Question

Problem 2 (25 pts.) A thermocouple is to be used with a 16 bit A/D converter that has an input range of ov to sv. The voltage output of the thermocouple varies linearly with the temperature over the temperature range of 50°C to 70°C; the corresponding voltages at these temperatures are 2.585 mV and 3.649 mV respectively. Parta) 5 pts. Compute the quantization error of the A/D convertor and the temperature resolution for the entire system. Part b) 10 pts. A non-inverting OP amp is to be used to amplify the voltage signal to reduce the relative quantization error (Gain = * = 1 +). Practical considerations limit the resistances on the OP amp from 1 k12 to 100 k12. Only using one non-inverting OP amp, what is the lowest relative quantization error achievable if the input to the OP amp is 2.585mV? Parte) 10 pts. Using the gain found in part b), if the binary readout of the A/D convertor is 0000111101110110, what is the indicated voltage at the input to the OP amp and the corresponding temperature being sensed by the thermocouple in this experimental system? Assume the binary readout of the A/D is an exact representation of the analog signal.

Answer 1

Given that A thermocouple is to be used with a 16 bit A/D converter that has an input range of OV to 5V. Given 16 bit A/D converter input range = O to 5V (Tmin) at 50°C = 2.585 mV (Tmar) at 70°C = 3.649 mV (a) Quantization error eQ = + 1 / EFSR 2 2m m = 16 bit 1 6 Efsr = 5V = eq = + 216 eq = 3.8146 x 10-5V Temperature resolution Trev = (Tmar - Tmin) 2- Trev (70 – 50) 2-16 = = 0.000305 (b) Gain of no inverting amplifier Gain R2 1+ R1 100K 1+ 1K 101

The lowest relative Quantization error is eRQ = Gain x eo = 101 % (-3.8146 x 10-5) ERQ = -3.8527mV (c) The indicated voltage at the input to the OP amp is 1 1 1 1 Gain = V 0+0+0+ =v [0+0+0+3 + + + + +0+ 210 1 +0+ 212 214 + 1 215 + +0 211 101 V(0.06039) V = 1672.462V Note: Hope you like the solution, please upvote.