Op-amps have nearly zero output impedance. So what happens if you accidentally short the output of a working op-amp to ground? Ohm's law says the op-amp will try to deliver infinite current. It will fail, but it will supply a big enough current to fry itself, and so die trying.
The earliest successful op-amps (the Fairchild uA702 and uA709, both creations of Bob Widlar's genius) were susceptible to instant death if the output was short-circuited. The uA702 cost $300, a lot of money in 1964, so this was a rather serious problem.
The next generation of op-amps fixed this by adding internal protection circuitry that monitors the current drawn from the output pin. If that current exceeds a safe limit, the protection circuitry kicks in, and clips current peaks to a safe maximum value, so that the entire IC doesn't die an instant death, as the older op-amps did.
So when you tried to drive a too-low load with an op-amp, and it started clipping, be thankful - that clipping is what saved you from yourself, and kept the op amp alive in spite of the abuse you were inflicting on it!
Yes, quite correct. The plate is connected to B+ voltage, which (ideally) has no AC voltage on it at all. In fact, as far as AC is concerned, B+ is the same as signal ground - there is zero volts (AC) on it.
In your cathode follower, there will, however, be an AC voltage between plate and cathode, and also an AC voltage between cathode and ground. And a third AC voltage between grid and cathode.
It might help to remember that a triode acts like a voltage-controlled current source (though not a very good current source.) In other words, the (AC or DC or both) voltage between grid and cathode decides how much current flows between anode and cathode. AC voltage at anode or cathode or both is entirely optional!
To illustrate that last point, you could in principle ground the cathode, and wire the anode directly to B+ through a (nearly) zero-ohm milliammeter. The voltage you apply between grid and cathode (AC, DC, or both) will then set the current through the meter - but there will be no DC or AC voltage at the cathode, and only DC - no AC - voltage at the anode as well!
-Gnobuddy
Like most comments about all op-amps, true and false in equal measure.
Last tricky op-amp problem I had to solve was caused by the op amp's 340 ohm output impedance, probably a real resistor to avoid blowing up problems.
Can't remember which one it was, nor whose version but I can remember that number. By the way, the Ti version of the TL-072, which I used to use in buckets, has about 160ohms of Zout (see page 26)
So YMMV, folks!
My apologies, I was imprecise in the interests of brevity.Like most comments about all op-amps, true and false in equal measure.
To put it in context, 160 ohms open-loop Ro is still low enough to deliver nearly 100 mA from the usual 15 volt supply rail, producing more than a chip-melting watt of dissipation. And normal usage for an op-amp involves, say, 60 dB of negative feedback, which would reduce 160 ohms to 0.16 ohms at low frequencies.
YMMV applies to some degree to all engineering. And analogue electronics may be the most imprecise of all engineering fields, where it's routine to have components with tens of percent tolerance on their values. Amazingly, we still manage to make working circuits with them. (Just try making a working internal combustion engine with pistons that vary +/- 10% in size!)
-Gnobuddy
Now that brings the nightmare screaming back!My apologies, I was imprecise in the interests of brevity.
To put it in context, 160 ohms open-loop Ro is still low enough to deliver nearly 100 mA from the usual 15 volt supply rail, producing more than a chip-melting watt of dissipation. And normal usage for an op-amp involves, say, 60 dB of negative feedback, which would reduce 160 ohms to 0.16 ohms at low frequencies.
The design engineer had done just this, lots of feedback. And then was trying to drive a relatively low-Z load.
The circuit worked fine in spice but misbehaved in the real world with signal swings of more than a few tens of millivolts. Why? The output stage
was running out of headroom as it was powered by single 5V rail.
Yet the output pin was barely moving at all!
Well, at least he (or she) got one thing right.
This sounds like a bad case of not reading the datasheet. There are specialist op-amps that work happily off a single 5V rail, and specialist op-amps that happily deliver lots of output current, but a device that can do both those at the same time is much more of a specialist.
I know they do exist, as every large brushed-motor radio control servo has one inside, driving the relatively beefy servo motor off a +5V or +6V single supply rail.
I have read plenty of confused forum posts when it comes to the issue of output impedance vs output current delivery. Usually the question is about pentodes, and the confusion is about the apparent contradiction that a device with an 80 kilo ohm output resistance can deliver 150 mA from a 400 volt power supply.
It seems your design engineer found the opposite corner of the same confusion, this time about a device with low output impedance but quite limited output current capability. It's quite an unexpected combination if one is used to just thinking about Thevenin equivalent circuits. Only the nuances of negative feedback explain why you can end up with this combination.
I hope we are not confusing Max999 with this discussion of a fairly unusual corner-case.
If we look at the forest rather than the trees, what I said earlier remains true: speaking quite generally, a 12AX7 runs off roughly ten times more supply voltage, and can deliver roughly ten times less output current, than a typical op-amp circuit, which means the tube can only drive a load that's at least a hundred times stiffer than an op-amp can.
But when you look close-up at the details, of course there are a tonne of differences between tubes and op-amps.
-Gnobuddy
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