Objective: Determine the diode voltage and current in the circuit shown in Fig-ure 1.28, using a piecewise linear model. Also determine the power dissipated in the diode. Assume piecewise linear diode parameters of Vγ = 0.6 V and rf = 10 Ω

Question

Objective: Determine the diode voltage and current in the circuit shown in Fig-ure 1.28, using a piecewise linear model. Also determine the power dissipated in the diode. Assume piecewise linear diode parameters of [latex]V_{γ} = 0.6 V[/latex] and [latex]r_{f} = 10 Ω[/latex]

Accepted Answer

With the given input voltage polarity, the diode is forward biased or “turned on,” so [latex]I_{D} > 0 [/latex]. The equivalent circuit is shown in Figure 1.31(a). The diode current is determined by

[latex]I_{D} = \frac{V_{PS} - V_{\gamma}}{R + r_{f} } = \frac{5 - 0.6}{2 imes 10^{3} + 10 } ⇒ 2.19 mA[/latex]

and the diode voltage is

[latex]V_{D} = V_{\gamma} + I_{D} r_{f} = 0.6 + (2.19 imes 10^{-3})(10) = 0.622 V[/latex]

The power dissipated in the diode is given by

[latex]P_{D} = I_{D} V_{D}[/latex]

We then find

[latex]P_{D} = (2.19) (0.622) = 1.36 m W[/latex]

Comment: This solution, obtained using the piecewise linear model, is nearly equal to the solution obtained in Example 1.8, in which the ideal diode equation was used. However, the analysis using the piecewise-linear model in this example is by far easier than using the actual diode I–V characteristics as was done in Example 1.8. In general, we are willing to accept some slight analysis inaccuracy for ease of analysis.

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