--- title: "4-bit inverted R-2R DAC and why better" subject: "Advanced Electronics" chapter: 3 key: "semester_4/advanced_electronics:inverted-r2r-4bit-better" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/inverted-r2r-4bit-better site: BCT Engineering --- # 4-bit inverted R-2R DAC and why better Advanced Electronics · Chapter 3: Digital-to-Analog and Analog-to-Digital Conversion - Asked in: 2069 Chaitra, 2+4+1 marks - Video: [Inverted R 2R Ladder DAC | Analog to Digital and Digital to Analog Convertors in EXTC](https://www.youtube.com/watch?v=7SYyStFQDyg) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/inverted-r2r-4bit-better - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-3 ## Question Draw circuit for 4-bit inverted R-2R ladder network type DAC. Find expression for output voltage and explain why an inverted R-2R ladder DAC is better than R-2R ladder DAC. ## Short answer Inverted R-2R DAC output: Vo = −(Rf/R)·Vref·(∑ bᵢ/2ⁱ⁺¹). It is better because node voltages stay constant (switches at virtual/true ground), stray capacitances do not charge, giving fast settling, no glitches, and constant reference load. ## Answer An inverted (current-mode) R-2R ladder DAC applies Vref to the ladder input. The bits switch each 2R leg's current either to the op-amp's virtual ground (bit 1) or to true ground (bit 0). The op-amp converts the total current into a voltage. #### Circuit (4-bit) ```text Vref | I I/2 I/4 I/8 +---R----+----R----+----R----+ | | | | 2R 2R 2R 2R 2R |I/2 |I/4 |I/8 |I/16 | S3 S2 S1 S0 GND (b3) (b2) (b1) (b0) 1 -> to Iout line 0 -> to GND Iout line --> (-) op-amp, Rf from output to (-) (+) to GND Vo = −Iout·Rf ``` #### Derivation Both switch positions are at 0 V (virtual ground or ground), so the ladder always sees the same load. The resistance looking into the ladder from Vref is R, so: ```text I = Vref / R ``` At each node the current divides equally (2R down, 2R to the right), so the branch currents are I/2, I/4, I/8, I/16. The current reaching the op-amp: ```text Iout = (Vref/R)(b3/2 + b2/4 + b1/8 + b0/16) = (Vref/R)(D/16) ``` The op-amp output: ```text Vo = −Rf · Iout = −(Rf/R) · Vref (b3/2 + b2/4 + b1/8 + b0/16) = −(Rf/R) · Vref · D / 2⁴ ``` With Rf = R: Vo = −Vref·D/16. For example, 1000 gives −Vref/2. #### Why the inverted R-2R ladder is better - **Node voltages stay constant** whatever the code, because each switch only moves current between two 0 V points. Stray capacitances do not charge and discharge, so the DAC settles fast with fewer glitches. In the non-inverted ladder the node voltages change with every bit. - The reference source sees a constant load (R), so it stays steady. - The switches always work at ground potential, so simple MOS switches give accurate, equal behaviour. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the inverted R-2R ladder DAC, where does each 2R leg's current go? A. Always to the op-amp output B. To virtual ground when the bit is 1 and to true ground when the bit is 0 C. To Vref when the bit is 1 D. To -Vref when the bit is 0 Answer: B. Both switch positions are at 0 V, so the current only moves between virtual ground and ground. (Key point: Switch positions) 2. For a 4-bit inverted R-2R DAC with Rf = R, what is Vo? A. Vo = Vref x D/16 B. Vo = -Vref x D/16 C. Vo = -Vref x D/8 D. Vo = -Vref x D/15 Answer: B. Vo = -(Rf/R) Vref D/2^4, which is -Vref D/16 for Rf = R. (Key point: Output expression) 3. What is the resistance looking into the ladder from Vref, and the resulting input current? A. 2R, I = Vref/2R B. R, I = Vref/R C. R/2, I = 2Vref/R D. 4R, I = Vref/4R Answer: B. The ladder always looks like R to the reference, so I = Vref/R, which then halves at each node. (Key point: Resistance seen by Vref) 4. Why is the inverted R-2R ladder DAC better than the ordinary R-2R ladder? A. It uses fewer resistors B. Node voltages stay constant, so stray capacitances do not charge and the DAC settles faster with fewer glitches C. It needs no op-amp D. It gives a positive output Answer: B. Switching only moves current between two 0 V points, so node voltages do not change with the code. 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