Compact, Dual-Output Charge Pump
MAX865
converter (RS+), where I
LOAD+
is the combination of I
V-
and the external load on V+ (I
V+
):
V
DROOP+
= I
LOAD+
x RS+ = I
V+
+ I
V -
x RS+
Determine V+ and V- as follows:
V+ = 2V
IN
- V
DROOP+
V - = (V+ - V
DROOP
) = -(2V
IN
- V
DROOP+
- V
DROOP-
)
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:
V
RIPPLE
=
1
2
Efficiency 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:
LOSS
CYCLE
= LOSS
POS
+ LOSS
NEG
=
2
C1
(
V
+
)
−
2
(
V
+
)
(
V
IN
)
2
1
2
2
C2
(
V
+
)
−
(
V
−
)
+
2
1
(
)
I
LOAD
(1 / f
PUMP
) (1 / C
RESERVOIR
)
The average power loss is simply:
P
LOSS
= LOSS
CYCLE
x f
PUMP
Resulting in an efficiency of:
η =
Total Output Power / Total Output Power
−
P
LOSS
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-
).
(
)
V
IN
3.3µF
1
2
3.3µF
3
4
C1-
C2-
MAX865
C2-
V-
C1+
V+
IN
GND
8
7
3.3µF
1
2
3.3µF
6
5
3
4
C1-
C2+
MAX865
C2-
V-
C1+
V+
IN
GND
8
7
6
5
3.3µF
IN
GND
OUT+
3.3µF
OUT-
Figure 3. Paralleling MAX865s
6
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