1
\$\begingroup\$

I'm trying to make an input protection stage where the input to an ADC should be between 0V to 5V. And the following circuit seems working in simulation:

enter image description here

And here are the Vin and voltages as node a and node b(where node b will be coupled to an ADC which has range 0V to 5V):

enter image description here

And node a in detail looks like following:

enter image description here

As you see the clamping at node a starts from -0.7V and exceeds 5.6V. So the buffer takes care of this problem and produces a 0V to 5V for a -10V to +10V input.

But it also means that the node a input to the non-inverting input of the opamp has to be able to handle -0.7V and can go up to 5V. So if the opamp is choses as rail to rail it can go up to 5V since the Vcc is 5V.

But how about handling -0.7V input at node a? What parameter is this opamp datasheet shows the maximum negative input voltage with respect to ground. I couldn't find in ratings section.

\$\endgroup\$
3
  • \$\begingroup\$ The big question is not whether it can protect your ADC but whether it is fast enough and accurate enough so as not to degrade your precious signal. You need to specify what you can tolerate in terms of speed, bandwidth and accuracy. \$\endgroup\$
    – Andy aka
    Dec 10, 2022 at 17:06
  • \$\begingroup\$ What's R1 intended to be doing? \$\endgroup\$
    – brhans
    Dec 10, 2022 at 18:07
  • \$\begingroup\$ @brhans I think that generally is a good question and worthy of a formal question (I don't recall seeing one on that subject). I never use a resistor in the unity gain feedback but some folk suggest it's good for stabilization. I think it's likely to cause HF noise. \$\endgroup\$
    – Andy aka
    Dec 10, 2022 at 18:21

2 Answers 2

2
\$\begingroup\$

The opamp specifies "Maximum input current: 25mA" in its absolute maximum ratings. That means you can apply any voltage to its inputs as long as the current is limited to 25mA.

You'll have to increase the value of your series resistor to at least 1k Ohms.

\$\endgroup\$
8
  • \$\begingroup\$ Hi thanks for the answer. What is the reason for min 1k series resistor? \$\endgroup\$
    – pnatk
    Dec 11, 2022 at 16:50
  • \$\begingroup\$ Your voltage source can go down to -10V while the OpAmp's input is clamped to 0V. That's 10V across the 100 Ohm resistor, which results in 100mA. That's too much. 1k Ohms gives you a maximum clamping current of 10mA, which is still high, but not destructive. \$\endgroup\$ Dec 11, 2022 at 18:34
  • \$\begingroup\$ 100mA is high for what? For diodes? \$\endgroup\$
    – pnatk
    Dec 11, 2022 at 19:13
  • \$\begingroup\$ No, for your chip. It lists 25mA as the absolute maximum current, after all, so you shouldn't go anywhere near that current. 10mA should be safe. \$\endgroup\$ Dec 11, 2022 at 22:15
  • \$\begingroup\$ Im sorry but I still dont understand. Why do you think opamp inputs will pass 100mA? Wouldnt the current follow diode and resistor path? \$\endgroup\$
    – pnatk
    Dec 15, 2022 at 18:45
1
\$\begingroup\$

There is a difference between the maximum negative voltage that will damage the input, and the maximum negative voltage that the OP-amp will still process correctly. Many OP-amps suddenly experience "phase reversal": they invert the sign on the output when the input voltage is outside a specified range, leading to unexpected results.

In particular, with the OP-amp you chose, LT14090, the spec sheet says "There is no output phase reversal for inputs up to 22V below V–. " (page 8, top of right-side column).

\$\endgroup\$
3
  • 1
    \$\begingroup\$ I haven't seen an op amp that exhibits phase reversal that's newer than the 741 or so; modern ones are designed to avoid it (though there's some distortion near where phase reversal would happen on a lot of them). \$\endgroup\$
    – Hearth
    Dec 10, 2022 at 17:23
  • 1
    \$\begingroup\$ @Hearth The popular LM358 exhibits phase reversal. I've had it happen to me. \$\endgroup\$ Dec 11, 2022 at 18:35
  • 1
    \$\begingroup\$ @JonathanS. The LM358 is pretty much on par with the 741 in my mind: very obsolete and far too overused. \$\endgroup\$
    – Hearth
    Dec 11, 2022 at 23:20

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.