# Why is the BJT in this circuit in active mode?

I don't know why the BJT is in Active mode ant not in saturation mode.

My calculations are:

$$V_{TH} = 10 * \frac{10}{10+12} = 4.54V$$

$$R_{TH} = \frac{12*10}{12+10} = 5.45k\Omega$$

$$V_{B} = V_{TH} - 0.7 = 3.84V$$

$$i_{B} = \frac{V_{B}}{R_{TH}} = 0.7 mA$$

$$i_{c} < \beta i_{b}$$

$$2.48mA < 70mA$$

I don't know where I went wrong in my calculations

Edit: Forgot to add iC calculation

$$i_{C} = \frac{10 - 0.3}{3.9} = 2.48 mA$$

• I agree with your $V_{TH}$ and $R_{TH}$ for the base. But I get $V_B\approx 4.502\:\textrm{V}$, $V_E\approx 3.802\:\textrm{V}$, and $V_C\approx 6.877\:\textrm{V}$. $I_B\approx 8.009\:\mu\textrm{A}$. Your Thevenin base voltage is in series with the Thevenin base resistance before reaching the base.
– jonk
Apr 14, 2017 at 21:39

The KVL loop around the base resistor is $V_{th}-I_bR_{th}-V_{be}-I_ER_E=0$. You can solve for the base current from there. Assume active region, for example, and use the equation $I_b=\dfrac{I_E}{\beta+1}$ somehow, to find out if the transistor is indeed in the active region.