The NX2119 controller IC is a synchronous Buck controller IC designed for step down DC to DC converter
applications. It is optimized to convert bus voltages from
2V to 25V to outputs as low as 0.8V voltage. The
NX2119 operates at fixed 300kHz, while NX2119A operates at fixed 600kHz, making it ideal for applications
requiring ceramic output capacitors. The NX2119 employs fixed loss-less current limiting by sensing the Rdson
of synchronous MOSFET followed by latch out feature.
Feedback under voltage triggers Hiccup.
Other features of the device are: 5V gate drive, Adaptive
deadband control, Internal digital soft start, Vcc
undervoltage lock out and shutdown capability via the
comp pin.
Vin
HI=SD
+5V
M3
C7
27pF
C3
1uF
37.4k
C2
2.2nF
C4
100uF
R4
7
6
5
Vcc
COMP
FB
R5
10
NX2119
Gnd
1
BST
3
D1
MBR0530T1
Hdrv
SW
Ldrv
FEATURES
n Bus voltage operation from 2V to 25V
n Fixed 300kHz and 600kHz
n Internal Digital Soft Start Function
n Prebias Startup
n Less than 50 nS adaptive deadband
n Current limit triggers latch out by sensing Rdson of
Synchronous MOSFET
n No negative spike at Vout during startup and
shutdown
n Pb-free and RoHS compliant
APPLICATIONS
n Graphic Card on board converters
n Memory Vddq Supply
n On board DC to DC such as
12V to 3.3V, 2.5V or 1.8V
n ADSL Modem
TYPICAL APPLICATION
L2 1uH
C5
1uF
C6
0.1uF
2
8
4
M1
L1 1.5uH
M2
C1
4.7nF
R1
4k
10k
R2
Cin
280uF
18mohm
Vout
Co
2 x (1500uF,13mohm)
+1.8V 9A
Rev.3.2
04/10/08
R3
8k
Figure1 - Typical application of 2119
ORDERING INFORMATION
Device Temperature Package Frequency Pb-Free
NX2119CSTR 0 to 70oC SOIC - 8L 300kHz Yes
NX2119ACSTR 0 to 70oC SOIC - 8L600kHz Yes
NX2119ACUTR 0 to 70oC MSOP - 8L600kHz Yes
1
Page 2
NX2119/2119A
130CW
θ≈/
216CW
θ≈/
ABSOLUTE MAXIMUM RATINGS
VCC to GND & BST to SW voltage .................... -0.3V to 6.5V
BST to GND Voltage ........................................ -0.3V to 35V
SW to GND ...................................................... -2V to 35V
All other pins .................................................... -0.3V to VCC+0.3V or 6.5V
Storage Temperature Range ............................... -65oC to 150oC
Operating Junction Temperature Range ............... -40oC to 125oC
CAUTION: Stresses above those listed in "ABSOLUTE MAXIMUM RATINGS", may cause permanent damage to
the device. This is a stress only rating and operation of the device at these or any other conditions above those
indicated in the operational sections of this specification is not implied.
PACKAGE INFORMATION
8-LEAD PLASTIC SOIC(S) 8-LEAD PLASTIC MSOP
BST
HDrv
Gnd
LDrv
JA
1
2
3
4
o
8
7
6
SW
Comp
Fb
BST
HDrv
Gnd
LDrv
5
Vcc
JA
1
2
3
4
o
8
SW
7
Comp
6
Fb
5
Vcc
ELECTRICAL SPECIFICATIONS
Unless otherwise specified, these specifications apply over Vcc = 5V, and TA= 0 to 70oC. Typical values refer to Ta
= 25oC. Low duty cycle pulse testing is used which keeps junction and case temperatures equal to the ambient
temperature.
PARAMETERSYMTest ConditionMinTYPMAXUnits
Reference Voltage
Ref VoltageV
Ref Voltage line regulation0.2%
Supply Voltage(Vcc)
VCC Voltage RangeV
VCC Supply Current (Static)ICC (Static) Outputs not switching 3mA
VCC Supply Current
(Dynamic)
Supply Voltage(V
V
Supply Current (Static)I
BST
V
Supply Current
BST
(Dynamic)
BST
)
REF
CC
I
CC
(Dynamic)
(Static) Outputs not switching0.2mA
BST
I
BST
(Dynamic)
C
=3300pF FS=300kHzTBDmA
LOAD
C
=3300pF FS=300kHzTBDmA
LOAD
4.5
0.8
5
5.5
V
V
Rev.3.2
04/10/08
2
Page 3
NX2119/2119A
Soft Start
OCP
Ldrv going Low to Hdrv going
PARAMETERSYMTEST CONDITIONMINTYPMAXUNITS
Under Voltage Lockout
VCC-ThresholdVCC_UVLO VCC Rising
3.8
VCC-HysteresisVCC_HystVCC Falling0.2V
Oscillator (Rt)
FrequencyF
S
2119300kHz
2119A600kHz
Ramp-Amplitude VoltageV
Max Duty Cycle
Min Duty Cycle
Rise TimeTLdrv(Rise)10% to 90%50ns
Fall TimeTLdrv(Fall)90% to 10%50ns
Deadband TimeTdead(H to L)SW going Low to Ldrv going
30
High, 10% to 10%
OCP voltage320mV
Rev.3.2
04/10/08
ns
3
Page 4
PIN DESCRIPTIONS
PIN # PIN SYMBOL PIN DESCRIPTION
Power supply voltage. A high freq 1uF ceramic capacitor is placed as close as
5
1
VCC
BST
possible to and connected to this pin and ground pin.
This pin supplies voltage to high side FET driver. A high freq 0.1uF ceramic capacitor is placed as close as possible to and connected to these pins and respected
SW pins.
NX2119/2119A
3
6
7
8
2
4
GND
FB
COMP
SW
HDRV
LDRV
Ground pin.
This pin is the error amplifier inverting input. This pin is connected via resistor divider
to the output of the switching regulator to set the output DC voltage. When FB pin
voltage is lower than 0.6V, hiccup circuit starts to recycle the soft start circuit after
2048 switching cycles.
This pin is the output of the error amplifier and together with FB pin is used to
compensate the voltage control feedback loop. This pin is also used as a shut down
pin. When this pin is pulled below 0.3V, both drivers are turned off and internal soft
start is reset.
This pin is connected to source of high side FET and provides return path for the
high side driver. It is also used to hold the low side driver low until this pin is brought
low by the action of high side turning off. LDRV can only go high if SW is below 1V
threshold .
High side gate driver output.
Low side gate driver output.
Rev.3.2
04/10/08
4
Page 5
BLOCK DIAGRAM
NX2119/2119A
VCC
FB
COMP
GND
Bias
Generator
COMP
0.3V
START
Digital
start Up
0.8V
1.25V
ramp
START
0.8V
OSC
UVLO
0.6V
CLAMP
S
R
1.3V
CLAMP
FB
0.6V
POR
Q
OC
START
latch out
PWM
Hiccup Logic
OC
BST
HDRV
SW
Control
Logic
VCC
LDRV
320mV
OCP
comparator
Rev.3.2
04/10/08
Figure 2 - Simplified block diagram of the NX2119
5
Page 6
NX2119/2119A
RIPPLEINS
1
IVF
0.39A5V300kHz
=2.56A
SOUT
==Ω
ESR=7.8m
ERIPPLE
12m2.56A
APPLICATION INFORMATION
Symbol Used In Application Information:
VIN - Input voltage
VOUT - Output voltage
IOUT - Output current
DVRIPPLE - Output voltage ripple
FS - Working frequency
DIRIPPLE - Inductor current ripple
Design Example
The following is typical application for NX2119, the
schematic is figure 1.
VIN = 5V
VOUT=1.8V
FS=300kHz
IOUT=9A
DVRIPPLE <=20mV DVDROOP<=100mV @ 9A step
Output Inductor Selection
The selection of inductor value is based on inductor ripple current, power rating, working frequency and
efficiency. Larger inductor value normally means smaller
ripple current. However if the inductance is chosen too
large, it brings slow response and lower efficiency. Usually the ripple current ranges from 20% to 40% of the
output current. This is a design freedom which can be
decided by design engineer according to various application requirements. The inductor value can be calculated by using the following equations:
V-VV
OUT
∆
INOUT OUT
××
×
5V-1.8V1.8V1
OUT
OUT
××
×
...(1)
L=
I=kI
RIPPLEOUTPUT
where k is between 0.2 to 0.4.
Select k=0.3, then
L=
L=1.4uH
Choose inductor from COILCRAFT DO5010P152HC with L=1.5uH is a good choice.
Current Ripple is recalculated as
V-VV
∆××
I=
RIPPLE
INOUT OUT
LVF
OUTINS
5V-1.8V1.8v1
1.5uH5v300kHz
××=
1
...(2)
Output Capacitor Selection
Output capacitor is basically decided by the
amount of the output voltage ripple allowed during steady
state(DC) load condition as well as specification for the
load transient. The optimum design may require a couple
of iterations to satisfy both condition.
Based on DC Load Condition
The amount of voltage ripple during the DC load
condition is determined by equation(3).
∆
I
∆=×∆+
VESRI
RIPPLERIPPLE
Where ESR is the output capacitors' equivalent
series resistance,C
Typically when large value capacitors are selected
such as Aluminum Electrolytic,POSCAP and OSCON
types are used, the amount of the output voltage ripple
is dominated by the first term in equation(3) and the
second term can be neglected.
For this example, POSCAP are chosen as output
capacitors, the ESR and inductor current typically determines the output voltage ripple.
∆
tiple capacitors in parallel are better than a big capacitor. For example, for 20mV output ripple, POSCAP
2R5TPE220MC with 12mΩ are chosen.
integer. Choose N =2.
desire
If low ESR is required, for most applications, mul-
ESRI
N
=
Number of Capacitor is calculated as
Ω×
=
N
N =1.5
The number of capacitor has to be round up to a
If ceramic capacitors are chosen as output ca
20mV
is the value of output capacitors.
OUT
V
RIPPLE
∆
I2.56A
RIPPLE
20mV
×∆
V
∆
RIPPLE
RIPPLE
××
8FC
...(5)
...(3)
...(4)
Rev.3.2
04/10/08
6
Page 7
NX2119/2119A
8300kHz100uF
DROOPTRAN
2
∆=×∆+×τ
OUTcrit
ESRCifLL
OUTOUTEEOUT
crit
LL
2
=+×τ
EEcrit
ESRCifLL
0.56H
=µ
12m220F4.86us
2
1.7×∆=+×τ
pacitors, both terms in equation (3) need to be evaluated to determine the overall ripple. Usually when this
type of capacitors are selected, the amount of capacitance per single unit is not sufficient to meet the transient specification, which results in parallel configuration of multiple capacitors .
For example, one 100uF, X5R ceramic capacitor
with 2mΩESR is used. The amount of output ripple is
∆=Ω×+
V2m2.56A
RIPPLE
15mV
=
Although this meets DC ripple spec, however it
needs to be studied for transient requirement.
Based On Transient Requirement
Typically, the output voltage droop during transient
is specified as:
V<V∆∆ @ step load DI
During the transient, the voltage droop during the
transient is composed of two sections. One Section is
dependent on the ESR of capacitor, the other section is
a function of the inductor, output capacitance as well as
input, output voltage. For example, for the overshoot,
when load from high load to light load with a DI
transient load, if assuming the bandwidth of system is
high enough, the overshoot can be estimated as the following equation.
VESRI
overshootstep
where τ is the a function of capacitor, etc.
0ifLL
LI
×∆
τ=
V
OUT
≤
crit
step
−×≥
where
ESRCVESRCV
××××
==
II
∆∆
stepstep
L
crit
where ESRE and CE represents ESR and capacitance of each capacitor if multiple capacitors are used
in parallel.
The above equation shows that if the selected output inductor is smaller than the critical inductance, the
voltage droop or overshoot is only dependent on the ESR
2.56A
××
STEP
V
OUT
2LC
××
OUT
...(6)
...(7)
STEP
...(8)
of output capacitor. For low frequency capacitor such
as electrolytic capacitor, the product of ESR and ca-
pacitance is high and
≤ is true. In that case, the
transient spec is dependent on the ESR of capacitor.
In most cases, the output capacitors are multiple
capacitors in parallel. The number of capacitors can be
calculated by the following
ESRI
×∆
N
Estep
V2LCV
∆×××∆
tranEtran
V
OUT
...(9)
where
0ifLL
LI
×∆
τ=
V
OUT
≤
crit
step
−×≥
...(10)
For example, assume voltage droop during transient is 100mV for 9A load step.
If the POSCAP 2R5TPE220MC(220uF, 12mΩ ) is
used, the critical inductance is given as
ESRCV
××
EEOUT
==
I
∆
step
Ω×µ×
L
crit
12m220F1.9V
9A
The selected inductor is 1.5uH which is bigger than
critical inductance. In that case, the output voltage transient not only dependent on the ESR, but also capacitance.
number of capacitors is
LI
×∆
step
τ=−×
V
OUT
1.5H9A
µ×
=−Ω×µ=
1.8V
ESRI
N
12m9A
Ω×
=+
100mV
21.5H220F100mV
×µ×µ×
ESRC
EE
Estep
V2LCV
∆×××∆
tranEtran
1.8V
V
OUT
(4.86us)
×
2
=
The number of capacitors has to satisfied both ripple
and transient requirement. Overall, we can choose N=2.
Rev.3.2
04/10/08
7
Page 8
NX2119/2119A
F ...(11)
F ...(12)
F ...(13)
F ...(14)
OUT minin1
V1gZZ/R
f
OUT in
Z
VZ
−
It should be considered that the proposed equation is based on ideal case, in reality, the droop or overshoot is typically more than the calculation. The equation gives a good start. For more margin, more capacitors have to be chosen after the test. Typically, for high
frequency capacitor such as high quality POSCAP especially ceramic capacitor, 20% to 100% (for ceramic)
more capacitors have to be chosen since the ESR of
capacitors is so low that the PCB parasitic can affect
the results tremendously. More capacitors have to be
selected to compensate these parasitic parameters.
Compensator Design
Due to the double pole generated by LC filter of the
power stage, the power system has 180o phase shift ,
and therefore, is unstable by itself. In order to achieve
accurate output voltage and fast transient
response,compensator is employed to provide highest
possible bandwidth and enough phase margin.Ideally,the
Bode plot of the closed loop system has crossover frequency between1/10 and 1/5 of the switching frequency,
phase margin greater than 50o and the gain crossing
0dB with -20dB/decade. Power stage output capacitors
usually decide the compensator type. If electrolytic
capacitors are chosen as output capacitors, type II compensator can be used to compensate the system, because the zero caused by output capacitor ESR is lower
than crossover frequency. Otherwise type III compensator should be chosen.
1
2RC
×π××
42
1
2(RR)C
×π×+×
233
1
2RC
×π××
33
1
CC
×
2R
×π××
4
CC
12
+
12
=
Z1
=
Z2
=
P1
=
P2
where FZ1,FZ2,FP1 and FP2 are poles and zeros in
the compensator. Their locations are shown in figure 4.
The transfer function of type III compensator for
transconductance amplifier is given by:
V1gZ
emf
=
−×
+×+
For the voltage amplifier, the transfer function of
compensator is
V
e
=
To achieve the same effect as voltage amplifier,
the compensator of transconductance amplifier must
satisfy this condition: R4>>2/gm. And it would be desirable if R1||R2||R3>>1/gm can be met at the same time.
Zin
Vout
Zf
C1
A. Type III compensator design
For low ESR output capacitors, typically such as
Sanyo oscap and poscap, the frequency of ESR zero
caused by output capacitors is higher than the crossover frequency. In this case, it is necessary to compensate the system with type III compensator. The following figures and equations show how to realize the type III
compensator by transconductance amplifier.
Rev.3.2
04/10/08
R3
C2
R4
R2
C3
Fb
gm
Ve
R1
Vref
Figure 3 - Type III compensator using
transconductance amplifier
8
Page 9
NX2119/2119A
21.5uH440uF
26m440uF
==Ω
R=8k
=(-)
210k6.2kHz60.3kHz
=2.3nF
=440uF
20.756.2kHz16.9k
×π×××Ω
216.9k150kHz
260.3kHz2.2nF
Case 1: FLC<FO<F
power stage
LC
F
Gain(db)
loop gain
compensator
F
F
Z1 Z2
ESR
40dB/decade
ESR
F
20dB/decade
F
O
P1
F
Choose R
3. Set zero FZ2 = FLC and Fp1 =F
4. Calculate R
=8kΩ.
1
ESR
and C3 with the crossover
4
.
frequency at 1/10~ 1/5 of the switching frequency. Set
FO=30kHz.
C=(-)
R=C
=16.9k
111
3
2RFF
×π×
×π×Ω
V2FL
OSCO
4out
VC
1.5V230kHz1.5uH
5V2.2nF
×
2z2p1
111
in3
×
×π××
××
×π××
××
Ω
Choose C3=2.2nF, R4=16.9kΩ.
5. Calculate C2 with zero Fz1 at 75% of the LC
F
P2
double pole by equation (11).
Figure 4 - Bode plot of Type III compensator
Design example for type III compensator are in
order. The crossover frequency has to be selected as
FLC<FO<F
1.Calculate the location of LC double pole F
and ESR zero F
and FO<=1/10~1/5F
ESR,
.
ESR
F
=
LC
2LC
×π××
=
s.
LC
1
OUTOUT
1
×π××
6.2kHz
=
F
=
ESR
2ESRC
×π××
=
×π×Ω×
60.3kHz
=
1
OUT
1
2. Set R2 equal to 10kΩ.
×
RV
2REF
1
V-V1.8V-0.8V
OUT REF
Ω×
10k0.8V
1
2FR
×π××
Z14
1
C
=
2
=
2nF
=
Choose C2=2.2nF.
6. Calculate C1 by equation (14) with pole Fp2 at
half the switching frequency.
C
=
1
=
63pF
=
1
2RF
×π××
4P2
1
×π×Ω×
Choose C1=68pF.
7. Calculate R3 by equation (13).
1
2FC
×π××
P13
1
×π××
R
=
3
=
1.2k
=Ω
Choose R3=1.2kΩ.
Rev.3.2
04/10/08
9
Page 10
NX2119/2119A
21.5uH3000uF
26.5m3000uF
==Ω
R=8k
=(-)
210k2.3kHz8.2kHz
=4.76nF
28.2kHz4.7nF
1.5V230kHz1.5uH10k4k
5V6.5m10k4k
×π××Ω×Ω
ΩΩ+Ω
20.752.3kHz37.4k
×π×××Ω
237.4k150kHz
Case 2: FLC<F
power stage
Gain(db)
loop gain
compensator
F
Z1Z2
ESR<FO
LC
F
ESR
F
F
40dB/decade
F
F
P1
O
20dB/decade
F
P2
2. Set R2 equal to 10kΩ.
RV
2REF
1
V-V1.8V-0.8V
OUT REF
10k0.8V
×
Choose R1=8.06kΩ.
3. Set zero FZ2 = FLC and Fp1 =F
4. Calculate C3 .
C=(-)
111
3
2RFF
×π×
×π×Ω
×
2z2p1
111
×
Choose C3=4.7nF.
5. Calculate R3 .
1
2FC
×π××
P13
1
×π××
R
=
3
=
4.1k
=Ω
Choose R3 =4kΩ.
6. Calculate R4 with FO=30kHz.
Ω×
ESR
.
Figure 5 - Bode plot of Type III compensator
(FLC<F
ESR<FO
)
If electrolytic capacitors are used as output
capacitors, typical design example of type III
compensator in which the crossover frequency is
selected as FLC<F
and FO<=1/10~1/5Fs is shown
ESR<FO
as the following steps. Here two SANYO MV-WG1500
with 13 mΩ is chosen as output capacitor.
1. Calculate the location of LC double pole F
and ESR zero F
F
=
LC
=
.
ESR
1
2LC
×π××
OUTOUT
1
LC
×π××
2.3kHz
=
F
=
ESR
2ESRC
=
8.2kHz
=
1
×π××
OUT
1
×π×Ω×
R=
=
=37.3k
V2FLRR
4
VESRRR
×π×××
OSCO23
××
in23
+
××
Ω
Choose R4=37.4kΩ.
7. Calculate C2 with zero Fz1 at 75% of the LC
double pole by equation (11).
1
2FR
×π××
Z14
1
C
=
2
=
2.4nF
=
Choose C2=2.2nF.
8. Calculate C1 by equation (14) with pole Fp2 at
half the switching frequency.
C
=
1
=
28pF
=
1
2RF
×π××
4P2
1
×π×Ω×
Choose C1=27pF.
Rev.3.2
04/10/08
10
Page 11
B. Type II compensator design
Gain=gR ... (15)
F= ... (16)
F ... (17)
21.5uH3000uF
26.5m3000uF
==Ω
R=800
If the electrolytic capacitors are chosen as power
stage output capacitors, usually the Type II compensator can be used to compensate the system.
Type II compensator can be realized by simple RC
circuit without feedback as shown in figure 6. R3 and C
introduce a zero to cancel the double pole effect. C
introduces a pole to suppress the switching noise. The
following equations show the compensator pole zero location and constant gain.
R
1
××
m3
R+R
12
1
2RC
×π××
31
1
2RC
×π××
32
power stage
40dB/decade
20dB/decade
z
≈
p
Gain(db)
loop gain
NX2119/2119A
Vout
R2
Fb
1
2
R1
Vref
Figure 7 - Type II compensator with
transconductance amplifier
For this type of compensator, FO has to satisfy
FLC<F
tor design. Input voltage is 5V, output voltage is 1.8V,
output inductor is 1.5uH, output capacitors are two
1500uF with 13mΩ electrolytic capacitors.
and ESR zero F
<<FO<=1/10~1/5F
ESR
s.
The following is parameters for type II compensa-
1.Calculate the location of LC double pole F
.
ESR
F
=
LC
2LC
×π××
=
1
OUTOUT
1
×π××
2.3kHz
=
gm
Ve
R3
C2
C1
LC
compensator
F
Figure 6 - Bode plot of Type II compensator
Rev.3.2
04/10/08
F
=
ESR
2ESRC
Gain
F
F
LC
Z
ESR
P
F
F
O
8.2kHz
×π××
=
×π×Ω×
=
1
OUT
1
2.Set R2 equal to 1kΩ.
RV
2REF
1
V-V1.8V-0.8V
OUT REF
Ω×
1k0.8V
×
Choose R1=800Ω.
3. Set crossover frequency at 1/10~ 1/5 of the
swithing frequency, here FO=30kHz.
4.Calculate R3 value by the following equation.
11
Page 12
4.Calculate R3 value by the following equation.
5V6.5m2.0mA/V
214.7k0.752.3kHz
p
F
p14.7k300kHz
OUT
REF
2REF
OUT REF
IID1-D
V2FLV
R=
3
VRgV
1.5V230kHz1.5uH1
=
=14.6k
×π××
OSCOOUT
×××
inESRmREF
×π××
××
1.8V
×
0.8V
1
Ω
Ω
Choose R3 =14.7kΩ.
5. Calculate C1 by setting compensator zero F
Z
at 75% of the LC double pole.
C=
1
=
1
2RF
×π××
3z
1
×π×Ω××
=6.3nF
Choose C1=6.8nF.
6. Calculate C2 by setting compensator pole
at half the swithing frequency.
C=
2
=
1
RF
π××
3s
×Ω×
1
=72pF
Choose C1=68pF.
Output Voltage Calculation
Output voltage is set by reference voltage and
external voltage divider. The reference voltage is fixed
at 0.8V. The divider consists of two ratioed resistors
so that the output voltage applied at the Fb pin is 0.8V
when the output voltage is at the desired value. The
following equation and picture show the relationship
between
value of R1 value can be set by voltage divider.
Rev.3.2
04/10/08
V,
Vand voltage divider..
RV
R=
1
×
V-V
...(18)
where R2 is part of the compensator, and the
See compensator design for R1 and R2 selection.
NX2119/2119A
Vout
R2
Fb
R1
Vref
Voltage divider
Figure 8 - Voltage divider
Input Capacitor Selection
Input capacitors are usually a mix of high frequency
ceramic capacitors and bulk capacitors. Ceramic capacitors bypass the high frequency noise, and bulk capacitors supply switching current to the MOSFETs. Usually 1uF ceramic capacitor is chosen to decouple the
high frequency noise.The bulk input capacitors are decided by voltage rating and RMS current rating. The RMS
current in the input capacitors can be calculated as:
=××
RMSOUT
V
OUT
D
=
V
IN
VIN = 5V, VOUT=1.8V, IOUT=9A, using equation (19),
the result of input RMS current is 4.3A.
For higher efficiency, low ESR capacitors are recommended. One Sanyo OS-CON 16SP270M 16V 270uF
18mΩ with 4.4A RMS rating are chosen as input bulk
capacitors.
Power MOSFETs Selection
The power stage requires two N-Channel power
MOSFETs. The selection of MOSFETs is based on
maximum drain source voltage, gate source voltage,
maximum current rating, MOSFET on resistance and
power dissipation. The main consideration is the power
loss contribution of MOSFETs to the overall converter
efficiency. In this design example, two IRFR3706 are
used. They have the following parameters: VDS=30V, I
=75A,R
loss:conduction loss, switching loss.
=9mΩ,Q
DSON
GATE
=23nC.
There are two factors causing the MOSFET power
Conduction loss is simply defined as:
...(19)
12
D
Page 13
NX2119/2119A
P=I(1D)RK
×−××
SWINOUTSWS
PVITF
=××××
gateHGATEHGSLGATELGSS
P(QVQV)F
=×+××
I
KR
I23.7A
P=IDR
HCONOUTDS(ON)
LCONOUTDS(ON)
P=PP
TOTALHCONLCON
2
×××
2
K
+
...(20)
where the RDS(ON) will increases as MOSFET junction temperature increases, K is RDS(ON) temperature
dependency. As a result, RDS(ON) should be selected for
the worst case, in which K approximately equals to 1.4
at 125oCaccording toIRFR3706 datasheet. Conduc-
tion loss should not exceed package rating or overall
system thermal budget.
Switching loss is mainly caused by crossover conduction at the switching transition. The total switching
loss can be approximated.
1
2
...(21)
where IOUT is output current, TSW is the sum of T
and TF which can be found in mosfet datasheet, and FS
is switching frequency. Switching loss PSW is frequency
dependent.
Also MOSFET gate driver loss should be considered when choosing the proper power MOSFET.
MOSFET gate driver loss is the loss generated by discharging the gate capacitor and is dissipated in driver
circuits.It is proportional to frequency and is defined as:
...(22)
where QHGATE is the high side MOSFETs gate
charge,QLGATE is the low side MOSFETs gate charge,VHGS
is the high side gate source voltage, and V
is the low
LGS
side gate source voltage.
This power dissipation should not exceed maximum power dissipation of the driver device.
320mV
=
SET
If MOSFET R
×
DSON
=9mΩ, the worst case thermal
DSON
consideration K=1.5, then
320mV320mV
===
SET
KR1.59m
××Ω
DSON
Layout Considerations
The layout is very important when designing high
frequency switching converters. Layout will affect noise
pickup and can cause a good design to perform with
less than expected results.
Start to place the power components, make all the
connection in the top layer with wide, copper filled areas. The inductor, output capacitor and the MOSFET
should be close to each other as possible. This helps to
R
reduce the EMI radiated by the power traces due to the
high switching currents through them. Place input capacitor directly to the drain of the high-side MOSFET, to
reduce the ESR replace the single input capacitor with
two parallel units. The feedback part of the system should
be kept away from the inductor and other noise
sources,and be placed close to the IC. In multilayer PCB
use one layer as power ground plane and have a control
circuit ground (analog ground), to which all signals are
referenced.
The goal is to localize the high current path to a
separate loop that does not interfere with the more sensitive analog control function. These two grounds must
be connected together on the PC board layout at a single
point.
Over Current Limit Protection
Over current Limit for step down converter is
achieved by sensing current through the low side
MOSFET. For NX2119, the current limit is decided by
the R
FET is on, and the voltage on SW pin is below 320mV,
the over current occurs. The over current limit can be
calculated by the following equation.
Rev.3.2
04/10/08
of the low side mosfet. When synchronous
DSON
13
Page 14
SOIC8 PACKAGE OUTLINE DIMENSIONS
NX2119/2119A
Rev.3.2
04/10/08
14
Page 15
NX2119/2119A
Rev.3.2
04/10/08
15
Page 16
MSOP8 PACKAGE OUTLINE DIMENSIONS
NX2119/2119A
Rev.3.2
04/10/08
16
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