MAX680C/D View Datasheet(PDF) - Maxim Integrated
Part Name
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MAX680C/D Datasheet PDF : 8 Pages
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+5V to ±10V Voltage Converters
22µF
22µF
1 C1- MAX680 V+ 8
2 C2+
C1+ 7
3 C2-
4 V-
6
VCC
GND 5
22µF
1 C1- MAX680
8
V+
2 C2+
C1+ 7
22µF
3 C2-
4 V-
6
VCC
GND 5
22µF
22µF
V+ OUT
VCC IN
GND
V- OUT
Figure 4. Paralleling MAX680s For Lower Source Resistance
The MAX680/MAX681 are not voltage regulators: the
output source resistance of either charge pump is
approximately 150Ω at room temperature with VCC at
5V. Under light load with an input VCC of 5V, V+ will
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 (VDROP-) equals the
current drawn from V- - (IL-) times the source resistance
of the negative converter (RS-):
VDROP - = IL- x RS-
Likewise, the positive supply droop (VDROP+) equals
the current drawn from the positive supply (IL+) times
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 (IL+) plus the current drawn by the negative
converter (IL-):
(VDROP+) = IL+ x RS+ = (IL+ + IL-) x RS+
The positive output voltage will be:
V+ = 2VCC – VDROP+
The negative output voltage will be:
V- = (V+ - VDROP) = - (2VCC - VDROP + - VDROP-)
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:
VRIPPLE = 1⁄2IOUT (1⁄ fPUMP)(1⁄ CR)
For the nominal fPUMP of 8kHz with 10µF reservoir
capacitors, the ripple will be 30mV with IOUT at 5mA.
Remember that in most applications, the positive
charge pump’s IOUT is the load current plus the current
taken by the negative charge pump.
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