In this tutorial, we will describe the Switching Buck Regulator circuit. We already described the Buck Regulator Design in the previous tutorial. Here we will discuss different aspects of Buck converter and how to improve its efficiency.
The difference between the buck and boost regulator is, in the buck regulator the placement of inductor, diode and the switching circuit is different than the boost regulator. Also, in case of boost regulator the output voltage is higher than the input voltage, but in buck regulator, the output voltage is lower than the input voltage. A buck topology or buck converter is one of the most used basic topology used in SMPS. Same as the boost regulator, a buck converter or buck regulator consist of an inductor, but the connection of the inductor is in output stage rather than the input stage used in boost regulators.
So, in many cases, we need to convert lower voltage to the higher voltage depending on the requirements. Buck regulator converts the voltage from higher potential to lower potential. In the above image, a simple Buck regulator circuit is shown where an Inductor, diode, Capacitor and a switch are used.
The input is directly connected across the switch. The Inductor and capacitor are connected across the output, thus the load gets smooth output current waveform.
The diode is used for blocking the negative current flow. In case of switching boost regulators, there are two Phases, One is Inductor Charge phase or the Switch-on phase Switch is closed actually and the other one is Discharge phase or the switch-off phase Switch is open. If we assume that the switch has been in open position for a long time, the current in the circuit is 0 and there is no voltage present.
In this situation, if the switch gets close then the current will increase and the inductor will create a voltage across it. This voltage drop minimizes the source voltage at the output, after a few moments the rate of current change decrease and the voltage across the inductor also decrease which eventually increase the voltage across the load.
The current through the inductor rises at linearly with time. The linear current rising rate is proportional to the input voltage less output voltage divided by the inductance.
The upper graph showing the Charging phase of the inductor. The x-axis denotes t time and the Y-axis denotes i current through the inductor. The Current is increasing linearly with time when the switch is closed or ON. Now if the switch will open during this time while the current is still changing, there will always a voltage drop occurring across the inductor.
The voltage across the load will be lower than the input voltage. During the off state, while the switch is open, input voltage source gets disconnected, and the inductor will transfer the stored energy to the load.
The inductor will become the current source for the load. The Diode D1 will provide a return path of the current flowing through the inductor during the switch off-state. I love to solve your technical queries via comment section. Hi, thank you for the schematic very interesting. I thouht that powering an inverter from solar panels through a dc converter on correct voltage , could drive an inverter, with no need of using a battery during day time, for lighting use at first.
What about using solar panels with output voltage of 38V — no load. Which should be the changes on the schematic? Hi, You need to find a powerful buck converter that can accept 38 volt and convert it to 14VDC, rest of the circuit is the same. Or some converters in parallel, adjusted on the same Vout?
Is that correct or will cost in power loss? Thanks for the answer. Thanks very much for your hard work. Please kindly help as i really want to build the system. Thanks once again. Thankyou very much for finding the mistake in the content, we will fix it very soon. Your email address will not be published. Skip to content. We will see: Features of the solar inverter.
Circuit Diagram of Solar Inverter. Block Diagram and Description. Program Code of Solar Inverter. Transformer and Solar Panel Power Calculations. How to Test and Operate the Solar Inverter. The solar inverter sport features like under voltage and over voltage DC input cut-off so that the inverter can shut-off smoothly when the solar panels generates voltage lower than an acceptable level and start automatically when the solar panel generates voltage with in the acceptable level.
We can customize the output frequency to 50 Hz or 60 Hz from the program code. The inverter generates its frequency from Arduino which guarantees output frequency stability. 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.
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