2
\$\begingroup\$

I'm building a high-voltage regulator whose error amplifier is an op-amp.

I'm relatively new to op-amps and currently using cheap single-supply LM324 so mistakes don't cost much. The error amplifier section is fed from a separate 5 V DC supply. I am trying to implement proper protection of the op-amp and got mixed up a bit while looking at the datasheet.

In the maximum ratings section there is this footnote:

For supply voltages less than 32 V, the absolute maximum input voltage is equal to the supply voltage.

There is another footnote in the DC characteristics section:

The upper end of the common mode voltage range is VCC −1.7 V, but either or both inputs can go to +32 V without damage, independent of the magnitude of VCC.

I thought that footnote simply stated that common-mode voltage can go to +32 V without damage, but then the second part "either or both inputs can go to +32 V without damage independent of the magnitude of VCC" seems to contradict the first footnote in cases where Vcc is less than 32 V.

What am I missing?

I need to fully understand these concepts to implement proper protection in boundary conditions, like power-on, power-off, VCC supply failure or sag while the HV is still on, etc. The error amplifier input is connected to a voltage divider chain whose top voltage is in the range 150..300 V and maybe more, once I correctly tackle the design.

For completeness here's the relevant section of the regulator - it is a shunt design. What you don't see at the left is a constant-current source feeding the shunt. The missing part on the op-amp input is a possible bias current offset compensation resistor, I'm still undecided as to whether it's necessary since the voltage divider tap can be adjusted over a small range. Shunt regulator

March 8: conclusions

Key take-aways after going through recommended readings and a chapter on protection from a Walter Jung book on op-amp protection. As hinted in replies it appears that best practices are to NOT rely on built-in protection schemes or maximum ratings and rather bind the IC into a well-defined input and output voltage and current range. Integrated protections should be considered only as last-ditch protection.

This allows flexibility in op-amp selection and replacement as one is no longer relying on integrated protection that may be absent in other replacement parts. It mainly involves a couple series resistors and a bundle of diodes of all types, Schottkys in places where reverse voltage limitation has to be less than 0.3V, low-leakeage regular ones otherwise, and back-to-back zeners for output voltage limitation.

In my project there's also high voltage present at power-on due to capacitor C1 charging so the need for a shunt Vcc regulator that can sink current : TL431 and an extra 18V zener to protect it. The design now looks like the image below and so far boundary condition simulations seem to keep the op-amp in a secure operating environment at all times. Now I'm looking to see if back-to-back zeners could be replaced by MOVs.

Latest schematic revision

Thank to all who replied, this was an interesting thread.

-Joris

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Please edit the post and add in a link to the datasheet you're using. Table number references would save us some time too. \$\endgroup\$
    – Transistor
    Commented Mar 4 at 23:22
  • \$\begingroup\$ @Transistor that is embarassing, I'm using a 2005 datasheet from my own library and the first footnote has been removed from the newer versions online... That may be that modern versions of the IC don't have that limitation but my stock is NOS. Do you want me to link my copy online from say Dropbox? \$\endgroup\$
    – Joe
    Commented Mar 4 at 23:44
  • 1
    \$\begingroup\$ Don't worry. I probably have a couple of 40 year-old LM324s in my garage. I was just wondering what the context was and if you might have missed something. Spehro is much more knowledgeable. I used them for hobby work. He uses them for a living. \$\endgroup\$
    – Transistor
    Commented Mar 4 at 23:53
  • \$\begingroup\$ Semantically there is a difference between the common mode and the signal. So technically I don't see a contradiction. Does that help? \$\endgroup\$
    – HarryH
    Commented Mar 5 at 0:42
  • \$\begingroup\$ Input common mode range can go up to 32V without damaging doesn't mean it can function properly \$\endgroup\$
    – Willis Lin
    Commented Mar 5 at 3:20

1 Answer 1

5
\$\begingroup\$

Either input (or both) can go to 32V without damage regardless of Vcc. This is very often not the case with op-amps, many require the input to be no more (or not much more) than Vcc to prevent damage, so don't get used to it. Sometimes they don't like differential voltages more than a diode drop or two. Depends on the op-amp and not always obvious from a first glance at the datasheet.

If you want it to work as an op-amp then both inputs should be 1.7V (or thereabouts, check the detailed data over temperature etc.) below Vcc (and higher than about GND).

If you want it to work as a comparator then at least one input must be within the above mentioned common-mode range. I don't know if the datasheet says that, but it's true.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Thanks Spehro for your reply. The device is used as an op-amp and the error signal is offset at the same voltage as the reference, 2.5V common-mode voltage and Vcc is 5V . Currently there are the typical reverse-connected parallel diodes between the inputs and a diode from the HV input to Vcc to protect against the error voltage going higher than the op-amp supply. The reference is fed from Vcc so no protection needed there since it can't get higher that its supply. \$\endgroup\$
    – Joe
    Commented Mar 4 at 23:41
  • 1
    \$\begingroup\$ It should work fine then, but keep in mind that if you connect a diode to the 5V supply and the input goes higher than a diode drop above 5V, that supply rail would need to sink current to keep the voltage from rising. Most regulators don't actively sink current. So you could burn out everything connected to that 5V rail that could not withstand the higher voltage. I hope that is clear. It will, however, work in simulation with a 5V SPICE source because they can sink current very nicely. \$\endgroup\$ Commented Mar 5 at 0:58
  • \$\begingroup\$ perfectly clear, thanks. Boy do I still have a lot to learn, didn't thought about the external DC supply ability to sink current... That would probably ended badly. I think I'll go for an idea I had brewing for a while which is implement Vcc supply on-board as a 2nd TL431 set for 5V - then it can sink up to 100mA current. This will also allow for more flexibility for the external DC supply. \$\endgroup\$
    – Joe
    Commented Mar 5 at 4:21

Your Answer

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

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