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LT1996

1996f

Consider Figure 1. This shows the LT1996 configured as

a gain of 117 difference amplifier on a single supply with

Figure 3. Calculating Additional Voltage Range of

Inverting Inputs

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

Figure 1. Difference Amplifier Cannot Produce 0V on a Single

Supply. Provide a Negative Supply, or Raise Pin 5, or Provide

400µV of VDM

the output REF connected to ground. This is a great circuit,

but it does not support VDM = 0V at any common mode

because the output clips into ground while trying to

produce 0VOUT. It can be fixed simply by declaring the

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 RF and RG are

dictated by how the P inputs and REF pin are connected.

By superposition we can write:

VINT = VEXT • (RF/(RF + RG)) + VREF • (RG/(RF + RG))

Or, solving for VEXT:

VEXT = VINT • (1 + RG/RF) – VREF • RG/RF

But valid VINT voltages are limited to VCC – 1.2V and VEE +

1V, so:

MAX VEXT = (VCC – 1.2) • (1 + RG/RF) – VREF • RG/RF

and:

MIN VEXT = (VEE + 1) • (1 + RG/RF) – VREF • RG/RF

Figure 2. Calculating CM Input Voltage Range

VMORE = 0, the op amp output is at VREF. From the max

VEXT (the high cm limit), as VMORE goes positive, the op

amp output will go more negative from VREF by the amount

VMORE • RF/RG, so:

VOUT = VREF – VMORE • RF/RG

Or:

VMORE = (VREF – VOUT) • RG/RF

The most negative that VOUT can go is VEE + 0.04V, so:

Max VMORE = (VREF – VEE – 0.04V) • RG/RF

(should be positive)

The situation where this function is negative, and therefore

problematic, when VREF = 0 and VEE = 0, has already been

dealt with in Figure 1. The strength of the equation is

demonstrated in that it provides the three solutions

APPLICATIO S I FOR ATIO

4pF

–

1996 F01

450k/81

450k/27

450k/9

450k/81

450k/27

450k/9

450k

REF

VCM

2.5V

VDM

–

LT1996

VOUT = 117 • VDM

4pF

–

VREF

1996 F02

VEXT

VINT

VCC

VEE

–

VREF

1996 F03

VEXT

MAX OR MIN

VINT

VMORE

VCC

VEE

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