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Is it possible to hardwire battery to opamp input, considering that most of the time normal power on opamp will be off ? Battery positive wire is connected to non-inverting input of opamp workiing as a buffer (inverting input is connected to output). In opamp specs the input current is in picoampere range, but is it true for the unpowered state ?

Thank you

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4 Answers 4

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If the input current is in the pA range, it's probably a CMOS opamp. For this class of opamps input bias current is often a few fA typical, a few pA maximum. (for instance National LMC660). Since the gates of the MOSFETs are insulated, input bias current won't increase if the device isn't powered.

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Most ICs have ESD protection diodes on their inputs. That means there is a diode from the input to the positive supply and from the negative supply to the input. These are reverse biased in normal operation. Their purpose is to clip out of range voltages that could damage the rest of the IC.

When the opamp is unpowered, the positive supply is supposedly at zero, so applying more than one diode drop to any input will cause significant current to flow in that input and also sortof possibly apply power to the device thru the diode. That's not a reliable way to power the device, and it could result in latchup and other bad things.

Another point is why are you buffering a battery voltage? Most likely the battery can deliver significantly more current than the opamp can. The buffered output might be a little lower impedance if not exceeding the current limit, but in a battery powered device this current will still come from the battery. The buffer may hold the output voltage close to the input voltage, but the drain on that output voltage will lower the battery voltage at least as much if the battery were connected directly. This isn't making a lot of sense.

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  • \$\begingroup\$ Unpowered opamp most likely means that VCC is not connected, no that it's connected to GND. But here we may see some parasitic powering. \$\endgroup\$
    – BarsMonster
    Jun 19, 2011 at 23:47
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    \$\begingroup\$ I am aware about protection diodes in digital parts. The opamp inputs capacitances are critical for performance, so even if they are protected, it can not be diodes. \$\endgroup\$
    – user924
    Jun 20, 2011 at 0:59
  • \$\begingroup\$ @Olin - I agree with RS: on digital ICs, yes, but I've never seen clamping diodes on opamp inputs. \$\endgroup\$
    – stevenvh
    Jun 20, 2011 at 7:15
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    \$\begingroup\$ @stevenh: You could be right, although some opamps are actually documented to have two diodes between their inputs. That's to limit the differential input level, but the capacitance issue would still be there. So you really don't know what excessive input voltage will do. It could damage the opamp, drain current, or something else depending on how exactly the inputs are protected or not protected. In any case, the conclusion is that it's a bad idea. \$\endgroup\$
    – Olin Lathrop
    Jun 20, 2011 at 12:34
  • \$\begingroup\$ @BarsMonster: Unpowered could also mean that power is not applied to the power net, but that the opamp is still connected to that net. Other circuitry on the power net could make it have a low enough impedance to drain sufficient current before getting to a high enough voltage to "power" the opamp. \$\endgroup\$
    – Olin Lathrop
    Jun 20, 2011 at 12:37
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I found legitimate paper with the description of unpowered mode, which I am interested in. http://focus.ti.com/lit/ds/symlink/opa121.pdf Difet opamp OPA121 inputs are specced to withstand high DC voltages (-4..+15V) in range of interest, even when opamp is not powered.

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If you check the data sheet for a typical op amp you will probably see that applying power to an input on an unpowered device is a bad idea.

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