Datasheet

LTC3789
23
3789fa
applicaTions inForMaTion
Design Example
V
IN
= 5V to 18V
V
OUT
= 12V
I
OUT(MAX)
= 5A
f = 400kHz
Maximum ambient temperature = 60°C
Set the frequency at 400kHz by applying 1.2V on the FREQ
pin (see Figure 7). The 10µA current flowing out of the
FREQ pin will give 1.2V across a 120k resistor to GND. The
inductance value is chosen first based on a 30% ripple cur-
rent assumption. In the buck region, the ripple current is:
∆I
L,BUCK
=
V
OUT
f • L
• 1–
V
OUT
V
IN






I
RIPPLE,BUCK
=
∆I
L,BUCK
• 100
I
OUT
%
The highest value of ripple current occurs at the maximum
input voltage. In the boost region, the ripple current is:
∆I
L,BOOST
=
V
IN
f • L
• 1 –
V
IN
V
OUT






I
RIPPLE,BOOST
=
∆I
L,BOOST
• 100
I
IN
%
The highest value of ripple current occurs at V
IN
= V
OUT
/2.
A 6.8µH inductor will produce 11% ripple in the boost region
(V
IN
= 6V) and 29% ripple in the buck region (V
IN
= 18V).
The R
SENSE
resistor value can be calculated by using the
maximum current sense voltage specification with some
accommodation for tolerances.
R
SENSE
=
2 • 140mV • V
IN(MIN)
2 • I
OUT(MAX,BOOST)
• V
OUT
+ ∆I
L,BOOST
• V
IN(MIN)
Select an R
SENSE
of 10mΩ.
Output voltage is 12V. Select R1 as 20k. R2 is:
R2 =
V
OUT
• R1
0.8
– R1
Select R2 as 280k. Both R1 and R2 should have a toler-
ance of no more than 1%.
Selecting MOSFET Switches
The MOSFETs are selected based on voltage rating and
R
DS(ON)
value. It is important to ensure that the part is
specified for operation with the available gate voltage am-
plitude. In this case, the amplitude is 5.5V and MOSFETs
with an R
DS(ON)
value specified at V
GS
= 4.5V can be used.
Select QA and QB. With 18V maximum input voltage MOS-
FETs with a rating of at least 30V are used. As we do not yet
know the actual thermal resistance (circuit board design and
airflow have a major impact) we assume that the MOSFET
thermal resistance from junction to ambient is 50°C/W.
If we design for a maximum junction temperature, T
J(MAX)
= 125°C, the maximum R
DS(ON)
value can be calculated.
First, calculate the maximum power dissipation:
P
D(MAX)
=
T
J(MAX)
− T
A(MAX)
R
(j−a)






P
D(MAX)
=
(125 − 60)
50
= 1.3W
The maximum dissipation in QA occurs at minimum input
voltage when the circuit operates in the boost region and
QA is on continuously. The input current is then:
V
OUT
• I
OUT(MAX)
V
IN(MIN)
, OR 12A
We calculate a maximum value for R
DS(ON)
:
R
DS(ON)
(125°C) <
P
D(MAX)
I
IN(MAX)
2
R
DS(ON)
(125°C) <
1.3W
(12A)
2
= 0.009Ω