Voltage converter buck

Shotkey diodes are preferred for the discharging of buck converters due to fast recovery action. These diodes are known as fast recovery diodes so preferable for high frequency operation i. This section will discuss the current and voltage rating of shotkey diode for buck converter. It is given in the data sheet of the component. Where, the value of V sw is calculated at maximum load current. The same approach is adopted for calculating current rating for diode as it was adopted for the calculation of switch.

The average forward diode current is calculated from the current waveform of the diode. As diode conduct during off time of the switch therefore, t off is considered in the calculation.

By putting values in the above equation, the result will become. Simplifying the above equation will result. By putting previously discussed values and simplifying, the result will become. This will lead to one of most important result that is given below. The following wave form is the diode current wave form.

It provides ease in finding average value. This section will discuss the important parameter for a capacitor under which the capacitor can be operated in safe mode. Furthermore, the capacitor is designed such that the required functioned is performed.

The capacitor is designed and chosen such that the maximum capacitor voltage must withstand the maximum output voltage. Ideally the maximum capacitor voltage V cmax is. As given as. The case is a bit different for particle capacitors.

This contribution made by ESR can be suppressed by using following methods. The designed capacitor will provide a path for AC ripples of inductor current while pure DC current will flow into the load. That is how the capacitor will act as a filter. The waveform of the capacitor current will look as shown below. It can be seen form the below given waveform capacitor current w.

By putting the values and simplifying the result will become. The final result for minimum capacitance will be as shown below. The capacitor is designed such that the maximum input voltage will be with stand by capacitor voltage. Ideally both are considered equal i. The ESR factor contributes to capacitor loss. The ESR factor can be reduced for better efficiency by two methods. Either by paralleling capacitors or by choosing capacitor with low ESR.

By putting the value of q in this equation, the result will be as given below. I o is considered as I o max.

Therefore, the above equation become. This is the required capacitance value while the RMS current rating equation can be found as. The following table shows all the important equations required for the designing of Buck converter.

To design a buck converter that will convert volt input DC to 2. For such conversion we have some known data and some parameters are required. Proper selection of components is must for successful conversion from 12v to 2. This example will help to design buck converter for any conversion ratio. The duty cycle D can be found from the output input voltage ratio.

Critical inductance can be found from previously found equation. The critical inductance can be chosen as. The peak current rating can be found according to the equation. I Lmax 1. Forward diode current according to the given equation will be.

Maximum switch voltage according to the equation derived above. While maximum switch current. Minimum capacitance required for the converter according to the equation will be. Near value for this required capacitance can be. Voltage rating of capacitor. Losses for the buck converter must be considered when the efficiency estimation is required for it. Several major losses that are to be considered are given below and discussed briefly one by one.

This on resistance greatly contributes in over losses. The two graphs given below show the exponential increase of on state resistance. The other graph shows the increase of on-state resistance with increase in temperature. Drain current in this case 7. Switching losses are related with the transition time of the switch. During the transition time, both current and voltage are non-zero. Therefore, the main switching losses are due to overlapping of current and voltage.

The given graph show that how losses occurs in transition states. The voltage across the switch approaches to zero with a specific slope while current across it increase. During this time losses occur. The same case is with turning OFF the switch. During this time current approaches to zero with a specific slope while voltage drop across it increases. This is how transition losses occur during transition time.

According to the above discussion the overall power losses P loss is equal to the power losses during turn-on time and turn-off time. We know that losses during turning-on time is. While losses during turn-off time is.

By putting both these values in the above equation, we will get the following result. By taking common term, the final form for the overall switching losses will become.

Whereas the gate drive losses come from two parameters i. By considering both, the mathematical form for gate drive losses is. Losses that occur when diode is completely on or when diode is completely in off state. Static losses that occur when diode is in on state are known as forward static losses. In contrast, the losses occur in off state is known as reverse static losses.

For more precise value of the diode forward loss, the rms loss that occurs due to diode dynamic resistance , rd is added. All these calculations were for forward losses. While losses for reverse state are. This section will discuss the losses associated with diode connected in practical buck converter. The same is the case with diode as it is for switch discussed previously. This section will discuss losses associated in both turn-on time and turn-off time.

The losses associated in turn-on time are characterized by forward recovery time t fr and by low value of peak forward voltage V FP. By knowing above two value from the data sheet, the on-loss P ON can be calculated from the given equation. The losses associated in turn-off time are associated with the time for which diode voltage and current overlaps. This overlap manly contributes in reverse recovery time. This is really important equation for calculating turn-off losses in non-ideal case.

Some of unknown values required for the above equation can be found from the following equations. There can be at most three inductors in buck converter that are storage inductor, coupled inductor and filter inductor. Therefore, the losses of all these inductors are considered in buck converter. In most of buck converters, the coupled inductor is not used but storage inductor and filter inductor are must.

Therefore, losses of their two inductors are considered. Some of the losses that occur in magnetic components are as given as. Above is the general form of Steinmetz equation whereas the modified form of this equation is as. There are two main type of losses associated with inductor i. This section will discuss inductor copper loss while core losses are discussed separately. Inductor copper loss, as its name suggests that these losses are associated with the winding of the inductor.

As the winding is made of copper wire therefore, it is known as inductor copper losses. These losses are resistive in nature because the winding have some resistance. These losses are not significant that is why these are ignored for ideal buck converter while considered for more precise calculations.

Inductor copper losses occur due to the resistance of the winding. Core losses of an indicator in buck converter are mainly affected by three factors i. The general form of the formula for inductor core loss is given as. Values required for these constants need be as low as possible for low core losses.

Some of well-known manufacturers and companies provide with very low values of these coefficients for better efficiency. This resistance contributes to power loss in buck converter known as Capacitor ESR loss. R ESR. A buck converter has already been designed in this article. But that example was for pure ideal buck converter which does not exist in practical life.

This section will show how to use previously derived equations to compute the values of different components required for buck converter. This example will show that how to design non-ideal buck converter for given parameters. We will design a non-ideal buck converter for the given parameters according to the previous discussion and derived equations. Given parameters. Nominal output voltage of the system is.

Nominal input voltage of the system is. Maximum output power is. Switching frequency is. Maximum ripple percentage is. Minimum percent CCM is. Calculations for Designing. Nominal duty cycle for the non-ideal buck converter is.

Inductor value selection: The critical inductance formula is a little bit different that is. Conclusion: We have already discussed that the value of inductor must not be chosen less than Critical inductance. Therefore, the value of inductor can be any value greater than this critical value. Hence choosing Lo as. Calculating peak inductor current according to the above discussion is. Switch Selection:. The switch voltage is. Switch current is as. Diode selection:. For proper diode selection, we need to know the values of two parameters i.

We know that. Diode Forward Current according to the given equation is as. Conclusion: according to the above values, we came to know that Shotkey diode MBR is the best solution. For further detail, see the data sheet of MBR Choosing capacitor. For choosing the value of capacitor, we need to know the values of three parameters I.

We know that the value of capacitor voltage rating according to the formula is. The capacitance C o is. RMS Current Rating can be found according to the given formula as. Conclusion: looking at the above results, we came to know that 25V 50uF capacitor is the best value of the capacitor to be chosen.

Output voltage of the system is. Input voltage of the system is. Output Vpp-ripple percentage is. Percent minload-CCM is. Full load current.

Time period Ts is the reciprocal of the switching frequency. Here we will calculate all the important parameters required for the designing of buck converter. First of all, it is required to find ideal on time and ideal off time.

The ideal on time of the switch is the ratio of output voltage and input voltage with the product of total periodic time. Minimum current required to maintain continuous conduction mode CCM is.

I omax. Inductor value selection:. We will find the value of critical inductance value which will decide the minimum value for which buck converter can be operated in CCM mode. The value must be chosen bigger than critical inductance value for operation in CCM mode. I omin. The minimum output current value can be computed according to the chosen inductor value that is.

The next thing to be determined is I Lmin and I Lmax at minimum load. The given two equations in terms of I Lmin and I Lmax are. Hence, by simplifying I Lmin and I Lmax at minimum load are. Hence the difference between both. Calculating output capacitor value. The approximate value of the ESR according to its equation is. If we assume that Electrolytic capacitor is used, then. By putting their values, the equivalent result is. Total ripples calculation.

For calculating total ripples, we need to find ripples due to two factors i. These calculations are performed individually, and their addition gives total ripples.

For calculating all these values, we need to recalculate the value of ESR for chosen capacitor. By putting values and simplifying, we get the result.

The value of ripples due to capacitor charge and discharge is. The result of above equation, by putting values in it is. The efficiency can be improved by choosing the correct value of components used in buck converter. The efficiency can be improved further by applying the discussed strategies to the following components.

As it is discussed previously in this article that equivalent series resistance ESR of the capacitor directly contributes in power losses.

Therefore, the losses can be reduced by reducing ESR. In other words, reducing ESR will reduce power losses which will increase efficiency. Another way of reducing power losses and increasing efficiency is to use paralleling capacitors method. Using this method, over all capacitance will be increased while it will reduce ESR.

Reducing ESR is the alternate way of saying that efficiency is increased. The efficiency can be increased by reducing rms losses occurs both in inductor and output capacitor. These are magnificent losses and must be reduced for better efficiency. There are a lot of strategies for reducing inductor current ripples. The more common solution to reject ripples is to use reservoir capacitor.

This solution is for rejecting ripples from going to the output but the losses due to ripples still occurs. The more practical solution is to increase switching frequency or to increase inductance. This will reduce inductor ripple, but it will increase losses due to frequency. The reason is that frequency is also directly proportional to losses.

There is always tradeoff between both. Now, if inductor value is increased instead of increasing frequency then the DC resistance of the wire increases. As the resistance value is directly proportional to the length of the wire of the winding.

Therefore, choosing big value of inductor will increase the losses due to DCR. By reducing the gate drive voltage, the power losses that take place there can be reduced. In other words, the efficiency of the system can be increased. This improves the efficiency much better for low voltage and high current.

This procedure is especially beneficial for high current and low duty cycle applications. The final problem can be solved by replacing the body diode with Schotkey diode. The reason is because body diode is slow due to its dead time. Soft switching prevents hard switching which means that it prevents the overlapping of current and voltage of the switch during turn-off and turn-on transition.

This will significantly improve the converter efficiency because we have already discussed the relation of switching frequency and losses. Resonant circuit is used to change shape of the signal.

Using this circuit, current or voltage waveform goes to zero at the time when switch is initiated. The switch will be switched turned on and off by using a PWM signal. A small duty cycle means that the average voltage seen by the load is small and when the duty cycle is high the average voltage is high too. But average voltage is not what we need — a raw PWN signal oscillates between high voltage level and ground, something no delicate load like the microcontroller would like.

Of course, connecting an RC filter to a square wave source renders the output clean. The voltage level of the filter depends on the duty cycle of the PWM signal — the higher the duty cycle the higher the output voltage. So now we have a clean output voltage. The below graph shows the raw PWM signal in blue color and the filtered outputs in red and violet color. To fix this problem, we turn to another type of voltage filter, the LC filter, which does the same job as the RC filter but replaces the R with an L, in other words the resistor with an inductor.

The inductor resists changes in current and the capacitor resists changes in voltage, which results in the output being smooth DC. And now we have a converter that is capable of stepping down DC voltages and doing it efficiently! The switch turns on and lets current flow to the output capacitor, charging it up. Since the voltage across the capacitor cannot rise instantly, and since the inductor limits the charging current, the voltage across the cap during the switching cycle is not the full voltage of the power source.

The switch now turns off. Since the current in an inductor cannot change suddenly, the inductor creates a voltage across it. This voltage is allowed to charge the capacitor and power the load through the diode when the switch is turned off, maintaining current output current throughout the switching cycle. Determine the output power, that is, the product of the output voltage and current.

This is also the input power, by the law of conservation of energy though not exactly so — nothing is a hundred percent efficient! Now divide the output power by the selected switching frequency in order to get the power transferred per pulse.

Since it is easier to talk about inductors in terms of energy, we can assume now that the output power is simply the output energy per second. So if the output of our converter is 30 Watts, then we can say that the output energy is thirty Joules every second. Now that we have the energy per pulse, we can calculate the inductance using the input current and the energy:.

Using the values of the inductance, frequency and duty cycle, we can now get to work building a simple boost converter. Though this is possible, it is quite complicated. Using a P channel device in these circumstances would be recommended, they greatly simplify driving requirements, but remember that they turn on when the gate is low, so an inverted signal would be necessary.

However, there are many better devices available, the choice is entirely up to the designer based on the specific application. Since this diode does not have to handle very high voltages, rather high currents, it would be a good design choice to use a Schottky diode with a low forward voltage drop to keep things efficient.

The capacitor value depends on the output voltage ripple and can be calculated using the capacitor equation, but generally a value between uF to uF for low current applications should suffice.