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What is the circuit size complexity and / or circuit depth complexity of an op-amp?

This is for comparative purposes, so I'm really after a ballpark figure and any op-amp is fine (although the simpler circuit is better, as I guess I'm after a lower bound). My thoughts are that I would either count the transistors (I've found an op-amp design with 4 of them), but I want to know what the convention is. I'm a computer scientist rather than an electronic engineer, so I apologise if this is simple to work out, however I've not managed to find an answer with an internet search.

I'm actually after the circuit size / circuit depth complexity of a current-to-voltage converter (or transimpedence converter which I think will do the job) and that seems to use a resistor and op-amp in parallel, so I think I would count the circuit depth complexity of just the op-amp.

Please help.

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  • \$\begingroup\$ Is this related to Kolmogorov complexity? \$\endgroup\$
    – pjc50
    Mar 10, 2015 at 12:36

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There is no actual convention as to a number of transistors- the designers themselves don't particularly care. What matters is the performance, the manufacturability, the processes used (extra layers, certain types of parts such as good PNP transistors, CMOS, higher voltage-handling ability, and precision resistors require special or extra process steps which add to the cost) and the area of silicon used in that process. A transistor may use 1/10 or 1/100 the area of a resistor so using more transistors and fewer resistors may make for a better design. Capacitors tend to be very expensive and there is typically only one per op-amp. Inductors of reasonable value are more expensive again (or impossible) so they're rarely seen at all.

Those factors influence the commercial considerations- what the op-amp is worth to the customer (performance) and what it will cost to manufacture (the other factors).

If you still want to know, you can simply count what parts matter to you for whatever reason (if a schematic is provided in a given data sheet). You'll have to take it with a grain of salt, however, since the schematic may show (say) a differential pair (two transistors) in one particular case, but I happen to know that in that case each transistor is actually something like 100 small transistors in parallel (they're interposed with each other in a tight matrix so as to minimize Vos, TCVos and changes in Vos due to gradients across the chip surface.

Many modern parts have datasheets with neither a schematic nor a die photo (from which you might be able to suss something out about complexity). You could decap a part and look at it under a microscope or find someone who did, otherwise it's hard to tell.

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  • \$\begingroup\$ Why are capacitances expensive to fabricate. A reverse biased diode yields a capacitor which can be easily fabricated \$\endgroup\$ Mar 10, 2015 at 12:19
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    \$\begingroup\$ Tiny capacitances are inevitable and usually unwanted. To get a useful value in the context of an op-amp compensation network (say 5pF to 20pF) requires a lot of acreage on the silicon. Reverse biased diodes create a capacitor that requires reverse bias and that has a capacitance that varies greatly with the voltage across the cap, so they're unsuitable for many purposes (but certainly not useless in general). Silicon dioxide dielectric is very high performance, but you can't get much capacitance in the space of a handful of transistors. \$\endgroup\$ Mar 10, 2015 at 12:28
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    \$\begingroup\$ The proper unit of area on silicon chips is the nanoacre :-) \$\endgroup\$
    – Dave Tweed
    Mar 10, 2015 at 13:18
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    \$\begingroup\$ @DaveTweed 1 nanoacre = 4mm^2 (within ~1%). Nice. \$\endgroup\$ Mar 10, 2015 at 13:22

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