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:

most applications, the total load on V+ is the V+ load

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:

The average power loss is simply:

Resulting in an efficiency of:

MAX865

C1+

1

2

3

4

C1-

GND

3.3

µF

3.3

µF

3.3

µF

3.3

µF

3.3

µF

3.3

µF

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-

V+

IN

GND

Figure 3. Paralleling MAX865s

Compact, Dual-Output Charge Pump

6

_______________________________________________________________________________________

V

= I

x RS+ = I

+ I

x RS+

DROOP+

LOAD+

V+

V -

(

)

P

= LOSS

x f

LOSS

CYCLE

PUMP

η =

−

(

)

Total Output Power Total Output Power

/

V

=

I

(1 / f

) (1 / C

)

RIPPLE

1

2

LOAD

PUMP

RESERVOIR

V+ = 2V

- V

V - = (V+ - V

) = -(2V

- V

- V

)

IN

DROOP+

DROOP

IN

DROOP+

DROOP -

LOSS

= LOSS

+ LOSS

=

C1

CYCLE

POS

NEG

1

2

1

2

V

V

V

C

V

V

IN

+

+

( ) ( )









+

+

( )









2

2

2

2

2