--- title: "Bipolar DAC output for 0000, 1000, 1111" subject: "Advanced Electronics" chapter: 3 key: "semester_4/advanced_electronics:bipolar-dac-inputs-0000-1111" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/bipolar-dac-inputs-0000-1111 site: BCT Engineering --- # Bipolar DAC output for 0000, 1000, 1111 Advanced Electronics · Chapter 3: Digital-to-Analog and Analog-to-Digital Conversion - Asked in: 2073 Chaitra, 2+4+1 marks - Video: [Bipolar Digital to Analog Converter|| Part 6 ||](https://www.youtube.com/watch?v=2ysBkcyVZbY) — Hack 999 - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/bipolar-dac-inputs-0000-1111 - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-3 ## Question List out the advantages of bipolar D/A converter over Binary weighted resistor. Express output voltage of bipolar D/A converter. What happens when input is 0000, 1000 and 1111? ## Short answer Advantages: handles bipolar signals; uses only two resistor values (easier IC); zero at mid-code. Bipolar output: Vo = Vref(2D/2^n − 1). For n=4: 0000 → −Vref, 1000 → 0, 1111 → +7/8 Vref. ## Answer A bipolar DAC gives both positive and negative output voltages. It does this by adding a fixed offset of half full scale to a normal (unipolar) ladder DAC, using offset-binary coding. #### Advantages over the binary weighted resistor DAC - It converts signed (bipolar) signals. The binary weighted DAC gives only one polarity. - It is usually built on an R-2R ladder, so it needs only two resistor values. The binary weighted DAC needs a wide spread (R to 2ⁿ⁻¹R), which is hard to make accurately. - It gives better accuracy and is easier to make as an IC for large n. - Zero output sits at mid-code, so it suits AC signals, audio, and servo control that needs both directions. #### Output voltage ```text Vref --[R-2R ladder]--> I_D --+ | Rf -Vref --[ R_off ]--> I_off ---+---/\/\/---+ | | (-)\ | >---------+--> Vo (+)/ GND ``` The ladder gives a current proportional to the code: I_D = (Vref/R)(D/2ⁿ). The offset resistor R_off = 2R draws a constant current I_off = −Vref/(2R), which is half of full scale. With Rf = 2R: ```text Vo = −Rf (I_D + I_off) = −2R [ (Vref/R)(D/2ⁿ) − Vref/(2R) ] = −Vref (2D/2ⁿ − 1) ``` A unity-gain inverter at the output (or swapping the polarities of the two references) gives the standard textbook form: ```text Vo = Vref · (2D/2ⁿ − 1) = Vref · (D − 2ⁿ⁻¹) / 2ⁿ⁻¹ ``` where D = b3·8 + b2·4 + b1·2 + b0 for a 4-bit converter. #### Outputs for 0000, 1000 and 1111 (n = 4) | Input | D | Vo = Vref(2D/16 − 1) | Meaning | |---|---|---|---| | 0000 | 0 | −Vref | Negative full scale | | 1000 | 8 | 0 | Mid-scale, zero | | 1111 | 15 | +7/8 Vref = 0.875 Vref | Positive full scale (1 LSB below +Vref) | So the output swings from −Vref to almost +Vref. Code 1000 gives zero, and the MSB works as a sign bit (1 = positive, 0 = negative). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is one advantage of the bipolar DAC over the binary weighted DAC? A. It gives only positive output B. It needs resistors up to 2^(n-1)R C. It handles signed signals, giving both polarities D. It is slower Answer: C. The weighted DAC gives only one polarity. (Key point: Advantage over weighted) 2. The 4-bit bipolar output is: A. Vo = Vref D/16 B. Vo = Vref(D - 16) C. Vo = Vref(2D + 1)/16 D. Vo = Vref(2D/16 - 1) Answer: D. A half-scale offset subtracted from twice the unipolar output. (Key point: Output formula) 3. What is the output for 1000? A. 0 V B. +Vref C. -Vref D. +Vref/2 Answer: A. D = 8 gives 2x8/16 - 1 = 0. (Key point: Input 1000) 4. Outputs for 0000 and 1111 are: A. 0 and +Vref B. -Vref and +0.875 Vref C. -Vref and +Vref D. -0.875 Vref and +Vref Answer: B. D = 0 gives -Vref; D = 15 gives 30/16 - 1 = 0.875 Vref, 1 LSB short of +Vref. 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