Lower and Upper bounds of a diode?
@morgs-AmLSKs
•
Oct 23, 2024
Oct 23, 2024
985
Hello, I just need some help with a question. It is as follows.
By Kirchoff’s Law:
E = VR + vd
E = I×R + vd
The voltage across the diode, vd, depends on the current through the diode, I, and is
non-linear. The current, I, through the diode is related to the diode voltage drop, vd,
by:
where: Is -The amount of reverse bias current. For silicon this is typically: Is = 1 nA.
k - Boltzman’s constant, k = 1.38047 × 10-23 J/K.
T - Absolute temperature of the diode, (assume room temperature T = 300 K).
Q - Electronic charge, Q = 1.60203 × 10-19 C.
Substituting the value for current, I, into the above Kirchoff’s equation gives:
Determine the lower and upper bounds for the Diode Voltage, vd, and state the reasons
for these bounds. Then solve for the diode voltage, vd.
Any one out there explains this to me. I got an answer but Im assuming it is incorrect until proven right. Thnx
By Kirchoff’s Law:
E = VR + vd
E = I×R + vd
The voltage across the diode, vd, depends on the current through the diode, I, and is
non-linear. The current, I, through the diode is related to the diode voltage drop, vd,
by:
where: Is -The amount of reverse bias current. For silicon this is typically: Is = 1 nA.
k - Boltzman’s constant, k = 1.38047 × 10-23 J/K.
T - Absolute temperature of the diode, (assume room temperature T = 300 K).
Q - Electronic charge, Q = 1.60203 × 10-19 C.
Substituting the value for current, I, into the above Kirchoff’s equation gives:
Determine the lower and upper bounds for the Diode Voltage, vd, and state the reasons
for these bounds. Then solve for the diode voltage, vd.
Any one out there explains this to me. I got an answer but Im assuming it is incorrect until proven right. Thnx
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