0
\$\begingroup\$

enter image description here

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

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

\$\endgroup\$
1
  • \$\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

2
\$\begingroup\$

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.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.