Data Sheet

Table Of Contents
MCP4821/MCP4822
DS21953A-page 22 © 2005 Microchip Technology Inc.
6.6 Bipolar Operation
Bipolar operation is achievable using the MCP482X
devices by using an external operational amplifier (op
amp). This configuration is desirable due to the wide
variety and availability of op amps. This allows a gen-
eral purpose DAC, with its cost and availability advan-
tages, to meet almost any desired output voltage
range, power and noise performance.
Example 6-3 illustrates a simple bipolar voltage source
configuration. R
1
and R
2
allow the gain to be selected,
while R
3
and R
4
shift the DAC's output to a selected
offset. Note that R4 can be tied to V
DD
, instead of AV
SS
,
if a higher offset is desired. Note that a pull-up to V
DD
could be used, instead of R
4
or in addition to R
4
, if a
higher offset is desired.
EXAMPLE 6-3: Digitally-Controlled Bipolar Voltage Source.
6.6.1 DESIGN A BIPOLAR DAC USING
EXAMPLE 6-3
An output step magnitude of 1 mV, with an output range
of ±2.05V, is desired for a particular application.
1. Calculate the range: +2.05V – (-2.05V) = 4.1V.
2. Calculate the resolution needed:
4.1V/1 mV = 4100
Since 2
12
= 4096, 12-bit resolution is desired.
3. The amplifier gain (R
2
/R
1
), multiplied by full-
scale V
OUT
(4.096V), must be equal to the
desired minimum output to achieve bipolar
operation. Since any gain can be realized by
choosing resistor values (R
1
+R
2
), the V
REF
value must be selected first. If a V
REF
of 4.096V
is used (G=2), solve for the amplifier’s gain by
setting the DAC to 0, knowing that the output
needs to be -2.05V. The equation can be
simplified to:
4. Next, solve for R
3
and R
4
by setting the DAC to
4096, knowing that the output needs to be
+2.05V.
MCP482X
V
DD
V
DD
SPI™
3
V
OUT
R
3
R
4
R
2
R
1
V
IN
+
G = Gain select (1x or 2x)
D = Digital value of DAC (0 – 4096)
0.1 µF
V
CC
+
V
CC
V
OUT
2.048 G
D
2
12
-------
=
V
IN+
V
OUT
R
4
R
3
R
4
+
--------------------
=
V
O
V
O
V
IN+
1
R
2
R
1
------
+
⎝⎠
⎛⎞
V
DD
R
2
R
1
------
⎝⎠
⎛⎞
=
R
2
R
1
---------
2.05
4.096V
-----------------
=
If R
1
= 20 kΩ and R
2
= 10 kΩ, the gain will be 0.5
R
2
R
1
------
1
2
-- -
=
R
4
R
3
R
4
+()
-----------------------
2.05V 0.5 4.096V()+
1.5 4.096V
-------------------------------------------------------
2
3
-- -
==
If R
4
= 20 kΩ, then R
3
= 10 kΩ