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MCP4822-E/MS датащи(PDF) 22 Page - Microchip Technology |
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MCP4822-E/MS датащи(HTML) 22 Page - Microchip Technology |
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22 / 36 page ![]() MCP4821/MCP4822 DS21953A-page 22 © 2005 Microchip Technology Inc. 6.6 Bipolar Operation Bipolar operation is achievable using the MCP482X devices by using an external operational amplifier (op amp). This configuration is desirable due to the wide variety and availability of op amps. This allows a gen- eral purpose DAC, with its cost and availability advan- tages, to meet almost any desired output voltage range, power and noise performance. Example 6-3 illustrates a simple bipolar voltage source configuration. R1 and R2 allow the gain to be selected, while R3 and R4 shift the DAC's output to a selected offset. Note that R4 can be tied to VDD, instead of AVSS, if a higher offset is desired. Note that a pull-up to VDD could be used, instead of R4 or in addition to R4, if a higher offset is desired. EXAMPLE 6-3: Digitally-Controlled Bipolar Voltage Source. 6.6.1 DESIGN A BIPOLAR DAC USING EXAMPLE 6-3 An output step magnitude of 1 mV, with an output range of ±2.05V, is desired for a particular application. 1. Calculate the range: +2.05V – (-2.05V) = 4.1V. 2. Calculate the resolution needed: 4.1V/1 mV = 4100 Since 212 = 4096, 12-bit resolution is desired. 3. The amplifier gain (R2/R1), multiplied by full- scale VOUT (4.096V), must be equal to the desired minimum output to achieve bipolar operation. Since any gain can be realized by choosing resistor values (R1+R2), the VREF value must be selected first. If a VREF of 4.096V is used (G=2), solve for the amplifier’s gain by setting the DAC to 0, knowing that the output needs to be -2.05V. The equation can be simplified to: 4. Next, solve for R3 and R4 by setting the DAC to 4096, knowing that the output needs to be +2.05V. MCP482X VDD VDD SPI™ 3 VOUT R3 R4 R2 R1 VIN+ G = Gain select (1x or 2x) D = Digital value of DAC (0 – 4096) 0.1 µF VCC+ VCC– V OUT 2.048 G D 2 12 ------- ⋅ = V IN+ V OUTR4 R 3 R 4 + -------------------- = VO V O V IN+ 1 R 2 R 1 ------ + ⎝⎠ ⎛⎞ V DD R 2 R 1 ------ ⎝⎠ ⎛⎞ – = R 2 – R 1 --------- 2.05 – 4.096V ----------------- = If R1 = 20 kΩ and R2 = 10 kΩ, the gain will be 0.5 R 2 R 1 ------ 1 2 --- = R 4 R 3 R 4 + () ----------------------- 2.05V 0.5 4.096V ⋅ () + 1.5 4.096V ⋅ ------------------------------------------------------- 2 3 --- == If R4 = 20 kΩ, then R3 = 10 kΩ |
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