参数资料
型号: LTC3708EUH#TRPBF
厂商: Linear Technology
文件页数: 22/32页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM CM 32-QFN
标准包装: 2,500
系列: PolyPhase®
PWM 型: 电流模式
输出数: 2
占空比: 90%
电源电压: 4 V ~ 36 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 85°C
封装/外壳: 32-WFQFN 裸露焊盘
包装: 带卷 (TR)
LTC3708
APPLICATIONS INFORMATION
R DS ( ON )_ DRV ?
? V V GS ( TH ) ? ?
2.TransitionLoss.Thislossarisesfromthebriefamount
of time the top MOSFET spends in the saturated region
during switch node transitions. It depends upon the input
voltage, load current, driver strength and MOSFET capaci-
tance, among other factors. The loss is signi?cant at input
voltages above 20V and can be estimated from:
Transition Loss ≈
( 0 . 5 ) ? V IN 2 ? I OUT ? C RSS ? f ?
? 1 1 ?
+
? DRV CC GS ( TH )
3. DRV CC and V CC Current. This is the sum of the MOSFET
driver and control currents. The driver current supplies the
gate charge Q G required to switch the power MOSFETs.
This current is typically much larger than the control circuit
current. In continuous mode operation:
I GATECHG = f(Q G(TOP) + Q G(BOT) )
a load step occurs, V OUT immediately shifts by an amount
equal to Δ I LOAD (ESR), where ESR is the effective series
resistance of C OUT . Δ I LOAD also begins to charge or dis-
charge C OUT generating a feedback error signal used by the
regulator to return V OUT to its steady-state value. During
this recovery time, V OUT can be monitored for overshoot
or ringing that would indicate a stability problems. The
I TH pin external components shown in Figure 13 will pro-
vide adequate compensation for most applications. For a
detailed explanation of switching control loop theory see
Linear Technology Application Note 76.
Design Example
As a design example, take a supply with the following
speci?cations: V IN = 7V to 28V (15V nominal), V OUT1
= 2.5V, V OUT2 = 1.8V, I OUT1(MAX) = I OUT2(MAX) = 10A,
f = 500kHz and V OUT2 to track V OUT1 .
First calculate the timing resistor:
R ON1 =
4. C IN Loss. The input capacitor has the dif?cult job of
?ltering the large RMS input current to the regulator. It
must have a very low ESR to minimize the AC I 2 R loss and
suf?cient capacitance to prevent the RMS current from
2 . 5 V
( 0 . 7 V )( 500 kHz )( 10 pF )
Select a standard value of 715k.
= 714 k
causing additional upstream losses in fuses or batteries.
The LTC3708 2-phase architecture typically halves this C IN
loss over the single phase solutions.
R ON2 =
1 . 8 V
( 0 . 7 V )( 500 kHz )( 10 pF )
= 514 k
=
– 1 = 3 . 17
Otherlosses,includingC OUT ESRloss,Schottkyconduc-
tion loss during dead time and inductor core loss generally
account for less than 2% additional loss.
When making any adjustments to improve ef?ciency, the
?nal arbiter is the total input current for the regulator at
your operating point. If you make a change and the input
Select a standard value of 511k.
Next, choose the feedback resistors:
R1  2 .5 V
R 2 0 . 6 V
Select R1 = 31.6k, R2 = 10k.
current decreases, then you improve the ef?ciency. If there
is no change in input current, then there is no change in
ef?ciency.
R3
R 4
=
1. 8 V
0 . 6 V
– 1 = 2
Checking Transient Response
The regulator loop response can be checked by looking
at the load transient response. Switching regulators take
several cycles to respond to a step in load current. When
Select R3 = 20k, R4 = 10k.
For V OUT2 to coincidently track V OUT1 at start-up, connect
an extra pair of R3 and R4 across V OUT1 with its midpoint
tied to the TRACK2 pin.
3708fb
22
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