6

TC7662B

CHARGE PUMP DC-TO-DC

VOLTAGE CONVERTER

TC7662B-8 9/11/96

© 2001 Microchip Technology Inc.

DS21469A

1

2

3

4

8

7

6

5

+

V

+

VOUT

C1

COSC

+

C2

TC7662B

Figure 8. Lowering Oscillator Frequency

Positive Voltage Doubling

The TC7662B may be employed to achieve positive

voltage doubling using the circuit shown in Figure 9. In this

application, the pump inverter switches of the TC7662B are

or twice the supply voltage minus the combined forward

of 10 mA, it will be approximately 60

Ω.

Combined Negative Voltage Conversion

and Positive Supply Multiplication

Figure 10 combines the functions shown in Figures 3

and 9 to provide negative voltage conversion and positive

voltage doubling simultaneously. This approach would be,

for example, suitable for generating +9V and –5V from an

perform the pump and reservoir functions, respectively, for

positive voltage. There is a penalty in this configuration

which combines both functions, however, in that the source

impedances of the generated supplies will be somewhat

higher due to the finite impedance of the common charge

pump driver at pin 2 of the device.

1

2

3

4

8

7

6

5

+

V +

VOUT =

(2 V +) – (2 VF)

C1

D1

+

+

C3

C4

VOUT =

– (V+– VF)

C2

TC7662B

D2

+

Figure 10. Combined Negative Converter and Positive Doubler

Figure 9. Positive Voltage Multiplier

Voltage Splitting

The bidirectional characteristics can also be used to

split a higher supply in half, as shown in Figure 11. The

combined load will be evenly shared between the two sides

and a high value resistor to the LV pin ensures start-up.

Because the switches share the load in parallel, the output

impedance is much lower than in the standard circuits, and

higher currents can be drawn from the device. By using this

circuit, and then the circuit of Figure 6, +15V can be

converted (via +7.5V and –7.5V) to a nominal –15V, though

with rather high series resistance (~250

Ω).

1

2

3

4

8

7

6

5

V+

VOUT =

(2 V+) – (2 VF)

+

C2

D1

D2

+

C1

TC7662B

+

RL1

RL2

VOUT =

V + –V –

2

50

µF

50

µF

V +

V –

50

µF

+

1

2

8

7

TC7662B

3

4

6

5

-

-

+

-

Figure 11. Splitting a Supply in Half