For the quad system shown below, calculate the improvement factor over a single diode The following data of the system is given: All diodes are identical. • = 0.01. • Probability of normal operation of a diode: P Probability of open circuit of a diode: Po Probability of short circuit of a diode: P The improvement factor is defined as follows: • 0.02. ● I.F.= = 0.97. Probability of failure of one diode Probability of failure of a quad
For the quad system shown below, calculate the improvement factor over a single diode The following data of the system is given: All diodes are identical. • = 0.01. • Probability of normal operation of a diode: P Probability of open circuit of a diode: Po Probability of short circuit of a diode: P The improvement factor is defined as follows: • 0.02. ● I.F.= = 0.97. Probability of failure of one diode Probability of failure of a quad
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Transcribed Image Text:For the quad system shown below, calculate the improvement factor over a single diode.
The following data of the system is given:
All diodes are identical.
●
●
●
●
Probability of normal operation of a diode: P = 0.97.
Probability of open circuit of a diode: Po
= 0.01.
0.02.
Probability of short circuit of a diode: Ps
The improvement factor is defined as follows:
I.F.=
Probability of failure of one diode
Probability of failure of a quad
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lb
+
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Step 1
Given a quad system, all diodes are identical.
Probability of the normal operation of the diode Pn=0.97
Probabilty of the open circuit of the diode Po=0.01
Probabilty of the short circuit of the diode Ps=0.02
To calculate the improvement factor over a single diode
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