LT1996 Linear Technology, LT1996 Datasheet - Page 10

• Manufacturer Part Number: LT1996
• Description: 100uA Gain Selectable Amplifier
• Manufacturer: Linear Technology

www.DataSheet4U.com
DataSheet U .com
LT1996
APPLICATIO S I FOR ATIO
Consider Figure 1. This shows the LT1996 configured as
a gain of 117 difference amplifier on a single supply with
the output REF connected to ground. This is a great circuit,
but it does not support V
because the output clips into ground while trying to
produce 0V
valid input differential range not to extend below +0.4mV,
or by elevating the REF pin above 40mV, or by providing
a negative supply.
Calculating Input Voltage Range
Figure 2 shows the LT1996 in the generalized case of a
difference amplifier, with the inputs shorted for the com-
mon mode calculation. The values of R
dictated by how the P inputs and REF pin are connected.
By superposition we can write:
Or, solving for V
But valid V
1V, so:
and:
10
Figure 1. Difference Amplifier Cannot Produce 0V on a Single
Supply. Provide a Negative Supply, or Raise Pin 5, or Provide
400µV of V
4
V
V
MAX V
MIN V
INT
EXT
V
2.5V
= V
= V
CM
EXT
V
0V
EXT
DM
–
+
INT
DM
EXT
INT
10
OUT
8
9
1
2
3
= (V
= (V
voltages are limited to V
450k/81
450k/27
• (R
450k/27
450k/81
• (1 + R
450k/9
450k/9
. It can be fixed simply by declaring the
EXT
EE
CC
F
U
/(R
:
+ 1) • (1 + R
– 1.2) • (1 + R
G
F
/R
+ R
5V
DM
U
F
7
–
+
4
) – V
G
= 0V at any common mode
)) + V
450k
450k
4pF
4pF
4pF
4pF
REF
G
/R
LT1996
REF
G
W
F
• R
1996 F01
/R
REF
) – V
CC
F
• (R
G
) – V
6
5
/R
– 1.2V and V
REF
V
F
G
F
OUT
/(R
REF
and R
• R
= 117 • V
F
U
• R
+ R
G
/R
DM
G
G
F
G
/R
EE
))
are
F
+
These two voltages represent the high and low extremes
of the common mode input range, if the other limits have
not already been exceeded (1 and 3, above). In most cases,
the inverting inputs M9 through M81 can be taken further
than these two extremes because doing this does not
move the op amp input common mode. To calculate the
limit on this additional range, see Figure 3. Note that, with
V
V
amp output will go more negative from V
V
Or:
The most negative that V
The situation where this function is negative, and therefore
problematic, when V
dealt with in Figure 1. The strength of the equation is
demonstrated in that it provides the three solutions
MORE
EXT
MORE
V
V
Max V
(should be positive)
OUT
MORE
(the high cm limit), as V
Figure 3. Calculating Additional Voltage Range of
Inverting Inputs
• R
= 0, the op amp output is at V
Figure 2. Calculating CM Input Voltage Range
= V
MORE
MAX OR MIN
= (V
V
F
EXT
/R
REF
V
MORE
G
V
REF
, so:
EXT
= (V
– V
– V
MORE
REF
R
R
R
R
G
G
REF
G
G
OUT
V
INT
V
– V
INT
= 0 and V
• R
) • R
OUT
EE
–
+
–
+
F
/R
– 0.04V) • R
G
can go is V
V
V
R
MORE
V
CC
EE
V
R
/R
F
G
CC
EE
F
F
EE
R
R
F
F
goes positive, the op
= 0, has already been
REF
V
V
REF
REF
REF
EE
. From the max
G
/R
by the amount
+ 0.04V, so:
1996 F02
1996 F03
F
1996f