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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

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

  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Apr 14, 2017 at 21:39

1 Answer 1


The emitter isn't tied straight to ground. That's why your equation for the base current is incorrect.

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.


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