MAX680CPA Maxim Integrated Products, MAX680CPA Datasheet - Page 6

IC VOLT CONV CMOS HI-EFF 8-DIP

MAX680CPA

Manufacturer Part Number
MAX680CPA
Description
IC VOLT CONV CMOS HI-EFF 8-DIP
Manufacturer
Maxim Integrated Products
Type
Step-Up (Boost), Switched Capacitor (Charge Pump), Invertingr
Datasheet

Specifications of MAX680CPA

Internal Switch(s)
Yes
Synchronous Rectifier
No
Number Of Outputs
2
Voltage - Output
±4 ~ ±12 V
Current - Output
10mA
Frequency - Switching
8kHz
Voltage - Input
2 ~ 6 V
Operating Temperature
0°C ~ 70°C
Mounting Type
Through Hole
Package / Case
8-DIP (0.300", 7.62mm)
Power - Output
727mW
Function
Inverting, Step Up
Output Voltage
+/- 10 V
Output Current
10 mA
Maximum Operating Temperature
+ 70 C
Minimum Operating Temperature
0 C
Mounting Style
Through Hole
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
MAX680CPA
Manufacturer:
MAXIM
Quantity:
5 510
Part Number:
MAX680CPA
Manufacturer:
MAXIM/美信
Quantity:
20 000
The MAX680/MAX681 are not voltage regulators: the
output source resistance of either charge pump is
approximately 150Ω at room temperature with V
5V. Under light load with an input V
approach +10V and V- will be at -10V. However both,
V+ and V- will droop toward GND as the current drawn
from either V+ or V- increases, since the negative con-
verter draws its power from the positive converter’s out-
put. To predict output voltages, treat the chips as two
separate converters and analyze them separately. First,
the droop of the negative supply (V
current drawn from V- - (I
of the negative converter (RS-):
Likewise, the positive supply droop (V
the current drawn from the positive supply (I
the positive converter’s source resistance (RS+),
except that the current drawn from the positive supply
is the sum of the current drawn by the load on the posi-
tive supply (I
converter (I
+5V to ±10V Voltage Converters
Figure 4. Paralleling MAX680s For Lower Source Resistance
6
_______________________________________________________________________________________
(V
DROP
22 F
22 F
L -
L +
):
+) = I
) plus the current drawn by the negative
1
2
3
4
V
L
C1-
C2+
C2-
V-
+ x RS+ = (I
DROP
MAX680
L -
) times the source resistance
- = I
GND
C1+
V
CC
V+
L
- x RS-
L
8
7
6
5
+ + I
DROP-
CC
L
DROP +
-) x RS+
of 5V, V+ will
) equals the
L +
) equals
) times
CC
at
22 F
22 F
The positive output voltage will be:
The negative output voltage will be:
The positive and negative charge pumps are tested
and specified separately to provide the separate values
of output source resistance for use in the above formu-
las. When the positive charge pump is tested, the neg-
ative charge pump is unloaded. When the negative
charge pump is tested, the positive supply V+ is from
an external source, isolating the negative charge
pump.
Calculate the ripple voltage on either output by noting
that the current drawn from the output is supplied by
the reservoir capacitor alone during one half-cycle of
the clock. This results in a ripple of:
For the nominal f
capacitors, the ripple will be 30mV with I
Remember that in most applications, the positive
charge pump’s I
taken by the negative charge pump.
1
2
3
4
V- = (V+ - V
C2+
C2-
C1-
V-
MAX680
V
RIPPLE
GND
C1+
V
V+
CC
DROP
OUT
V+ = 2V
8
7
6
5
=
PUMP
) = - (2V
1
is the load current plus the current
2
IOUT (
CC
of 8kHz with 10µF reservoir
– V
CC
1
⁄ f
22 F
22 F
DROP
- V
PUMP
DROP
+
)(
1
V+ OUT
V
GND
V- OUT
⁄ CR)
+ - V
CC
OUT
IN
DROP
at 5mA.
-)

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