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LT1996CDD Datasheet(PDF) 10 Page - Linear Technology |
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LT1996CDD Datasheet(HTML) 10 Page - Linear Technology |
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10 / 24 page  LT1996 10 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 4pF – + 1996 F01 450k/81 450k/27 450k/9 450k/81 450k/27 450k/9 450k 450k REF 5V VCM 2.5V VDM 0V – + 8 7 6 5 4 9 10 1 2 3 LT1996 VOUT = 117 • VDM 4pF 4pF – + VREF RF RF RG RG 1996 F02 VEXT VINT VCC VEE – + VREF RF RF RG RG 1996 F03 VEXT MAX OR MIN VINT VMORE VCC VEE |
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