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The diode offset model may be represented by an ideal diode in series with a 06V ideal battery, as shown in Figure 823 Use of the offset diode model is best described by means of examples
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Figure 823 Offset diode as an extension of ideal diode model
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EXAMPLE 83 Using the Offset Diode Model in a Half-Wave Recti er
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Compute and plot the recti ed load voltage, vR , in the circuit of Figure 824
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Known Quantities: vS (t) = 3 cos( t); V = 06 V Find: An analytical expression for the load voltage Assumptions: Use the offset diode model
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Semiconductors and Diodes
Analysis: We start by replacing the diode with the offset diode model, as shown in the
+ 3 cos t _ i
Actual circuit + vD _ + V _
+ _
lower half of Figure 824 Now we can use the method developed earlier for ideal diode analysis, that is, we can focus on determining whether the voltage vD across the ideal diode is positive (diode on) or negative (diode off) Assume rst that the diode is off The resulting circuit is shown in Figure 825(a) Since no current ows in the circuit, we obtain the following expression for vD : vD = vS 06
vs(t) i
+ vR _
To be consistent with the assumption that the diode is off, we require that vD be negative, which in turns corresponds to vS < 06 V Diode off condition
Circuit with offset diode model
With the diode off, the current in the circuit is zero, and the load voltage is also zero If the source voltage is greater than 06 V, the diode conducts, and the current owing in the circuit and resulting load voltage are given by the expressions: vS 06 vR = iR = vS 06 R We summarize these results as follows: i= vR = 0 vR = vS 06 for vS < 06 V for vS 06 V
06 V + vD _ + _ vS + _ + R vR _ (a) Diode off 06 V + vD _ + _ vS + _ i (b) Diode on + R vR _
The resulting waveform is plotted with vS in Figure 826
3 2 1 Volts 0 _1 _2 _3 0 0005 001 0015 Time 002 0025 003
Figure 826 Source voltage (dotted curve) and recti ed voltage (solid curve) for the circuit of Figure 824
Comments: Note that use of the offset diode model leads to problems that are very similar to ideal diode problems, with the addition of a voltage source in the circuit Also observe that the load voltage waveform is shifted downward by an amount equal to the offset voltage, V The shift is visible in the case of this example because V is a substantial fraction of the source voltage If the source voltage had peak values of tens or hundreds of volts, such a shift would be negligible, and an ideal diode model would serve just as well Focus on Computer-Aided Solution: The half-wave recti er of Figure 820 is simulated
using Electronics WorkbenchTM in the CD that accompanies the book The circuit is simulated by using an ideal diode model Replace the ideal diode with any of the other available options (physical diodes), and observe any differences in the result (Hint: The differences will be more dramatic for small peak source voltage values, say 5 volts)
Part II
Electronics
EXAMPLE 84 Using the Offset Diode Model
Problem
Use the offset diode model to determine the value of v1 for which diode D1 rst conducts in the circuit of Figure 827
+ R1
D1 + VB _ + vo _
Solution
Known Quantities: VB = 2 V; R1 = 1 k ; R2 = 500
v1 _
; V = 06 V
Find: The lowest value of v1 for which diode D1 conducts Assumptions: Use the offset diode model Analysis: We start by replacing the diode with the offset diode model, as shown in
Figure 828 Based on our experience with previous examples, we can state immediately that if v1 is negative, the diode will certainly be off To determine the point at which the diode turns on as v1 is increased, we write the circuit equation assuming that the diode is off If you were conducting a laboratory experiment, you might monitor v1 and progressively increase it until the diode conducts; the equation below is an analytical version of this experiment With the diode off, no current ows through R1 , and v1 = vD1 + V + VB According to this equation vD1 = v1 26 and the condition required for the diode to conduct is: v1 > 26 V Diode on condition
06 _ + vD1 _ + V + v1 _ 1 k D1 + 2V _ 500 + Vo _
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