ad space

Output Filter(L,C) Calculator Switching Power Supply

Design Inputs:
W
V
mV
V
A
% of IL
kHz
%
Results:



Output Filter(L,C) Calculator Switching Power Supply
 
Solved Example: How to calculate L and C values of Switching power supply converters

In power supply electronics, designing determining the inductor (L) and capacitor (C) values is crucial for achieving stable operation with minimal ripple. This post explains how to calculate the L and C values for a such as buck converter, flyback converter etc that employs low pass filter circuit step by step, assuming the ripple current (ΔIL) is percentage of the load current(IL), ripple output voltage, maximum output voltage tolerance to get minimum output capacitor value, duty cycle as a percentage of the total period of the desired switching frequency.


Key Design Assumptions

  1. Output Voltage (Vo): 5V.
  2. Maximum Output Voltage (Vp):6V
  3. Ripple Voltage (ΔVo): 50mV.
  4. Load Current (IL): 2A.
  5. Ripple Current (ΔIL): 30% of the load current (IL).
  6. Switching Frequency (fsw): 20kHz.
  7. Duty Cycle (D): 40% of switching period(tp).

Step 1: Inductor Design

The voltage across the inductor (VL) during the "on" period is:

VL=VsVo

Here:

  • Vs is the secondary voltage of the transformer, that is the input voltage to the output filter.
  • Vs=VoutD=50.3=16.66V, where D =30% / 100%(duty cycle).

Thus:

VL=16.665=11.66V

The inductance (L) is calculated using:

L=VLtonΔIL

Where:

  • ton=DT=0.3120,000=15μs,
  • ΔIL=0.3×IL=0.3×2A=600mA(30% of Load Current IL)

Substitute the values:

L=11.6615μs0.6=291.67μH

Step 2: Capacitor Design

The capacitor is designed to handle ripple current and voltage. For ripple voltage, we use:

C=ΔILtonΔVout

Substitute the values:

C=600mA15μs0.5=180μF

However, a small capacitance may lead to unacceptable overshoot during sudden load changes. To address this, consider energy transfer from the inductor to the capacitor during load transients.


Step 3: Adjusting Capacitance for Overshoot

The energy stored in the inductor is:

EL=12×291.67×106×(2)2=583.33μJ

The capacitor must absorb this energy without exceeding Vp=6V. Using:

EC=12C(Vp2Vo2)E_C

Rearrange for C:

Cmin=LI2Vp2Vout2

Substitute values:

Cmin=19.4106(20)26252=7.761033625=106.06 µF

Conclusion

For the given converter design:

  • Inductance (L): 291.67 μH
  • Capacitance (C): 180 μF
  • Min. Capacitance(Cmin):106 μF
  • Circuit Diagram

    Below is a basic schematic of the buck converter showing L and C placement.

    Power Supply Output Filter

    By carefully selecting these components, you can ensure efficient operation with minimal ripple and stable voltage.

    Related Web Calculators

    Post a Comment

    Previous Post Next Post