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(C) Given:Barrier voltage of the diode,$V_{D} = 0.5\,V$Maximum power rating of the diode,$P_{\max} = 10\,mW = 10 	imes 10^{-3}\,W$Supply voltage,$V =

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$100\,\Omega$

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(C) Given:Barrier voltage of the diode,$V_{D} = 0.5\,V$Maximum power rating of the diode,$P_{\max} = 10\,mW = 10 	imes 10^{-3}\,W$Supply voltage,$V = 2.5\,V$The maximum current that can flow through the diode is given by:$P_{\max} = I_{\max} 	imes V_{D}$$10 	imes 10^{-3} = I_{\max} 	imes 0.5$$I_{\max} = \frac{10 	imes 10^{-3}}{0.5} = 20 	imes 10^{-3}\,A = 20\,mA$Applying Kirchhoff's voltage law in the circuit:$V = I_{\max} 	imes R_{\min} + V_{D}$$2.5 = (20 	imes 10^{-3}) 	imes R_{\min} + 0.5$$2.0 = (20 	imes 10^{-3}) 	imes R_{\min}$$R_{\min} = \frac{2.0}{20 	imes 10^{-3}} = \frac{2.0}{0.02} = 100\,\Omega$Thus,the minimum value of resistance $R$ is $100\,\Omega$.

The diode used in the circuit has a barrier voltage of $0.5\,V$ and a maximum power rating of $10\,mW$. What should be the minimum value of resistance $R$ connected in series with the diode?

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