A diode operates at a constant potential drop | vedclass

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(C) The maximum current $I$ that the diode can handle is given by the power rating $P$ and the voltage drop $V_d$ across the diode:$I = \frac{P}{V_d}

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$5$

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(C) The maximum current $I$ that the diode can handle is given by the power rating $P$ and the voltage drop $V_d$ across the diode:$I = \frac{P}{V_d} = \frac{100 	imes 10^{-3} \, W}{0.5 \, V} = 0.2 \, A$To limit the current to this maximum value when connected to a source voltage $V_s = 1.5 \, V$,we use Ohm's law for the series resistor $R$:$V_s = V_d + I 	imes R$$1.5 \, V = 0.5 \, V + (0.2 \, A) 	imes R$$1.0 \, V = 0.2 \, A 	imes R$$R = \frac{1.0}{0.2} \, \Omega = 5 \, \Omega$

$A$ diode operates at a constant potential drop of $0.5\, V$ for all currents and has a maximum power rating of $100\, mW$. If the source voltage is $1.5\, V$,what resistance must be connected in series with the diode so that the current in the circuit is at its maximum safe value? (in $\Omega$)

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