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I've read in various places (such as here) that you can't safely feed an input voltage to an op-amp / comparator that's above its supply voltage, even when within maximum ratings.

However, the datasheet for the MIC6270 states a 2-36V supply, and also states (general note 6, page 2):

Positive excursions of input voltage may exceed the power supply level. As long as the other voltage remains within the common-mode range, the comparator will provide a proper output state.

This seems pretty clear, but I'd like to double check - am I correct in thinking that this means the input voltage can above the supply voltage?

(Aside: my potential use case here is a signal that will be fed into the comparator of around 18V on one input and ground on the other. Supply will be 3V and the comparator output fed directly to a microcontroller pin. I was going to put a voltage divider on the signals going into the comparator, but if I don't need to do that then great.)

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    \$\begingroup\$ the data sheet also says abs max input voltage 36v, ie not dependent on power supply, so it appears so. However, if this is a design feature of this comparator, that it doesn't have diodes input to +ve supply to allow significant excess input voltage, it would be nice to spell it out more clearly. Supply a high input voltage via a current limiting resistor, see if the comparator draws significant input current or not \$\endgroup\$
    – Neil_UK
    Sep 16, 2017 at 20:25
  • \$\begingroup\$ @Neil_UK Sorry, I missed the maximum input voltage off the question (supply voltage is also listed as 2-36V though.) But yes, if it was written clearly in various parts of the datasheet I'd be more confident this was the case, the fact it's reasonably unusual and buried in a general note made me a bit wary. I'll order some and try the experiment you suggest, then see what the result is! \$\endgroup\$
    – berry120
    Sep 17, 2017 at 8:43

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In addition to note 6, note 4 of the "Electrical characteristics" shown in the datasheet also explicitly states:

The input common-mode voltage, VIN+, or VIN– must not go below –0.3V. The upper end of the common-mode voltage range is V+ – 1.5V at 25°C, but either or both inputs can go to +36Vdc without damage, independent of VV+.

Therefore, if the inputs are within 36V, there won't be damage, but the common mode range must be within V+-1.5V, otherwise the part might not work correctly.

In your case, if I understood correctly, you want to power it with 3V, and put one input to 18V. Regardless the other input, the common-mode voltage will be larger or equal to 9V, therefore it will exceed 3V - 1.5V = 1.5V. Therefore your solution might not work. Use a resistor divider if possible, to bring the VCM within the specified range.

EDIT: (following what @berry120 got from Microchip as an answer).

Despite note 4, Microchip answered that it's enough that at least one of the two inputs is within the common mode voltage range.

In fact, here's what Microchip replied:

Most comparators (this device included) only require that one input remains in common mode range for it to function properly. As long as one input is in the common range mode, the other input can accept a voltage greater than Vcc. If you power the device with 3V, you must ensure that one of the inputs does not go over 1V or below -0.3V for the whole temperature range.

Still, if your signal goes from 0V to 18V, then comparing that signal (or better, its divided version) with 0 will lead to unpredictable results due to offsets. To check if 18V is above a certain value, compare it to a value larger than 0 (e.g. 0.5V or so), so that, when the input voltage goes to 0, the comparator will compare 0.5 to 0, and not 0 to 0 (plus offsets).

By the way, is not so uncommon that in some IC you can feed a voltage higher than the supply: think about a 5-V tolerant IC, powered at 3.3V.

In addition, on some ICs, you can also safely tie an open drain/collector output above Vcc/Vdd. For instance the output of comparator LM311 can be tied up to 40V (abs maximum rating), even if you power it at 5V.

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  • \$\begingroup\$ Thanks for the answer! I did look at note 4, but note 6 seems to specifically state that only one of the inputs needs to be within the common-mode voltage range for it to operate correctly. In my case one of the signals will always be at 0V (it's a differential signal with each input swapping between around 18V to 0V), so would this be ok? Or have I misunderstood something (highly possible!) \$\endgroup\$
    – berry120
    Sep 17, 2017 at 14:34
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    \$\begingroup\$ Actually note 6 might seem to actually say what you stated. But note 4 explicitly say "common mode voltage" ... Which is basically the average value of the two inputs. Still note 6 might have its sense, i.e. if with both inputs within the VCM range you have a correct output, then if you rise the higher input above the VCM range, the output will be still correct (no phase reversal). \$\endgroup\$
    – next-hack
    Sep 17, 2017 at 15:16
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    \$\begingroup\$ Hmm, I agree - it's really not too clear! I might try firing off a question to Microchip for clarification. If I hear back I'll let you know. \$\endgroup\$
    – berry120
    Sep 18, 2017 at 9:55
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    \$\begingroup\$ Microchip replied: "Most comparators (this device included) only require that one input remains in common mode range for it to function properly. As long as one input is in the common range mode, the other input can accept a voltage greater than Vcc. If you power the device with 3V, you must ensure that one of the inputs does not go over 1V or below -0.3V for the whole temperature range." Bolding is mine, and I still agree that their use of common mode voltage range seems odd, but it would appear this validates that the comparator should be ok for this application. \$\endgroup\$
    – berry120
    Sep 19, 2017 at 12:02

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