Question

Please help with Transistor circuit question!

Answer 1

By definition, we have:
\[\Large \frac{{{I_C}}}{{{I_B}}} = {h_{FE}}\]
furthermore, if we apply the second Kirchhoff law, we can write these subsequent equations:
\[\Large \left\{ \begin{gathered}
{V_{BE}} + {R_E}{I_E} - 10 = 0 \hfill \\
{R_C}{I_C} + {V_{CE}} + {R_E}{I_E} - 20 = 0 \hfill \\
\end{gathered} \right.\]
Please solve that system

Answer 2

Or another approach is using Ohms Law. For example Ic is given as .91 ma. You can now compute the voltage drop across Rc, .91 ma times 5 KOhm = 4.55 volts. Your now know Vce, 10 - 4.55 = 5.45. You are given Vbe of .7 volts, so that also tells you the Ve is also .7 volts with respect to ground or base of the transistor. Knowing that, you also know the voltage drop across Re is 10.7 volts and can now calculate Ie, 10,7/10 KOhm = 1.07 ma. Now since you know Ic and Ie you can easily compute Ib.

Now this is based on assuming that the voltages provided in the drawing is referenced to ground (or the base)

Looking at this drawing the transistor appears to be connected in the Common Base Configuration.

I would suggest solving the problem using the info provided by the previous post and then check for a fairly close agreement in the solved values.

Answer 3

Thank you!