This does not make sense to me. Can you explain?
I think there might be misunderstanding of how analog computing is used to build a neural network. First, a weight is stored as some analog physical property, typically as charge on a floating gate, or on a capacitor in a DRAM type cell. Second, the multiplication operation is performed by modulating the analog input signal going through the floating gate transistor by the charge on the floating gate (weight). Third, the summation is done via simple summation of the currents. Finally, activation function is performed by an opamp.
Regarding power consumption: 1. A digital computer needs a thousand of transistors to perform multiplication, analog circuit can do it with a single one. 2. Analog NN stores parameters (weights) locally, right where they are needed to perform computation. Digital NN will need lots of memory transfers to bring weights from RAM to ALU, and to store intermediate results.
That's why a properly implemented analog NN will always consume much less power.
That's interesting. What would the circuit be?
> Digital NN will need lots of memory transfers to bring weights from RAM to ALU, and to store intermediate results.
That's not necessarily the case. Cellular neural networks were proposed long ago, for example, and they're digital -- how multiplication happens is independent from the data flow architecture.
> That's why a properly implemented analog NN will always consume much less power.
How do you know that the I^2 cost of operating in the linear regime isn't excessive? I'm totally ignorant on the matter -- I'd love to see a ballpark calculation to understand why it isn't important.
Cellular neural networks were proposed long ago, for example, and they're digital
What is so inherently digital about cellular networks? Can you provide a link to an implementation of a cellular net in digital hardware? How the weights are stored? Where the multiplication happens?
I understood the reasoning to be that to increase the range of accurately representable values in a circuit, you either need to increase the voltage or current used in an analog circuit (to achieve a certain accuracy versus a noise baseline), or devote more bits in a digital circuit. The first gives a linear dependence (or quadratic for I^2 losses) of power on range, the second logarithmic.
To give a bit of a dramatic illustration, if you circuit has on the order of 1 nV of thermal noise and you wanted to do the linear analog equivalent of 64bit integer arithmetic, you would need a signal on the order of 10,000,000,000 V to have enough precision. In fact, in terms of power consumption it's even worse. If the 1 nV signal consumes something like 1 pW, you would need something like the total power output of the Sun (on the order of 10^26 W) -- a bit of an expensive multiplication, no :) ? That's how crazy it is!
Again, if you can get away with less than 8 bits of precision and imperfect linearity the picture changes, but I wouldn't declare it superior a priori without looking at the numbers.
But yes, I understand your point. Both analog and digital implementations have their strengths and weaknesses. If you value power over precision, go with analog. If the opposite - go with digital.
The reason NNs don't exhibit strong error propagation is because of the non-linearities between linear layers that perform operations analogous to threshold/majority voting or the like, which have error correction properties.
Be careful with jumping to conclusions: I never even cited ReLUs or Sigmoids in my post! I don't have any opinion on which non-linearity is better, I only know both are dramatic non-linearities. My claims were about linear circuits. You should use whatever nonlinear element works best in your Neural Network, of course (and I've heard ReLUs have good advantages).
That's true, I feel stupid for not having thought of that!
I'm not an electrical engineer, but with the FETs that modern Intel chips are using, what fraction of their power consumption comes from parasitic gate capacitance, versus other losses?
And if you operated in the linear region, what's the ballpark steady-state I_SD current you'd need on one FET to drive the gate of the next FET?
I think that's what this comes down to: if gate capacitance dominates other losses even in the linear regime, you still win by not having a clock and lots of digital transitions.
You could even imagine exploiting that: apply 'slow' augmentations of the input data that get you the equivalent of a bunch of iterations on a single example batch, while incurring a much smaller fraction of that initial cost because activations aren't going to change nearly as much as switching to a whole new example batch.
Power is mainly lost via leakage (the smaller the transistor, the more it leaks), and via interconnect capacitance, which dominates all other capacitances in modern circuits.
Of course, but the 'ideal' FET has zero gate capacitance, despite that being the way they work.
> Power is mainly lost via leakage (the smaller the transistor, the more it leaks), and via interconnect capacitance, which dominates all other capacitances in modern circuits.
Interconnect meaning things like the buses? There's no reason to want a von Neumann architecture for an analog chip. If that leaves leakage, I suppose an analog chip would be the beneficiary of needing a lot fewer transistors per op.
I don't understand this statement. What do you mean? A FET is a capacitor (gate to channel). If a gate has no capacitance, you have no transistor.
Interconnect means wire. This has nothing to do with von Neumann architecture. If you have wires in your circuit, then you have wire capacitance. As transistors get smaller, that capacitance starts to dominate internal transistor capacitances.