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(D) The circuit is a constant current source where the emitter current $I_E \approx I_C = I = 1\ mA = 10^{-3}\ A$. The voltage across the emitter resi

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$2$

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(D) The circuit is a constant current source where the emitter current $I_E \approx I_C = I = 1\ mA = 10^{-3}\ A$. The voltage across the emitter resistor $R_E = 2\ k\Omega$ is $V_E = I_E 	imes R_E = 10^{-3}\ A 	imes 2 	imes 10^3\ \Omega = 2\ V$. The base-emitter voltage is $V_{BE} = 0.5\ V$. Therefore,the base voltage is $V_B = V_E + V_{BE} = 2\ V + 0.5\ V = 2.5\ V$. The total supply voltage is $V_{CC} = 7.5\ V$. The voltage across $R_1$ is $V_{R_1} = V_B = 2.5\ V$. The voltage across $R_2$ is $V_{R_2} = V_{CC} - V_B = 7.5\ V - 2.5\ V = 5\ V$. The ratio of the voltage across $R_2$ to $R_1$ is $\frac{V_{R_2}}{V_{R_1}} = \frac{5\ V}{2.5\ V} = 2$.

In the transistor circuit shown,assume that the voltage drop between the base and the emitter is $0.5\ V$. What will be the ratio of the voltage across resistances $R_2$ and $R_1$ in order to make this circuit function as a source of constant current,$I = 1\ mA$?

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