The peak current rating of inductor can be found using maximum value of the inductive current I Lmax. The maximum inductor current occurs at maximum load. The peak current rating of inductor can be found using the equation of maximum inductor current as given as. The simplified form is given below which will define the rating the rating of inductor current. This section will discuss both current and voltage rating of a switch for buck converter.
For ideal diode, the V switch-max is equal to V dcmax while for non-ideal diode; V switch-max is equal to V dcmax plus V F. The current rating for a switch is calculated based on average current.
By drawing the switch current waveform, the average value of the current can be calculated. The average current for the switch is calculated here. Overall inductor current is equal to switch current and diode current using KCL.
During the turn ON time, the inductor current is equal to switch current while the inductor current is equal to diode current during turn OFF time. By putting the values in above equation, the result will become as given below. Further simplifying and putting the previously discussed value. The final form of the equation of current rating for the switch become.
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. Here's a brief description of the timing and operation of the buck converter power stage, starting from an off condition and not from a steady-state condition.
Please refer to the first schematic, below. During this time, the current in L1 increases linearly with a linear ramp shape from zero to a peak value Ipk. During this interval, both the C1 and load voltage increase but probably won't get to their steady-state value on the first switching interval. Next, FET M1 is switched off. Now, the current in L1 continues to flow in the same direction toward the load, but it's decreasing.
Instead of current flowing thru M1, in Interval-2 it flows thru D1. Note that in Inverval-1, the voltage polarity on L1 is positive on the D1 side. But on Interval-2, the voltage polarity across L1 reverses since L1 becomes a power source. Since the polarities of V1 and L1 are opposite, they get subtracted, or "bucked-out" and this is why it's called a Buck Converter.
Capacitor C1 stores energy and releases it, thereby smoothing the output voltage. One bad point about buck converters is that if FET M1 shorted, the full input voltage would appear across the load and depending on the particulars, may damage the load.
This is especially true for high-voltage input buck converters. For example, a VDC input with a 3. Most any 3. But with well-designed and well protected buck converters, these kinds of failures are rare. I think I'll end today's blog here. The main question when designing a converter is what sort of inductor should be used.
In most designs the input voltage, output voltage and load current are all dictated by the requirements of the design, whereas, the Inductance and ripple current are the only free parameters. It can be seen form Equation 1, that the inductance is inversely proportional to the ripple current.
In other words, if you want to reduce the ripple, then use a larger inductor. Thus, in practice a ripple current is decided upon which will give a reasonable inductance. There are tradeoffs with low and high ripple current. Large ripple current means that the peak current is i pk greater, and the greater likelihood of saturation of the inductor, and more stress on the transistor.
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