# Analysis of voltage regulator circuit

I have found out the output voltage to be 9V.

To find power, we need $$\I_CV_{CE}\$$

We can find $$\I_E\$$ from the output circuit.

$$\ I_E=\frac{6}{24k}+\frac{9}{10} \$$

But can we find $$\I_C\$$? Or do we have to approximate $$\I_C\$$ approximately equal to $$\I_E\$$?

Will increase of $$\ V_{in}\$$ by 20% anyway impact the transistor currents?

• The output current is not 9V, current is measured in amps... – Solar Mike Oct 28 '18 at 10:59
• Since beta is not given you can assume the base current is negligible. That's the safer assumption (from a dissipation point of view anyway). Increasing the input voltage will increase the dissipation. The effect on the transistor currents will be small. – Spehro Pefhany Oct 28 '18 at 11:03

For this theoretical circuit, the output voltage will be around: $$V_O = 6V \cdot (1 + \frac{12\textrm{k}\Omega}{24\textrm{k}\Omega}) = 9V$$
Hence the output current is: $$\I_O = \frac{9V}{10\Omega} = 0.9A\$$
The power dissipation is $$\P \approx (V_{IN} - V_O) \cdot I_O = 5.4W\$$
I assumed that $$\I_C \approx I_E\$$ because $$\\frac{I_C}{I_E} = \frac{I_C}{I_B+I_C} = \frac{ \beta I_B}{I_B +\beta I_B} = \frac{ \beta I_B}{(\beta +1) I_B} =\frac{ \beta}{\beta +1} \$$
So if the beta is large we can assume that $$\I_C \approx I_E\$$