MAX865 Maxim, MAX865 Datasheet - Page 6

• Manufacturer Part Number: MAX865
• Description: Compact / Dual-Output Charge Pump
• Manufacturer: Maxim

converter (RS+), where I
and the external load on V+ (I
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
cal f
ripple is 25mV when I
most applications, the total load on V+ is the V+ load
current (I
charge pump (I
Compact, Dual-Output Charge Pump
Figure 3. Paralleling MAX865s
6
V+ = 2V
V - = (V+ - V
V
_______________________________________________________________________________________
PUMP
V
RIPPLE
DROOP+
LOAD
V
V+
of 30kHz with 3.3µF reservoir capacitors, the
IN
IN
=
) and the current taken by the negative
is the load on either V+ or V-. For the typi-
- V
3.3 F
3.3 F
1
2
= I
V-
DROOP
I
DROOP+
LOAD
).
LOAD+
1
2
3
4
LOAD
C1-
C2-
C2-
V-
(1 / f
) = -(2V - V
LOAD+
x RS+ = I
MAX865
PUMP
is 5mA. Remember that, in
V+
IN
is the combination of I
):
) (1 / C
GND
C1+
V+
DROOP+
IN
V+
+ I
8
7
6
5
RESERVOIR
V -
- V
x RS+
DROOP-
)
V-
)
3.3 F
3.3 F
Theoretically, a charge-pump voltage multiplier can
approach 100% power efficiency under the following
conditions:
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:
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.
1
2
3
4
Total Output Power Total Output Power
C1-
C2+
C2-
V-
LOSS
MAX865
P
CYCLE
LOSS
GND
C1+
V+
IN
=
= LOSS
= LOSS
8
7
6
5
1
2
1
2
Efficiency Considerations
C1
C
/
2
POS
CYCLE
V
V
+ LOSS
2
2
x f
3.3 F
3.3 F
2
PUMP
V
V
NEG
GND
2
IN
OUT+
OUT-
V
IN
P
LOSS