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Here's a question from my homework:

Find \$ I_C \$ and \$ V_{CE} \$ for the circuits below, assuming \$ |V_{BE}| = 0.7V \$ and \$ \beta = 100 \$.

The BJT circuit. There are two voltage sources connected to the base: +15 V with a 470k resistor, and -15 V with a 1M resistor. Another voltage source of +15 V with a 6.8k resistor is connected to the collector.

I understand circuit (a), but for (b), I don't understand how to incorporate the \$ -15V \$ branch into calculation.

Below is my attempt:

$$ R_{TH} = \frac {R_1 R_2} {R_1 + R_2} = 3.197 \times 10^5 \Omega $$

$$ V_{TH} = \frac {V_1 R_2 + V_2 R_1} {R_1 + R_2} = 5.408 V $$

$$ \therefore I_B = \frac{V_{TH} - V_{BE}} {R_{TH}} = 1.47 \times 10^{-5} A $$

$$ I_C = \beta I_B = (100) (4.69 \times 10^{-5}) = 1.47 mA $$

Edit: Thank you all so much for the helppppppppp~

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  • 2
    \$\begingroup\$ Do you know the Thevenin's theory? electronics.stackexchange.com/questions/471906/… \$\endgroup\$ – G36 Mar 11 at 7:40
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    \$\begingroup\$ According to Kirchhoffs current law: Ib1=Ib+Ib2. \$\endgroup\$ – LvW Mar 11 at 8:38
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    \$\begingroup\$ Disconnect the base. What is the voltage at the 2-resistor node? What is the Thevenin-equivalent resistance? Re-connect the base. \$\endgroup\$ – AnalogKid Mar 11 at 12:29
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    \$\begingroup\$ -15V means that the voltage at the lower side of 1M resistor is 15V lower than the reference voltage. electronics.stackexchange.com/questions/392010/… \$\endgroup\$ – G36 Mar 11 at 13:06
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    \$\begingroup\$ I see the mistake the base current is equal to \$I_B = \frac{V_{TH} - V_{BE}}{R_{TH}}\$ \$\endgroup\$ – G36 Mar 11 at 13:08
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https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

What is the Thevenin-equivalent voltage at the 2-resistor node? Hint: Disconnect the base.

What is the Thevenin-equivalent resistance?

Re-connect the base.

Solve.

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