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Fig. P4.30, Dj is a large-area, high-current diode whose reverse leakage i ge, while D2 is a much smaller, low-current diode. At an ambient temperatun ke VR1 = V2 = 520 mV. Subsequent measurement indicates that R1 is 520 kl.

Fig. P4.30, Dj is a large-area, high-current diode whose reverse leakage i ge, while D2 is a much smaller, low-current diode. At an ambient temperatun ke VR1 = V2 = 520 mV. Subsequent measurement indicates that R1 is 520 kl.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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14 4.30 In the circuit shown in Fig. P4.30, D is a large-area, high-current diode whose reverse leakage is high and independent of applied voltage, while D2 is a much smaller, low-current diode. At an ambient temperature of 20°C, resistor R1 is adjusted to make VRI = V2 = 520 mV. Subsequent measurement indicates that R1 is 520 k. What do you expect the voltages VRI and V2 to become at 0°C and at 60°C? +10 V R VRI D, D, SZ Figure P4.30
Transcribed Image Text:14 4.30 In the circuit shown in Fig. P4.30, D is a large-area, high-current diode whose reverse leakage is high and independent of applied voltage, while D2 is a much smaller, low-current diode. At an ambient temperature of 20°C, resistor R1 is adjusted to make VRI = V2 = 520 mV. Subsequent measurement indicates that R1 is 520 k. What do you expect the voltages VRI and V2 to become at 0°C and at 60°C? +10 V R VRI D, D, SZ Figure P4.30
Expert Solution
Step 1

GIVEN,

         VR1=V2=520mV

           R1=520kohm

The voltage across the resistance is,

     VR1=520mV= IR1=520X10-3

Put  the value of R1,

 I(520X103)=520X10-3

I= 10-6μA

  =1μA

The  reverse-biased diode, the diode current is equal to the saturation current.

I=IS=1μA

The reverse voltage across the diode D1 is,

V1=10-(VR1+V2)

    =10-0.52-0.52=8.96V

As the reverse current doubles for every 10°C rise in temperature , the reverse current at 0°C is,

IS(0°C) = IS/4= 1/4=0.25μA

Calculate the voltage across the resistance, VR1

VR1=R1IS0°C

       =(520x103)(0.25x10-6)

      =0.13V

      =130mV

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