Datasheet LT1996 (Analog Devices) - 10

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