Thermodynamics of A/D Conversion

1. Thermodynamics of A/D Conversion

2. Problem • What is the minimum power required to perform n-bit A/D conversion on a signal with a certain bandwidth? • Would like a bound independent of technology and A/D architecture • But the flip side is that THIS BOUND MAY NOT BE PRACTICAL • Only would like to know what’s physically possible and what’s not • Eventually need to consider power required to overcome: • Thermal noise • Mismatch • Clock jitter • and power required for discretize the analog voltage

3. Thermal Noise • Most obviously causes power dissipation in the sample and hold circuits • Have to spend energy to overcome <Vn2> = kT/C noise • If the stream of signal samples is uncorrelated and has variance of <Vin2>= SNR·<Vn2> … • …we spend average energy <E> = C·<Vin2> = kT·SNR per sample • But is kT/C noise unavoidable? • Even if it is, can we construct a more efficient sampler?

4. kT/C Noise • Capacitance is not the only memory element • Can use capacitance, inductance, mass on a spring – anything that can store energy • But as long as there is dissipation (equivalent of a resistor) in the system, there is still equivalent noise! • Thus <E> = kT·SNR per sample still has to dissipated • Note that that it’s independent of C, L, kspring, etc. • But does any sampler have to have a dissipative element (resistor or equivalent)?

5. Dissipation in a Sampler • For example, try to construct a capacitive sampler without a resistor: • Cannot transmit any meaningful information to the capacitor, because system has infinitely narrow bandwidth • Can charge and discharge the capacitor without any dissipation • But can not erase the previous sample without memorizing what it was somewhere else • Fundamentally need a “trash can” to dump memorized information • This follows from the 2nd Law of Thermodynamics

6. “Trash Can” • In electrical circuits, “trash can” is usually a thermal environment • We dump out old information by connecting to a dissipative element (resistor) • But then we always pick up thermal noise • Therefore, ANY system that memorizes input information must have thermal noise and dissipate energy to overcome this noise • Can we construct something more efficient than normal sampler? • Yes, by making it reversible • means at every step process can go either forward or back • then no energy is needlessly wasted and maximum efficiency is obtained • similar to Carnot cycle in thermodynamics

7. Carnot Sampler • Step 1 – connect C=C1 to voltage source and slowly ramp the voltage from 0V to Vsignal: • Can deposit signal charge qsignal=C1·Vsignal onto C with arbitrary low energy loss in R by making ramp rate >> RC1 • Voltage source only supplies average energy W1= <qsignal2> / ( 2C1 ) • Step 2 – disconnect C from the input and reset voltage source to 0V: • Signal charge is memorized on the capacitor together with thermal noise <qnoise2> • Average energy stored on C=C1 is • <E1>= (<qsignal2>+ <qnoise2>) / (2C1) • SNR = <qsignal2> / <qnoise2>

8. Carnot Sampler • Step 3 – let C increase from C1 to C2 and use the energy released in this process: • Charge remains the same, but average energy stored reduces to • <E2>= (<qsignal2>+ <qnoise2>) / (2C2) • Want to make <E2> equal to energy of the thermal environment kT/2 • The excess energy W2=<E1>-<E2> can be recovered as useful work • Step 4 – connect C=C2 back to the signal source: • Average energy <E2> stored on the capacitor is equal to the average energy of thermal noise kT/2 • Thus, no energy exchange takes place, and no energy is wasted

9. Carnot Sampler • Step 5 – prepare for the next sample by decreasing C from C2 back to C1: • Requires external work • W3 = 0.5 · kT · ln( C2 / C1 ) • This work is dissipated in R and cannot be recovered • The total average energy spent per sample is only • Qwaste = W1-W2+W3 = 0.5 · kT · ln( C2 / C1 ) • But C1 = <qnoise2> / kT, and C2 = (<qsignal2>+<qnoise2>) / kT • Thus, Qwaste=0.5 · kT · ln((<qsignal2>+<qnoise2>)/<qnoise2> )= • Qwaste=0.5 · kT · ln( SNR + 1 ) is the absolute lower bound on sampler energy dissipation per sample

10. Quantization • So far we only considered analog sampling • Similarly, some energy fundamentally has to be spent for memorizing and erasing digital outputs as a function of bit error rate • But do we really have to “memorize” analog value? • Do we really have to “memorize” digital outputs? • Does energy fundamentally have to be spent on any other intermediate process? • Not sure, but have some ideas…

11. Conclusions • Found theoretical lower bound on energy required to store and erase analog signal with certain SNR • Still need to figure out what’s the equivalent bound on energy for quantization • What are the other possible fundamental sources of energy dissipation? Jitter? Mismatch?