BD6066GU,BU6066EKN
Technical Note
Ton=(Iave × (1-VIN/VOUT) × (1/fsw) × (L/VIN) × 2)
5) Over Current Limit
Over current flows the current detection resistor that is connected to switching transistor source and between GND,
SENSP pin voltage turns more than detection voltage, over current protection is operating and it is prevented from flowing
more than detection current by reducing ON duty of switching Tr without stopping boost.
As over current detector of BD6066GU/EKN is detected peak current, current more than over current setting value does
not flow.
And, over current value can decide freely by changing over current detection voltage.
<Derivation sequence of detection resistor>
Detection resistor =Over current detection voltage / Over current setting value
TYP value of over current detection voltage is 100mV, MIN = 70mV and MAX = 130mV and after the current value which
was necessary for the normal operation was decided, detection resistor is derived by using MIN value of over current
detection value.
For example, detection resistor when necessary current value was set at 1A is given as shown below.
Detection resistor =70mV / 1A = 70m ?
MAX current dispersion of this detection resistor value is
MAX current = 130mV / 70m ? = 1.86A
<The estimate of the current value which need for the normal operation >
As over current detector of BD6066GU/EKN is detected the peak current, it have to estimate peak current to flow to the
coil by operating condition.
In case of, Supply voltage of coil = VIN
Reactance value of coil = L
Switching frequency = fsw MIN=0.8MHz, Typ=1MHz, MAX=1.2MHz
Output voltage = VOUT
Total LED current = IOUT
Average current of coil = Iave
Peak current of coil = Ipeak
Efficiency = eff (Please set up having margin, it refers to data on p.4.)
ON time of switching transistor = Ton
Ipeak = (VIN / L) × (1 / fsw) × (1-(VIN / VOUT))
Iave=(VOUT × IOUT / VIN) / eff
1/2
Each current is calculated.
As peak current varies according to whether there is the direct current superposed, the next is decided.
(1-VIN/VOUT) × (1/fsw) < Ton → peak current = Ipeak /2 + Iave
(1-VIN/VOUT) × (1/fsw) > Ton → peak current = Ipeak
(Example 1)
In case of, VIN=6.5V, L=4.7μH, fsw=1MHz, VOUT=39V, IOUT=80mA, Efficiency=85%
Ipeak = (6.5V / 4.7μH) × (1 / 1MHz) × (1-(6.5V / 39V)) =1.08A
Ton = (0.61A × (1-6.0V / 39V) × (1 / 1MHz) × ( 4.7μH /6.0V) × 2)
Iave = (39V × 80mA / 6.0V) / 85% = 0.61A
1/2
= 0.90μs
(1-VIN/VOUT) × (1/fsw)=0.85μs < Ton
Peak current = 1.08A/2+0.61A = 1.15A
(Example 2)
In case of, VIN=12.0V, L=4.7μH, fsw=1MHz, VOUT=39V, IOUT=80mA, Efficiency=85%
Ipeak = (12.0V / 4.7μH) × (1 / 1MHz) × (1-(12V / 39V)) =1.77A
Ton = (0.31A × (1-12 V / 39V) × (1 / 1MHz) × ( 4.7μH /12 V) × 2)
Iave = (39V × 80mA / 12.0V) / 85% = 0.31A
(1-VIN/VOUT) × (1/fsw)=0.69μs > Ton
Peak current = 12V/4.7μH × 0.41μs = 1.05A
1/2
= 0.41μs
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8/15
2011.06 - Rev.B
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