A step-up chopper is used to feed a load at 400 V dc from a 250 V dc s...
Chopper has a duty cycle of 0.6, what is the value of the inductance required to maintain a ripple current of 2 A?
To solve this problem, we can use the following equation for the average output voltage of a step-up chopper:
Vout = (D/(1-D)) * Vin
where Vout is the average output voltage, Vin is the input voltage, and D is the duty cycle. We can rearrange this equation to solve for Vin:
Vin = Vout * (1-D)/D
Using the given values of Vout = 400 V, Vin = 250 V, and D = 0.6, we get:
Vin = 400 * (1-0.6)/0.6 = 266.67 V
Next, we can use the following equation for the average inductor current of a step-up chopper:
Iavg = (Vout - Vin)/L * T
where Iavg is the average inductor current, L is the inductance, and T is the period of the chopper. We can rearrange this equation to solve for L:
L = (Vout - Vin)/(Iavg * T)
The ripple current can be calculated as:
ΔI = Vout * D/(L * f)
where ΔI is the ripple current, f is the chopper frequency, and we assume a continuous inductor current. We can rearrange this equation to solve for L:
L = Vout * D/(ΔI * f)
Using the given values of Vout = 400 V, Vin = 266.67 V, D = 0.6, ΔI = 2 A, and assuming a chopper frequency of 10 kHz, we get:
L = 400 * 0.6/(2 * 10^4) = 0.012 H
Therefore, the inductance required to maintain a ripple current of 2 A is 0.012 H.