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参数资料
| 型号： | LM27262MTDX/NOPB |
| 厂商： | NATIONAL SEMICONDUCTOR CORP |
| 元件分类： | 稳压器 |
| 英文描述： | SWITCHING CONTROLLER, 345 kHz SWITCHING FREQ-MAX, PDSO48 |
| 封装： | TSSOP-48 |
| 文件页数： | 10/22页 |
| 文件大小： | 1105K |
| 代理商： | LM27262MTDX/NOPB |
Component Selection (Continued)
V
MAX = 1.526V
There will also be an initial step equal to
I
OUT x ESR, which
in our example will be approximately 41mV. The two effects
are not entirely additive since the voltage across the ESR is
decreasing as the capacitor gets charged. Therefore, the
actual peak voltage will be somewhat less than the 1.567V
that simple addition would predict. In general, it’s a good
idea to provide pads for a couple of extra capacitors on the
layout in case a little extra decoupling proves necessary.
The output ripple currents of multiphase regulators tend to
cancel to some degree. This greatly reduces the demand on
the output capacitors. Figure 6 allows a simple estimate to
be made of the total output ripple current based on the
number of phases and the nominal duty cycle. Simply pick
the worst case (highest ripple) operating point off the curve
and multiply by the single channel pk-pk ripple current. The
expected pk-pk output ripple voltage will be approximately:
Vrip = Irip x ESR
This simplified equation ignores the reactive term of the
capacitor’s impedance. With any kind of electrolytic capaci-
tor it’s generally safe to ignore the capacitive reactance term
since it will prove to be negligible compared to the ESR.
CALCULATING THE INDUCTOR RIPPLE CURRENT
The LM27262 operates at a switching frequency of 300kHz
per phase. The high side switch on time is therefore the
period, 3.33s, multiplied the duty factor, V
OUT/VIN. During
this time the inductor is connected between the input and the
output, so inductor current ramps positive during this time.
The peak-peak ripple current
I is approximately equal to:
I = 3.3s x (V
OUT -VOUT
2 /V
IN)/L
Continuing our example and assuming a maximum input of
12V:
I = 3.3s x (1.5V - 1.5V2 / 12V) / 0.50H
I = 8.66A
I
IN = 1.5A
With a maximum phase current of 17.5A this is a bit higher
than desired so a little larger inductor value may be in order.
Assuming a 35% ripple current and 17.5A/phase:
L = 3.3s x (1.5V - 1.5V
2 / 12V) / 6.125
L = 0.71H
If the larger value of inductor is used, it may be necessary to
go back and recalculate some of the early assumptions.
The peak current seen by the inductors will be the maximum
DC current plus one half of the ripple current. The maximum
DC current should be assumed to be approximately 10% to
15% greater than the maximum anticipated load current to
allow for short circuit current. The inductors must not hard
saturate in a fault. Again from our example, 110% of 17.5A
plus one half of the 6.125A peak-peak ripple current yields a
peak inductor current of 22.3A. There will usually be two
current ratings associated with an inductor. One is the aver-
age current rating and the second is the saturation current.
Only the saturation current need be considered for short
circuit limiting. The sustained DC current is the 110% of
17.5A or roughly 19A in this example. If over current latch off
is employed, only the 17.5A steady state current need be
considered since the inductor’s steady state current rating is
basically a thermal limitation.
MOSFET SELECTION
The choice of power FETs is driven primarily by efficiency or
thermal considerations. There are two main loss compo-
nents to consider, conduction losses and switching losses.
The switching losses are primarily due to parasitics in the
FETs and are very hard to estimate with any degree of
accuracy. The conduction losses are much easier to charac-
terize. The switching losses in the low-side FET are very low
since it’s essentially a zero-voltage switched device. How-
ever, the high-side device’s switching losses are usually
larger than its conduction losses. The primary contributor to
high-side FET switching losses is related to the reverse
recovery characteristics of the low-side FET’s body diode.
During the small dead band where both FETs are off every
cycle, the low-side FET’s body diode will carry the inductor
current. The problem is that the body diode exhibits a sig-
nificant reverse recovery time, trr. During this time, the FET
looks like a short circuit. When the high-side FET is subse-
quently turned on, there is a shoot through path from the
input supply to ground. A larger high-side FET will tend to
exhibit a larger shoot through current. Therefore, it is unde-
sirable to oversize the high-side device. Since the low-side
device looks like a short, the entire supply voltage is im-
pressed across the high–side device, along with a simulta-
neous high current. The result is very high momentary power
dissipation. The total power lost is a direct function of the
switching frequency.
As a starting point, assume that about 12 of the switcher’s
total losses will take place in the MOSFETs. If from our prior
example, we assume a desired 90% efficiency at full load,
the FET losses are therefore 5% of the full output power or
5.25W. Since we have four phases, that works out to 1.31W/
phase. This is divided between the upper and lower FETs.
Since the step down ratio is large, the low side FETs will be
on for most of the period. The low-side FET conduction loss
is I
2 x Rds(on) * (1-DF), where DF is the duty factor, V
OUT/
V
IN. The worst-case dissipation occurs at high input line
voltage. We’ve assumed 12V for our example. Allowing 50%
of the total FET loss to the low-side switch gives us a
dissipation of 0.66W x Rds(on) . This will usually result in a
conservative design. Solving for Rds(on) :
Rds(on) = P
dis/(I
2 x (1- V
OUT/VIN)) Rds(on) = 0.65W/(17 x
5A
2x (1-1.5V/12V))
Rds(on) = 2.4m
A pair of Si7356 MOSFETs in parallel is very close to this.
These are trench devices with excellent thermal character-
istics as well.
The high-side FET on-resistance is calculated similarly, but
allow one half of the dissipation to switching losses. There-
fore the allowable conduction loss is approximately 0.325W.
Rds(on) = Pdis/(I2 x (V
OUT/VIN))
Rds(on) = 0.325W/(17x 5A
2 x 1.5/12V))
Rds(on) = 8.5m
A pair of Si7392 FETs will meet this requirement.
GATE DRIVE RTEQUIREMENTS
Energy for the high side gate drives is stored in the boost
capacitors, which in turn are powered by the 5V supply. The
charge stored in each boost capacitor should be between 10
times and 20 times the Qg specified for the high-side FETs.
For the example given, the specified maximum Qg is 15nC.
There will be two of these devices in parallel so the com-
bined gate charge is 30nC. Therefore, the charge stored on
the boost capacitor should be a minimum of 300nC.
LM27262
www.national.com
18
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| LM2726M | 功能描述:功率驱动器IC RoHS:否 制造商:Micrel 产品:MOSFET Gate Drivers 类型:Low Cost High or Low Side MOSFET Driver 上升时间: 下降时间: 电源电压-最大:30 V 电源电压-最小:2.75 V 电源电流: 最大功率耗散: 最大工作温度:+ 85 C 安装风格:SMD/SMT 封装 / 箱体:SOIC-8 封装:Tube |
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