Figure 1 shows a typical buck converter. When power switch Q 1 turns on, the free-wheeling diode D 1 is reverse bias. The input current goes through LC filter to load directly.
When Q 1 turns off, D 1 is forward biased by inductor current i L. Switching voltage waveform shown as in Figure 1 b is pulsating rectangular. After LC filtering, assuming corner frequency of LC is much lower than switching frequency, output voltage appears almost pure dc. It can be understood the higher the inductance L is, the lower the capacitance C leads the same output voltage ripple. However, too big inductor causes high volume and high cost. While too low inductance causes big output capacitor.
It is not purely a design trade-off problem. When power switch is at ON state, the voltage across inductor is voltage difference between input and output voltage. When power switch is at OFF state, inductor voltage is the same output voltage with negative polarity. Form volt-sec balance of a inductor voltage 1 and 3 , one can get the voltage transfer ratio easily,.
Duty cycle, D, is defined as the switch turn-on time oven the entire switching period. Figure 2 shows the inductor current waveform. Because voltage waveform of inductor is pulsating rectangular, the inductor current will be triangular with certain dc level. When ripple factor is less than 2, the converter operates in continuous conduction mode C. M , otherwise discontinuous conduction mode D. The inductor voltage reverses in polarity because it will resist to a sudden change in current.
As a result, the diode will conduct. Then, the current path is described below. In my designs, I always use the RMS value to compute the component stress and power dissipation. Therefore, I can guarantee ample margin. This is the easiest part! As you can see on the buck converter circuit, the inductor is in series to the load. The load side is already a pure DC considering a very good control loop. This value is the average of the peak current and the minimum current that may be available across the load of a buck converter output.
OK, so the above were the various parameters and expressions essentially involved with a buck converter which could be utilized while calculating a buck inductor. Now let's learn how the voltage and current may be related with a buck inductor and how these may be determined correctly, from the following explained data:. Remember here we are assuming the switching of the transistor to be in the continuous mode, that is the transistor always switches ON before the inductor is able to discharge its stored EMF completely and become empty.
This is actually done by appropriately dimensioning the ON time of the transistor or the PWM duty cycle with regard to the inductor capacity number of turns. The above formula may be used for calculating the buck output current and it holds good when the PWM is in the form of an exponentially rising and decaying wave, or may be a triangle wave. However if the PWM is in the form of rectangular waveform or pulses, the above formula can be written as:. Here Vt is the voltage across the winding multiplied by the time for which it's sustained in micro-secs.
The above expression reveals that the current output from a buck inductor is in the form of a linear ramp, or wide triangle waves, when the PWM is in the form of triangular waves. Now let's see how one may determine the peak current within a buck inductor, the formula for this is:.
The AC part of that current mainly flows into C1 by design but the average DC part of the inductor current cannot flow in C1 hence All of the charge current flowing out of the inductor must eventually flow through the load, it's just that some of this charge current coming from the inductor flows to the load via the capacitor.
When the buck converter's switch closes, the inductor's current will start to rise. The inductor's current will rise to a level which will be greater than that being drawn by the load and at this point the capacitor will start to take charge current and it's voltage and that of the load will rise slightly. Next the buck converter's switch opens and the inductor's current will start to decrease.
At some point the inductor's current will be less than the current required by the load and at this point the capacitor will start to contribute to the load current and the capacitor's voltage and load voltage will fall slightly as the capacitor discharges.
So we can see how a small load ripple voltage is created. Note how the charge current which the capacitor supplies to the load during the switch-off period has previously been supplied to the capacitor by the inductor during the switch-on period.
Therefore the average current in the inductor must equal the average current in the load when calculated over a time period which is large compared to the time period of the buck converter's switching frequency. Sign up to join this community. The best answers are voted up and rise to the top.
Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Average inductor current in Buck converter Ask Question. Asked 6 months ago. Active 6 months ago. Viewed times. Edit: I get that steady-state operation implies zero capacitor current average.
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