, $I_E = -U_{out}/R_{load}$. – The Photon Dec 25 '16 at 17:46