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參數(shù)資料
| 型號(hào)： | MAX865 |
| 廠商： | Maxim Integrated Products, Inc. |
| 英文描述： | Compact, Dual-Output Charge Pump |
| 中文描述： | 緊湊的、雙輸出電荷泵 |
| 文件頁數(shù)： | 6/8頁 |
| 文件大小： | 89K |
| 代理商： | MAX865 |
converter (RS+), where I
LOAD+
is the combination of I
V-
and the external load on V+ (I
V+
):
V
= I
x RS+ = I
DROOP+
LOAD+
Determine V+ and V- as follows:
The output resistance for the positive and negative
charge pumps are tested and specified separately. The
positive charge pump is tested with V- unloaded. The
negative charge pump is tested with V+ supplied from
an external source, isolating the negative charge
pump.
Current draw from either V+ or V- is supplied by the
reservoir capacitor alone during one half cycle of the
clock. Calculate the resulting ripple voltage on either
output as follows:
where I
LOAD
is the load on either V+ or V-. For the typi-
cal f
PUMP
of 30kHz with 3.3μF reservoir capacitors, the
ripple is 25mV when I
LOAD
is 5mA. Remember that, in
most applications, the total load on V+ is the V+ load
current (I
V+
) and the current taken by the negative
charge pump (I
V-
).
Effic ienc y Considerations
Theoretically, a charge-pump voltage multiplier can
approach 100% power efficiency under the following
conditions:
The charge-pump switches have virtually no offset
and extremely low on-resistance.
The drive circuitry consumes minimal power.
The impedances of the reservoir and pump capaci-
tors are negligible.
For the MAX865, the energy loss per clock cycle is the
sum of the energy loss in the positive and negative
converters, as follows:
The average power loss is simply:
Resulting in an efficiency of:
MAX865
C1+
1
2
3
4
C1-
GND
3.3
μ
F
3.3
μ
F
3.3
μ
F
3.3
μ
F
3.3
μ
F
3.3
μ
F
OUT+
IN
OUT-
8
7
6
5
C2-
C2-
V-
V+
IN
GND
MAX865
C1+
1
2
3
4
C1-
8
7
6
5
C2+
C2-
V
IN
V-
V+
IN
GND
Figure 3. Paralleling MAX865s
M
Compac t, Dual-Output Charge Pump
6
_______________________________________________________________________________________
+ I
x RS+
V+
V-
(
)
P
= LOSS
x f
LOSS
CYCLE
PUMP
η =
)
Total Output Power Total Output Power
/
P
LOSS
V
= I
(1 / f
) (1 / C
)
RIPPLE
PUMP
RESERVOIR
V+ = 2V - V
V- = (V+ - V
)= -(2V
- V
- V
)
DROOP+
DROOP
IN
DROOP+
DROOP-
LOSS
= LOSS
1
2
1
2
+ LOSS
)
=
C1
CYCLE
POS
(
(
NEG
V
V
V
C
V
V
IN
+
+
(
)
(
)
+
+
)
(
)
2
2
2
2
2
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