Switching Regulator IC Series
Inductor Calculation for Buck Converter IC
No.12027ECY01
This application note covers the steps required in choosing the inductor and to calculate the value used in buck regulator IC
circuits.
● Buck (Step-Down) Converter
Switching regulators are used in a variety of applications to provide stable and efficient power conversion. A buck converter
is a specific type of switching regulator that steps down the input voltage to a lower level output.
Fig. 1 shows a typical buck converter circuit when switching element Q1 is ON. When N-ch MOSFET Q1 is ON, current
flowing from input VIN to coil L charges the output capacitor CO and supplies output current IO. In this scenario, the current
flowing through coil L produces a magnetic field and electric energy is converted to magnetic energy to be stored.
Fig. 2 illustrates the same circuit when switching element Q1 is in an OFF state. When Q1 is OFF, free-wheeling diode D1 is
activated, and the energy stored in coil L is released.
Q1 Q1
ON VSW OFF
VL VL
VIN L IO VIN L IO
D1 CO RL VOUT VD D1 CO RL VOUT
Fig. 1: Basic Buck Converter Circuit Fig. 2: Basic Buck Converter Circuit
Switching Element ON Switching Element OFF
● Inductor Current Waveform
Fig. 3 shows the inductor’s current waveform. IOUT is the average inductor current value. When switching element Q1 is
ON, current flow is shown during ON period tON of Q1, and voltage VL(ON) of coil L can be calculated by the following equation:
VL( ON ) ( VIN VSW VOUT ) (1)
VIN: Input Voltage (V)
VSW: Q1 ON-state Voltage Drop (V)
VOUT: Output Voltage (V)
The relation between current IL and voltage VL of coil L, which has self-inductance, can be calculated using the equation
below:
dI
VL L L (2)
dt VL(ON) VL(OFF)
dIL= dt dIL= dt
IL L L
From equation (2), it is clear that by applying additional voltage
to the inductor, the reverse-current direction increases by slope ILP
V/L.
Current flowing through the coil during tON can be calculated
using equation (1), (2), and by the following method; ILT IOUT ΔIL
represents current right before switching element Q1 turns ON,
ILP represents current right before switching element Q1 turns
OFF. ILT
ILP ILT
VSW VIN VOUT t ON
(3) t
L tON tOFF
The next step is to determine current flow in coil L when the
Fig. 3: Inductor Current Waveform
switching element is OFF.
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Inductor Calculation of Buck Converter Application Note
From Fig. 2, the coil voltage when Q1 in OFF- state is VL(OFF), can be calculated using the following method:
VL( OFF ) VD VOUT (4)
VD: Forward Voltage Drop across D1 (V)
VOUT: Output Voltage (V)
Using equations (2) and (4), the current flowing through coil L when Q1 is OFF is as follows:
ILP ILT
VD VOUT t OFF (5)
L
Current flow in the coil L is almost the same as the output current; consequently,
ILP ILT 2 IOUT (6)
From equations (3) and (6), ILP during Q1 ON-state is
ILP IOUT
VIN VSW VOUT t ON
(7)
2L
Using equations (5) and (6), ILP can be calculated during the period when the switching element is OFF:
ILP IOUT
VD VOUT t OFF (8)
2L
● On-Duty Calculation
On-duty D is the ratio of time the switching element is ON tON versus the switching oscillatory cycle TSW:
t ON t ON
D t ON fSW 1 t OFF f SW (9)
TSW t ON t OFF
Using (7), (8), and (9), the duty can be calculated using the expression below:
VD VOUT
D (10)
VIN VSW VD
By ignoring the voltage drop VSW of the switching element and voltage drop of the diode in equation (10), it is clear that on-
duty is fixed by the ratio of output voltage over input voltage:
VOUT
D (11)
VIN
● Maximum Coil Current Value
Use equations (9) and (10) to determine tON:
D VD VOUT
t ON (12)
f SW VIN VSW VD fSW
The equation below determines the maximum value of ILP by substituting (12) into (7):
ILP IOUT
VIN VSW VOUT VD VOUT
(13)
VIN VSW VD 2 L fSW
Equation (14) is used to determine the minimum value ILT by substituting (13) into (6):
ILT IOUT
VIN VSW VOUT VD VOUT
(14)
VIN VSW VD 2 L fSW
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Inductor Calculation of Buck Converter Application Note
Current-difference between max. and min. (ILP－ILT) is as follows:
ILP ILT VIN VSW VOUT VD VOUT (15)
VIN VSW VD L fSW
Equations (13) and (15) show that large inductance L and high switching frequency will reduce maximum current (ILP)and
current difference between max. and min. (ILP－ILT).
● Inductance L Value Calculation
Define the ratio of current-difference flowing in coil L (ILP-ILT) versus output current IOUT as current ripple-ratio “r”.
IL I I
r LP LT (16)
IOUT IOUT
Substitute (15) into (16):
r
VIN VSW VOUT VD VOUT
(17)
VIN VSW VD fSW L IOUT
Then, solve (17) for L to calculate the inductance value:
L
VIN VSW VOUT VD VOUT (H) (18-1)
VIN VSW VD fSW r IOUT
When the output voltage is high, the calculation can be simplified.
L
VIN VOUT VOUT (H) (18-2)
VIN fSW r IOUT
The inductor value can be increased to reduce the ripple-ratio; however, this will typically result in an inductor size that is
physically too big for practical use. Consequently, “r” is usually set between 0.2 and 0.5 for buck converters.
● Maximum Current Flow through the Coil
Maximum current flow through the coil can be calculated using the following equation:
r IOUT VOUT VIN VOUT
ILpeak IOUT (A) or IOUT (A) (19)
2 2 VIN fSW L
Current flowing through the coil is a combination of output current and ripple-current. When an abnormality occurs (ex.
output short) in a transient load condition and there will be a power surge due to the absence of soft-start feature, and it is
possible that the actual current flow through the inductor will exceed the maximum calculated current. Under transient
conditions, coil current may increase up to the IC’s switching current limit. Therefore, the safest approach is to select an
inductor with a saturation current rating that exceeds the switching current limit, and not the maximum coil current.
● Effective RMS current flowing through the coil
Effective current value of triangular wave can be calculated by the following method.
2 2
ILP ILT ILP ILT (20)
ILrms
3
By substituting equations (13) and (14) into (20), we arrive at the following:
1 VIN VSW VOUT VD VOUT
2
2
ILrms IOUT (A) (21)
12 VIN VSW VD L fSW
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Inductor Calculation of Buck Converter Application Note
● Example for Coil selection:
Determine the operating conditions of the buck converter:
- VIN = 12V Input Voltage
- VOUT = 3.3V Output Voltage
- IOUT = 2A Output Current
- r = 0.3 Output Current Ripple Ratio
- VSW = 0.30 ON State Voltage Drop of Switching Element Q1
- VD = 0.26 * Forward Voltage Drop of Free Wheel Diode
- fSW = 380 kHz Switching Frequency
* For synchronized rectifier type power conversion, specify ON-state voltage drop of lower side element Q2 (Fig. 4).
Calculate the inductance value of the coil by substituting the above parameters into equation (18-1) or (18-2).
As per equation (18-1), L
12 0.30 3.3 0.26 3.3 10.663 (µH)
12 0.30 0.26 380000 0.3 2
As per equation (18-2), L
12 3.3 3.3 10.493 (µH)
12 380000 0.3 2
Using equation (19), the maximum current flowing through the coil is as follows:
r IOUT 0.3 2
ILpeak IOUT 2 2.3 (A)
2 2
As per equation (21), the effective current value flowing through the coil is
22
ILrms
1
12 0.30 3.3 0.26 3.3
2.008 (A)
12 12 0.30 0.26 10.663 10 380000
6
Coil selection should be based on the results of the calculations. For this design, 10µH is chosen as the closest standard
inductance value. If the selected inductor value is different from the calculated result, adjust the current ripple value “r”
using equation (17), and substitute the revised value into equation (19) to recalculate the maximum current flowing through
the coil.
r
12 0.30 3.3 0.26 3.3 0.329
12 0.30 0.26 380000 10 10 6 2
0.329 2
ILpeak 2 2.33 (A)
2
Q1
OFF
VL
VIN L IO
Q2
VD CO RL VOUT
ON
Fig. 4: Basic circuit configuration of a buck converter using
synchronous rectification power conversion. In this
example, the upper switching element is in OFF-state.
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R1120A
