In the DC to DC converter part one we discussed about linear voltage regulation and switching voltage regulation. At the part two discussed about Buck converters. In this part we are going to design a buck converter for a given specification. If you not read part one and two, please go back and read them first. Because the design methodology and theory was discussed in the part two. Only the calculation part and simulation part will discuss here.
As we discussed before there two modes in buck converter operation. Mode one is when the switch close and other is when the switch open.
Mode 1: Switch Closed
Mode 2: Switch Open
But we know that Net change in inductor over one period is zero. In inductors, the current cannot change instantaneously because inductors tend to keep the current constant. An ideal inductor with infinite inductance act as a constant current source.
In the steady state output current and inductor current are same.
Find Inductor Value,
Using equation 2,
𝛥I (%) given as 10% of output current and Frequency F given as 1000kHz
𝛥𝑖 = 𝑖𝐿 ∗ 10% = 10𝑚𝐴 ∗ 10% = 0.001𝐴
Find Capacitor Value,
In the capacitor
𝛥𝑄 = 𝐶𝛥𝑉𝑜 —————– Equation 5
Using equation 3,4 and 5,
𝛥Vo (%) given as 10% of output voltage
Result and Observation
The simulation is done by using PSIM software,
The given voltage and current ripple is 10% and the voltage ripple that was observed from the simulation was lower at 6.93% and the current ripple was also a lower value at 7%.
When we analyzing the buck converter after the designing, some assumptions have to make for the accuracy of the circuit.
- The circuit operation in the steady state
- The inductor current is continuous (operate in continuous mode)
- The switching period is T, the switch is closed for DT and open for time (1-D) T.
- The components are ideal
The key to the analysis for determining the output voltage is to examine the inductor current and inductor voltage first for the switch closed and then for the switch open. So the net change in inductor current over one period must be zero for steady state operation. The average inductor voltage is zero.
As in the simulation result show, the calculations proved to be true, because at the load it was able to output an average of 3.3V voltage with the average current of 10mA. Also the capacitor current variant on ‘x’ axis therefore the average current through the capacitor is zero.
Buck converter applications
- use in self-regulating power supplies
- use as low-loss current sources to drive LED arrays (solid state lighting applications).
- use as Point-of-load (POL) converters in servers
- use in advanced telecom and Datacom systems
Buck converter Advantage
- very simple and it requires only one Power switch
- Efficiency of Buck regulator is about 90%
- The cost and size is low
Buck converter Disadvantage
- Slow transient response
- Normally Input filter is required
- High output ripple