A boost converter has the following specifications:Vin = 36 V, Vo= constant = 108 V, Fs = 51 kHz, and the output power may vary with a range of Po(range) = 107 W to 200 W. Assuming that the converter is ideal, calculate:a) the smallest value of the inductor, L, in order to have continuous current
Question
A boost converter has the following specifications:Vin = 36 V, Vo= constant = 108 V, Fs = 51 kHz, and the output power may vary with a range of Po(range) = 107 W to 200 W. Assuming that the converter is ideal, calculate:a) the smallest value of the inductor, L, in order to have continuous current
Solution
To calculate the smallest value of the inductor, L, for continuous current in a boost converter, we can use the following formula:
L = (Vin * D) / (Fs * Iripple)
Where:
- Vin is the input voltage
- D is the duty cycle
- Fs is the switching frequency
- Iripple is the ripple current
First, we need to calculate the duty cycle (D). The duty cycle for a boost converter is given by:
D = 1 - (Vin / Vo)
Substituting the given values:
D = 1 - (36V / 108V) = 0.67
Next, we need to calculate the ripple current (Iripple). The ripple current can be approximated as the change in output power divided by the change in input voltage.
Iripple = ΔPo / ΔVin
The output power range (ΔPo) is 200W - 107W = 93W. The input voltage (Vin) is constant at 36V, so ΔVin = 36V.
Iripple = 93W / 36V = 2.58A
Now we can substitute all the values into the formula for L:
L = (36V * 0.67) / (51kHz * 2.58A) = 0.19 mH
So, the smallest value of the inductor, L, in order to have continuous current is 0.19 mH.
Similar Questions
A boost converter has the following specifications: Vin = 36 V, Vo= constant = 108 V, Fs = 51 kHz, and the output power may vary with a range of Po(range) = 107 W to 200 W. Assuming that the converter is ideal, calculate: a) the smallest value of the inductor, L, in order to have continuous current b) the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V. Use the inductance you calculated in part (a)
In a Boost converter of Figure below, the inductor current has ΔiL = 2 A. It is operating in dc steady state under the following conditions: Vin = 5 V, Vo = 12 V, Po = 11 W, and fs = 200 kHz. Assuming ideal components, calculate the inductance value in Micro Henery
A Buck converter is charging a battery, with the following specifications: (Consider ideal switch and ideal diode) a) Vin = 25 V, Vb = 8 V, RL=20 Ω, b) fs= 100kHz (switching frequency), d=0.4 (duty ratio) c) Peak to peak inductor current ripple is 10% of the output current. d) Output voltage ripple across capacitor is 5% of the output voltage. What is the minimum inductor value L from the below Circuit in
A Single phase PFC Circuit is required to be designed for the following Specs:Rated output power Po: 2.5kW AC input voltage range: 100-270VAC Grid frequency range: 50±1HzOutput DC voltage 400VSwitch frequency fsw: 120kHz Efficiency h: >94% Therefore: The Peak Value of the boost inductor current is Select one:a. 25 Ab. 13 Ac. None of the given choicesd. 35.35 A
The calculation of the Inductor Value in all the Primary converters is based on : Select one:a.Energy Balance Equationsb.Power Balance equationsc.Inductor Voltage Equationd.Capacitor Current Equation
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.