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0 . 45 = I RAMP × R SLC × D + I L _ pk Limit × R CS
V OUT ? V IN MIN ) × R CS
R SLC =
Eq. (13)
L = IN
= 43 μ H
12 V × 0 . 56 × 2 μ s
R CS =
Eq. (13a)
I in _ PP =
V IN _ nom × D nom × T 12 v × 0 . 56 × 2 us
R CS =
( 28 v ? 8 v ) × ( 0 . 50 ) + 1 . 9 A = 179 m Ω
( 28 ? 8 ) × 150 m Ω
R SLC = = 511 Ω
( I IN _ RMS _ max ) 2 ? ( I IN 12 PP )
Eq. (13b)
I IN _ AVE _ max =
Micrel, Inc.
(see the current waveforms in Figure 5).
It can be difficult to find large inductor values with high
saturation currents in a surface mount package. Due to
this, the percentage of the ripple current may be limited
by the available inductor. It is recommended to operate
in the continuous conduction mode. The selection of L
described here is for continuous conduction mode.
V × D × T
I in _ PP
Using the nominal values, we get:
L =
0 . 31 A
Select the next higher standard inductor value of 47μH.
Going back and calculating the actual ripple current
gives:
= = 0 . 29 A PP
L 47 uh
The average input current is different than the RMS input
current because of the ripple current. If the ripple current
is low, then the average input current nearly equals the
RMS input current. In the case where the average input
current is different than the RMS, Equation 10 shows the
following:
2
_
MIC3230/1/2
Eq. (14a)
To calculate the value of the slope compensation resistance,
R SLC , we can use Equation 5:
MAX
L × 250 μ A × F SW
First we must calculate R CS , which is given below in
Equation 15:
Eq. (15) 0 . 45
( VOUT MAX ? VIN MIN ) × D max + I L _ pk Limit
L × F SW
Therefore;
0 . 45
47 μ H × 500 kHz
Using a standard value 150m ? resistor for R CS , we obtain
the following for R SLC :
47 μ H × 250 μ A × 500 kHz
Use the next higher standard value if this not a standard
value. In this example 511 ? is a standard value.
Check: Because we must use a standard value for Rcs and
R SLC; I L _ pk Limit may be set at a different level (if the calculated
value isn’t a standard value) and we must calculate the
actual I L _ pk Limit value (remember I L _ pk Limit is the same as
I IN _ AVE _ max =
( 1 . 64 ) 2 ? ( 0 . 29 ) 2 / 12 ≈ 1 . 64 A
I in _ pk Limit ).
I in _ pk Limit =
I in _ actual Limit =
= 2 . 34 A
Eq. (13c) P INDUCTOR = I in _ RMS _ max × DCR
The Maximum Peak input current I L_PK can found using
equation 11:
I L _ PK _ max = I IN _ AVE _ max + 0 . 5 × I L _ PP _ max = 1 . 78 A
The saturation current (I SAT ) at the highest operating
temperature of the inductor must be rated higher than
this.
The power dissipated in the inductor is:
2
Current Limit and Slope Compensation
Having calculated the I L_pk above, We can set the current
limit 20% above this maximum value:
I L _ pk Limit = 1 . 2 × 1 . 6 A = 1 . 9 A
The internal current limit comparator reference is set at
0.45V, therefore when V IS _ PIN = 0 . 45 , the IC enters
current limit.
Eq. (14) 0 . 45 = ( V A PK + Vcs PK )
Where V A PK is the peak of the V A waveform and
Vcs PK is the peak of the Vcs waveform
Rearranging Equation 14a to solve for I L _ pk Limit :
( 0 .45 ? I RAMP × R SLC × D )
R CS
( 0 . 45 ? 250 ua × 511 × 0 . 75 )
. 150
This is higher than the initial 1 . 2 × I L _ PK _ max = 1 . 9 A limit
because we have to use standard values for R CS and for
R SLC . If I in _ actual Limit is too high than use a higher value for
R CS . The calculated value of R CS for a 1.9A current limit was
179m ? . In this example, we have chosen a lower value
which results in a higher current limit. If we use a higher
standard value the current limit will have a lower value. The
designer does not have the same choices for small valued
resistors as with larger valued resistors. The choices differ
from resistor manufacturers. If too large a current sense
resistor is selected, the maximum output power may not be
able to be achieved at low input line voltage levels. Make
sure the inductor will not saturate at the actual current limit
I in _ actual Limit .
Perform a check at I IN =2.34Apk.
V IS _ PIN = 250 μ A × ( 0 . 78 ) × 511 Ω + 2 . 34 A × 150 m Ω = 0 . 45 V
March 2011
14
M9999-030311-D
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