4
\$\begingroup\$

I haven't done and studies in Op-Amp category yet but I watched a (trusted) video on Youtube and one thing is not quite logical as the next sentence: "No current flows in or out of the inputs of operational amplifier", as says Dave Jones from EEVblog on Youtube (this video: https://www.youtube.com/watch?v=7FYHt5XviKc).

So, I am wondering if he meant like there is a really small amount of current flowing in op-amp on the inputs, or there isn't any of current on inputs at all?

I am sure there must be at least some current on inputs (for base if BJT or for gate if FET/MOSFET for sure).

enter image description here

\$\endgroup\$
12
  • 2
    \$\begingroup\$ (The 741 shouldn't be used for any argument. It's old and not used except by builder hobbyists copying old schematics. But it can be and should be studied for how it worked because it's complexity is relatively low.) The ideal opamp isn't obtainable, of course, but doesn't have any input current. The ideal opamp also has infinite gain. Real opamps will have extremely low input currents and extremely high gain, with different technologies offering different trade-offs in offset voltage and input, bias, and offset current, etc. See: ti.com/lit/an/sloa011/sloa011.pdf \$\endgroup\$
    – jonk
    Jan 15, 2017 at 20:45
  • \$\begingroup\$ @jonk: I used to take solace in the fact that one day the million-pieces stocks of such ancient ICs will simply run empty, and TI et al. simply won't care to re-run them (if they even still have the masks to do so – compare the SA612/NE612 Philips fire disaster); but then I realized that even if that ever happens, the market for replacement in critical circuits is large enough to allow for smaller semiconductor suppliers to operate profitably by making these pricey (but the hobbiest will just assume it's normal a shitty Opamp costs 5€). \$\endgroup\$
    – Marcus Müller
    Jan 15, 2017 at 21:53
  • 2
    \$\begingroup\$ @MarcusMüller Yes. FABs need to keep running. Brokering is one way. Producing detector diodes (or anything, really, that doesn't require a lot of NRE) is another. (Diodes, Inc. was willing to make anything I wanted, if I could spec it for their processes.) I also worked with G-squared's RTP (rapid thermal processing) units, developing the temperature component of the multi-zone thermal control loop system, which are often used for "one-off" wafer production (and/or RTA -- annealing.) But I don't fully understand the marketing side, not caring so much about it. \$\endgroup\$
    – jonk
    Jan 16, 2017 at 13:11
  • 1
    \$\begingroup\$ @MarcusMüller That's kind of where I think things are headed! RTP pulls in the wafer process from weeks/months to mere days or less. It's hard to achieve 400 C per second rates on a 12" wafer without causing it to "potato-chip" itself. But hundreds of zones of control to deal with edge effects, etc., do make it possible. (I was working on these in the mid 1990's to late 1990's and was able to point out a cheaper way to deal with emissivity and surface variation issues that opened the door to economic means to get there.) UltraShuttle is headed in the direction I saw! Thanks for the link! \$\endgroup\$
    – jonk
    Jan 16, 2017 at 14:05
  • 1
    \$\begingroup\$ @MarcusMüller That was really cool. I'm imagining the day where anyone can use freely available (open source) tools on Linux (or Windows), that are also idiot proof and easy to use, to develop their own custom ASIC design and get it fabricated. Should be about as easy as using a schematic layout tool, today. What's sad is that we already have such fantastic tools today and too few bother to learn to use them. When I was growing up, there were 5 periodicals and dozens of suppliers just for telescope makers like me. Today there are zero of both. (No one makes raw glass supplies in hobby qtys.) \$\endgroup\$
    – jonk
    Jan 16, 2017 at 14:14

2 Answers 2

Reset to default
6
\$\begingroup\$

Well spotted! There is an input current, but it can be really small. It's usually specified as "bias current" in the datasheet. For older parts it'll be some number of microamperes. For specially designed JFET inputs for instrumentation amps, it can be as small as femtoamperes, which are really hard to measure. (The part mentioned is the LPC662 which claims 2fA input current!).

For normal analysis purposes you ignore it, but it can matter when you're trying to amplify a small signal.

\$\endgroup\$
4
  • \$\begingroup\$ What makes it so low? What is the purpose of such small amount of current? \$\endgroup\$
    – MucaGinger
    Jan 16, 2017 at 14:02
  • \$\begingroup\$ FET and JFET gates are insulating, which makes it low. They're just not perfect insulators, hence the very small current leaking through "insulators" inside the chip. Or do you mean "why bother reducing it to such a low level"? \$\endgroup\$
    – pjc50
    Jan 16, 2017 at 14:14
  • \$\begingroup\$ Yes, why bother reducing it to such a low level? I mean, if there are FETs in op-amp then the low current level is understandable but how does it go with BJT transistors from inner structure? \$\endgroup\$
    – MucaGinger
    Jan 17, 2017 at 13:55
  • \$\begingroup\$ See e.g. analog.com/media/en/training-seminars/tutorials/MT-038.pdf "Bias current is a problem to the op amp user because it flows in external impedances and produces voltages, which add to system errors. Consider a non-inverting unity gain buffer driven from a source impedance of 1 MΩ. If IB is 10 nA, it will introduce an additional 10 mV of error. This degree of error is not trivial in any system. " \$\endgroup\$
    – pjc50
    Jan 17, 2017 at 14:24
3
\$\begingroup\$

Ideally, op-amp is considered to have infinite input impedance. It is assumed to have zero source/sink current from/to it's input terminals.

However, practical op-amp do have some small input current, which is mentioned in the datasheets as input bias current. There is also some difference between input currents of inverting and non-inverting terminals of op-amp.

You can ignore it for small-signal operation.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.