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参数资料
| 型号： | ADP1874-0.3-EVALZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 33/44页 |
| 文件大小： | 0K |
| 描述： | BOARD EVAL FOR ADP1874 |
| 标准包装： | 1 |
| 系列： | * |
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Data Sheet
ADP1874/ADP1875
Assuming an overshoot of 45 mV, determine if the output
capacitor that was calculated previously is adequate.
C COMP =
1
2 π R COMP f ZERO
( L × I
LOAD )
C OUT =
2
( ( V OUT ? ? V OVSHT ) 2 ? ( V OUT ) 2 )
=
1
2 × 3 . 14 × 60 . 25 × 10 3 × 6 . 25 × 10 3
=
1 × 10 ? 6 × ( 15 A ) 2
( 1 . 8 ? 45 mV ) 2 ? ( 1 . 8 ) 2
= 423 pF
Loss Calculations
1 1 ( V IN , MAX ? V OUT ) V OUT
1 1 ( 13 . 2 V ? 1 . 8 V ) 1 . 8 V
= × × = 1 . 49 A
3 1 μF × 300 × 10
2
13 . 2 V
R T = 1 kΩ ×
= 2 kΩ
P N1,N2(CL) = [ D × R N1(ON) + ( 1 ? D ) × R N2(ON) ] × I LOAD
P BODY ( LOSS ) =
× I LOAD × V F × 2
= 1.4 mF
Choose five 270 μF polymer capacitors.
The rms current through the output capacitor is
I RMS = × ×
2 3 L × f SW V IN , MAX
3
The power loss dissipated through the ESR of the output
capacitor is
P COUT = ( I RMS ) 2 × ESR = (1.5 A) 2 × 1.4 mΩ = 3.15 mW
Feedback Resistor Network Setup
Choosing R B = 1 kΩ as an example, calculate R T as follows:
(1 . 8 V ? 0 . 6 V)
0 . 6 V
Compensation Network
To calculate R COMP , C COMP , and C PAR , the transconductance
parameter and the current-sense gain variable are required. The
transconductance parameter (G m ) is 500 μA/V, and the current-
sense loop gain is
Duty cycle = 1.8/12 V = 0.15
R ON (N2) = 5.4 mΩ
t BODY(LOSS) = 20 ns (body conduction time)
V F = 0.84 V (MOSFET forward voltage)
C IN = 3.3 nF (MOSFET gate input capacitance)
Q N1,N2 = 17 nC (total MOSFET gate charge)
R GATE = 1.5 Ω (MOSFET gate input resistance)
2
= (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A) 2
= 1.215 W
t BODY ( LOSS )
t SW
= 20 ns × 300 × 10 3 × 15 A × 0.84 × 2
= 151.2 mW
P SW(LOSS) = f SW × R GATE × C TOTAL × I LOAD × V IN × 2
= 300 × 10 3 × 1.5 ? × 3.3 × 10 ?9 × 15 A × 12 × 2
= 534.6 mW
[
P DR ( LOSS ) = V DR × ( f SW C upperFET V DR + I BIAS ) ] +
[ VREG × ( f SW C lowerFET VREG + I BIAS ) ]
G CS
=
1
A CS × R ON
=
1
24 × 0 . 005
= 8 . 33 A/V
= ( 4 . 62 × ( 300 × 10 3 × 3 . 3 × 10 ? 9 × 4 . 62 + 0 . 002 )) +
( 5 . 0 × ( 300 × 10 3 × 3 . 3 × 10 ? 9 × 5 . 0 + 0 . 002 ))
P DCR ( LOSS ) = DCR × I LOAD = 0.003 × (15 A) = 675 mW
where A CS and R ON are taken from setting up the current limit
(see the Programming Resistor (RES) Detect Circuit section
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12 the switching frequency.
300 kHz/12 = 25 kHz
The zero frequency is 1/4 the crossover frequency.
25 kHz/4 = 6.25 kHz
= 57.12 mW
P DISS ( LDO ) = ( V IN ? VREG ) × ( f SW × C total × VREG + I BIAS )
= ( 13 V ? 5 V ) × ( 300 × 10 3 × 3 . 3 × 10 ? 9 × 5 + 0 . 002 )
= 55 . 6 mW
P COUT = ( I RMS ) 2 × ESR = (1.5 A) 2 × 1.4 mΩ = 3.15 mW
2 2
1 + ( s × ESR × C OUT )
V
R COMP =
f CROSS
f CROSS 2 + f ZERO 2
×
1 2 + ( s ( R L + ESR ) C OUT ) 2
2 2
×
1
R L
× OUT ×
V REF
1
G M G CS
P CIN = ( I RMS ) 2 × ESR = (7.5 A) 2 × 1 mΩ = 56.25 mW
P LOSS = P N1,N2 + P BODY(LOSS) + P SW + P DCR + P DR + P DISS(LDO) +
P COUT + P CIN
= 1.215 W + 151.2 mW + 534.6 mW + 57.12 mW + 55.6 +
R COMP =
25 K
25 k 2 + 6 . 25 k 2
×
1 2 + ( 2 π × 25 k × ((1 . 8 15) + 0 . 0035) × 0 . 0011 ) 2
1 2 + ( 2 π × 25 k × 0 . 0035 × 0 . 0011 ) 2
×
3.15 mW + 675 mW + 56.25 mW
= 2.655 W
1 . 8 1 15
500 × 10 ? 6 × 8 . 3 1 . 8
× ×
0 . 6
= 60.25 kΩ
Rev. A | Page 33 of 44
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