Consider the circuit shown below, which is used to feed a 12-V dc output from a
ID: 2085155 • Letter: C
Question
Consider the circuit shown below, which is used to feed a 12-V dc output from a car battery, (which is typically charged to 13.8 V maximum). The function of the diode is to protect any device being plugged in from being damaged by a reverse polarity voltage. The fuse may be modeled by a 250-m resistor, and the diode needs to be modeled as specified below.
a. Specify the current rating of the fuse if its sole purpose is to protect the 3-W diode from overheating. Use the characteristic curve to obtain reasonably accurate answers.
F1 D1 -V2 T 13.8VExplanation / Answer
Answer:-a.) Here diode can work with maximum power of 3 W, let's assume a current " i " flows which causes power dissipation in diode equal to 3 W. Also output voltage will be 12 V always. Hence we can write a power equation as-
source power = fuse power + diode power + load power
=> 13.8 x i = (i2 x 0.25) + 3 + (12 x i)
=> 0.25i2 - 1.8i + 3 = 0
hence we get-
i = 2.62 A and i = 4.58 A
But from characteristic curve we can see when current is 4.58 A, voltage across diode is around 1.1 V. Hence this value can't be our answer since this gives power dissipated in diode as 4.58 x 1.1 W = 5 W. Hence our answer is i = 2.62 A, this should be the current rating of the fuse to prevent the diode.
b.) So for the circuit the rating can be witten as i = 2.62 A and P = 12 x 2.62 W = 31.44 W. Here the attached fuse provides the limit for the current in the circuit.