Sanjaya Maniktala, 2013
Turns Ratio and Primary Inductance
Calculator for LX7309 and IPS18
We will do this in several ways:
Example: Input 36V-57V, Output 12V 1A, f=150kHz using IPS18/LX7309
Duty cycle equation of Buck-Boost modified for Flyback:
refected output voltage (VOR) replaces output voltage (VO )
VOR = (ηVIN ) ×
D
1−D
Here η is estimated efficiency at the voltage VIN
Set min Input VINMIN
VINMIN = 36V
Set expected efficiency at VINMIN
ηMIN = 0.75
At this min voltage we must be
at max duty cycle limit of
0.66
controller and the reflected
VOR = 36V × 0.75 ×
= 52.4 V
1 − 0.66
output voltage must be VOR
For LX7309 just set DMAX = 0.44 instead
of 0.66 as for IPS18 here...rest remains
the same!
Set effective output (including
0.6V diode drop for nonsynchronous topology)
VO=12.6V
Turns ratio converts thereflected output voltage
to actual output
n=
NP VOR 52.4
=
=
= 4.16
NS VO 12.6
Turns ratio
required
In a Buck-Boost/Flyback, the center of ramp (“COR”) of
the current (on Secondary side in Flyback) is
ICOR ≡ Average Inductor Current = IL =
ICOR _SEC =
IO
(Buck-Boost)
1−D
IO
(Flyback)
1−D
So at minimum input condition we get
ICOR _SEC =
IO
1A
=
= 2.94 A
1 − D 1 − 0.66
Best operating condition (in CCM) is a current swing ΔI of about 0.3 to 0.4 at
minimum input (about 0.4 to 0.45 at nominal operating). Let us set to 0.35 here
∆I = 0.35 × ICOR
Copyright © 2013
Rev. 0.1, March 2013
Microsemi
Analog Mixed Signal Group
One Enterprise Aliso Viejo, CA 92656 USA
PRELIMINARY/ CONFIDENTIAL
Page 1
Sanjaya Maniktala, 2013
Turns Ratio and Primary Inductance
Calculator for LX7309 and IPS18
There are actually two conditions here, for the Primary and Secondary sides in a Flyback, but they are
related through turns ratio so we can use either the Secondary or Primary sides to do the calculations. Let
us use the Primary side here.
From turns ratio we know that
ICOR _SEC
2.94 A
= 0.247 A
n
4.16
Now all we have to do is to find Primary Inductance using V=Ldi/dt
∆IPRI = 0.35 ×
LPRI = VINMIN ×
= 0.35 ×
TON
D/f
= ( η× VINMIN ) ×
∆IPRI
∆IPRI
LPRI = ( 0.75 × 36V ) ×
0.66 / 150k
= 481µH
0.247
LPRI = VOR ×
(1 − DMAX ) / f
TOFF
= VOR ×
∆IPRI
∆IPRI
LPRI = VOR ×
(1 − 0.66 ) / 150k = 481µH
TOFF
= 52.4 ×
∆IPRI
0.247
One-step Solution (using equation for Buck-Boost):
LSEC =
VO
2
× (1 − D ) µH
IO × r × f
(12 + 0.6)
2
× (1 − 0.66 ) = 27.7µH
1A × 0.35 × 150k
Reflecting this to get Primary Side Inductance
LPRI = n2 × LSEC = 4.162 × 27.7µH = 480µH
Copyright © 2013
Rev. 0.1, March 2013
Microsemi
Analog Mixed Signal Group
One Enterprise Aliso Viejo, CA 92656 USA
PRELIMINARY/ CONFIDENTIAL
Page 2
Sanjaya Maniktala, 2013
Currents
Turns Ratio and Primary Inductance
Calculator for LX7309 and IPS18
I COR_SEC
ICOR _SEC = n × ICOR _ PRI
I COR_PRI
where n is turns ratio = NP/NS
Primary Secondary
IO
ICOR _SEC =
1−D
Time
Copyright © 2013
Rev. 0.1, March 2013
Microsemi
Analog Mixed Signal Group
One Enterprise Aliso Viejo, CA 92656 USA
PRELIMINARY/ CONFIDENTIAL
Page 3