Project 1 - Loss calculation for a DC/DC converter-MATLAB
EV Refresher
Batch-5
Losses in a converter
There are two types of losses
1. Conduction losses
2. Switching Losses
Conduction losses:
Pcond = Enegy consumed during Conduction * Switching frequency
Energy consumed during conduction is given by
Conduction power losses
Switching Losses:
Switching looses = Turn On losses + Turn OFF losses
Turn ON energy
Turn OFF energy
total losses is given by
Turn ON energy as a function of id, and gate resistance
Turn OFF energy as a function of id, and gate resistance
We know that efficiency is given by
Efficiency = Output power / input power
Input = Output + losses
Efficiency= Output power / output + losses
Efficiency = Vout*Iout / Vout*Iout + Pcond+Psw
Buck Converter
let us assume folling data
Vout = 12V
Iout = 5A
Output power = Vout * Iout = 12 * 5 = 60W
Load resistor is given by
Rl=VoutIout⇒125=2.4ohmsRl=VoutIout⇒125=2.4ohms
Load resistor is equal to 2.4 ohm
Specify the duty ratio of the converter.
The Duty cycle ratio(D) for the Buck converter is given by the ratio of Output voltage to Input voltage
D=VoutVinD=VoutVin
Given,
Vin = 48 V
Vout = 12V
D=1248D=1248
D = 0.25
The duty cycle ratio of the given condition is 0.25
Suggest a switching frequency of the converter.
The switching frequency is taken as 50000 Hz
Fs = 50000 Hz
Design the switching inductor such that the current is continuous and the inductor current ripple is less than 30% of the average inductor current.
Given,
Il = Io = 5A
Ripple is <30%
ΔI = Il * % of ripple = 5*0.3 = 1.5A
ΔI = 1.5A
Inductor for the buck converter is given by
`L=(Vout*fs⋅ΔI)⋅(1−D)L=(Vout*fs⋅ΔI)⋅(1-D)`
L = 1.2000e-04 = 0.12 mH
L = 0.12 mH
Inductor for the given data for the Buck converter is 0.12mH
Design the output capacitor such that the output voltage ripple is less than 5%.
Given
Output ripple is < 5%
`(Δ V) / (V o u t) = 0.05`
Capacitor of the buck converter is given by
`C = (1 − D)/ (8 ⋅ L ⋅ f s^ 2 ⋅ ( Δ V /V o u t ))`
C = 6.25 μF
Capacitor for the given data for the Buck converter is C = 6.25 μF
Model
Input Voltage = Vin = 48V
Input Current = Iin = 2.41A
Output Voltage = Vout = 11.35V
Output Current = Iout = 4.27A
Efficiency = Vout* Iout / Vin*Iin
`Efficiency = (11.35*4.27) / (48*2.41) *100`
Efficiency = 41.89%
Total losses = Input power - Output power
Toatal losses = Vin*Iin - Vout* Iout
Toatal losses = 48*2.41 - 11.35*4.27
Total losses = 67.215W
This toatal losses includes Conduction looses + Switching losses
Conduction losses
I min = 4.6A
Imax = 4.825A
Rds(On) = 0.1 ohm
Pcond = 1/3* (4.6^2 + 4.825 * 4.6+4.825^2)*0.1*0.25
Pcond = 6.03 W
Switching losses = total losses - Conduction losses
Switching looses = 67.215 - 6.03 = 61.177W
Switching looses = 61.177W
We know that the switching looses are almost 75% of total losses in this case 61%