How do I approach this question?
My attempt:
$$I_{C} = 1mA$$
$$I_{B} = \frac {I_{C}} {\beta} $$
$$10 - I_{B}R_{B} - 0.7 = 0 $$
$$R_{B} = \frac{\beta (10 - 0.7)}{I_{C}}$$
$$R_{B} = 930 k\Omega$$
The answer should be \$R_{B} = 186.3 k\Omega\$
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1\$\begingroup\$ You are on the right track. What is the voltage across the collector resistor? Use that to determine the collector current. \$\endgroup\$– Dwayne ReidCommented Apr 15, 2017 at 2:41
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\$\begingroup\$ I got the correct answer now. Why is the calculated collector current used instead of the given \$1mA\$? \$\endgroup\$– user3067059Commented Apr 15, 2017 at 2:49
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1 Answer
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I think the following includes all of the elements you were asked to achieve, including make adjustments on \$V_{BE}\$ because of the higher collector current than the nominal value given. It's implied by the question, I suspect, that you need to adjust the \$700\:\textrm{mV}\$ figure.
$$\begin{align*} I_C&=\frac{10\:\textrm{V}-5\:\textrm{V}}{1\:\textrm{k}\Omega}\\\\ &= 5\:\textrm{mA}\\\\ I_B&= \frac{I_C}{\beta}\\\\ &= 50\:\mu\textrm{A}\\\\ V_{BE}&\approx 700\:\textrm{mV} + 26\:\textrm{mV}\cdot\operatorname{ln}\left(\frac{5\:\textrm{mA}}{1\:\textrm{mA}}\right)\\\\ &\approx 742\:\textrm{mV}\\\\ R_B&=\frac{10\:\textrm{V}-V_{BE}}{I_B}\\\\ &\approx 185.2\:\textrm{k}\Omega \end{align*}$$
However, this doesn't agree with the answer you said is right. But there it is, anyway. By the way, \$\frac{k T}{q}\approx 26\:\textrm{mV}\$ at room temperature. That's where that number came from, above.
