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A4402
Constant On-Time Buck Converter
With Integrated Linear Regulator
Application Information
V OUT + V f + ( R S × I OUT )
D =
V OUT + V f + ( R S × I OUT ) (16)
V IN + V f + ( R S × I OUT ) – ( R DS × I OUT )
t ON =
( t ON – 60 × 10 –9 ) × V IN
SwitcherOn-TimeandSwitchingFrequency Inorderfor
the switcher to maintain regulation, the energy that is transferred
to the inductor during the on-time must be transferred to the out-
put capacitor during the off-time. This relationship must be main-
tained for stable operation and governs the fundamental operation
of a switching regulator. Each component along the current path
changes the voltage across the inductor and therefore the energy
that is transferred during each cycle. Summing the voltage from
V IN to V OUT during each cycle gives a relationship of the voltage
across the inductor during the on-time and during the off-time.
These terms are represented as V ON and V OFF .
Given a target operating frequency, represent t ON as:
t ON = T × D (10)
where T equals 1 / f SW , and D is the duty cycle.
Duty cycle can be represented as the voltage across the inductor
during the off-time, divided by the total voltage of the off-time
and on-time:
D = V OFF / ( V OFF + V ON ) (11)
Next, determine the voltage drops during the on cycle and the
off cycle. Figure 2 shows the current path during the on-time and
off-time.
Creating voltage summation during each cycle will give equa-
tions to represent V ON and V OFF :
V ON = V IN – V OUT – ( I OUT × R L ) – ( R DS × I OUT ) (12)
V OUT + V f + ( R S × I OUT ) + (15)
V IN – V OUT – ( R DS × I OUT )
Further simplification and grouping of terms yields:
D =
Substitute this simplified expression for duty cycle back into
equation 10. The following formula results in the on-time, given
a target switching frequency:
1 V OUT + V f + ( R S × I OUT ) (17)
f SW V IN + V f + ( R S × I OUT ) – ( R DS × I OUT )
The formulas above describe how t ON changes based on input and
load conditions. Because load changes are minimal, and the out-
put voltage is fixed, the dominant factor that effects on-time is the
input voltage. The converter is able to maintain a constant period
over a varying supply voltage because the on-time is proportional
to the input voltage. The current into the TON terminal is derived
from a resistor tied to VIN1, which sets the on-time proportional
to the supply voltage. Selecting the resistor value, based on the
t ON calculated above, is done using the following formula:
R TON = (18)
3.12 × 10 –12
After the resistor is selected and a suitable t ON is found, it must
be demonstrated that t ON does not, under worst-case conditions,
V OFF = V OUT + ( I OUT × R L ) + V f + ( R S × I OUT )
(13)
V OUT × ( I OUT × R L ) + V f + ( R S × I OUT )
D =
LX
Now substituting V ON and V OFF intoequation11givesacom-
plete formula for duty cycle as it relates to the voltage across the
inductor:
V OUT + ( I OUT × R L ) + V f + ( R S × I OUT ) + (14)
V IN – V OUT – ( I OUT × R L ) – ( R DS + I OUT )
VIN1
A4402
V RL Current path (on-cycle)
L1
V f V RL
Current path
V RS (off-cycle)
R LOAD
The effects of the voltage drop across the inductor resistance and
trace resistance do have an effect on the switching frequency.
However, the frequency variation due to these factors is small
and is covered in the variation of the switcher period, T SW , which
is ±25% of the target. Removing these current-dependent terms
simplifies the equation:
Star Ground
Figure 2. Current limiting during overload
Allegro MicroSystems, LLC
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
11
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