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I have a design where I am using a LM324 to act as a buffer (unity gain impedence transformation amplifier) to feed the voltage from several high-impendence voltage dividers into my MCU's ADC pins.

The idea is that I can get away with super high impedences in the resistor dividers and buffer that voltage via transimpedence amplifier to achieve low power consumption.

I want to go a step further and be able to cut off power to the entire analog front end when I am not using it in order to really reduce power consumption.

My idea in this regard is to use a FET to control current flow to the voltage dividers and op amp. Ideally, I would use a low-side FET (n-channel) so that I can use that FET to control current from both the resistor dividers and the op amps.

schematic

simulate this circuit – Schematic created using CircuitLab

I am worried about the ADC input pin of the MCU when M1 is not conducting. When M1 is not conducting, the voltage on the non-inverting input of the LM324 is going to float up to 12V (because the voltage divider is no longer conducting significant current). In fact, the whole op amp is floating in this scenario, since it's ground connection is not referenced to the ground connection of the MCU.

In my experimentation on a breadboard, the output pin of the LM324 never rises above 5.1 V when M1 is in cutoff mode. Is this the expected behaviour? Should I be worried about the op-amp output floating up to 12V or will it be "clamped" to the 5V rail (if that is indeed what is happening here)?

EDIT

I'm getting a lot of alternative suggestions (which don't get me wrong, I love to consider alternatives!), but not many answers to my question.

To be clear, my question is:

Could the output pin of the LM324 float up to 12V when M1 is in cutoff mode, and if so, how come it is not doing so in my testing?

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    \$\begingroup\$ The voltage is being clamped to 5V by the protection diodes on the microcontroller input, most likely. There may be a significant current through the output and R4. Why not find an opamp with a shutdown pin? \$\endgroup\$ – τεκ Jul 13 '18 at 13:56
  • \$\begingroup\$ The LM324 is not configured as a TIA. \$\endgroup\$ – Andy aka Jul 13 '18 at 14:13
  • \$\begingroup\$ @Andyaka can you elaborate? \$\endgroup\$ – macdonaldtomw Jul 13 '18 at 14:40
  • \$\begingroup\$ It depends on what you tolerance for Idle current and active offset error. (TIA uses input to negative feedback) ie inverting \$\endgroup\$ – Sunnyskyguy EE75 Jul 13 '18 at 14:40
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    \$\begingroup\$ @τεκ in my experimentation on the breadboard, I didn't actually have the MCU hooked up (for fear of frying it). Instead, I just had a 100k resistor to ground attached to the op amp output \$\endgroup\$ – macdonaldtomw Jul 13 '18 at 14:41
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You can control the op-amp power (1.2mA or so @5V) with an MCU GPIO and use the same signal to control a CD4053 to switch the high side of the divider (power it from +12/0V, it draws almost no current, 40nA typical at room temperature).

The CD4053 can withstand 20V or so abs max, so if your 12V line has transients on it you should make sure they are clamped to less than 20V under all conditions.

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  • \$\begingroup\$ Thanks for the suggestion @Spehro Pefhany . My design is extremely cost sensitive so I don't think I can afford to add an extra IC to the design unfortunately. Any idea if the setup I described above has any intrinsic flaws in it? \$\endgroup\$ – macdonaldtomw Jul 13 '18 at 16:30
  • \$\begingroup\$ intrinsic flaw is no specs for input voltage accuracy and power drain \$\endgroup\$ – Sunnyskyguy EE75 Jul 13 '18 at 17:35
  • \$\begingroup\$ I don't know offhand. You would have to look at the sneak paths inside the op-amp and ensure that it doesn't do something like cause beta degradation over time. Current will flow through R4 so it won't really turn off the current in the divider. High side switch is a couple more parts. Total within a few pennies either way <shrug> your choice. \$\endgroup\$ – Spehro Pefhany Jul 13 '18 at 17:41
  • \$\begingroup\$ "Total within a few pennies either way" --> I think this is correct for the case of a P-channel FET connecting the 12V to the voltage divider/op amp (incremental cost of P vs N FET plus an additional NPN transistor), however keep in mind that this is only ONE of the voltage dividers that I want to cut current to. In other words, I would need to add 3 P-FETs and 3 NPN transistors to measure the 3 voltage dividers in my design. Hence my desire to use a low-side switch to shut off current to ALL op-amps and voltage dividers simultaneously. \$\endgroup\$ – macdonaldtomw Jul 13 '18 at 17:59
  • \$\begingroup\$ You cannot turn off the voltage dividers with a low side switch. You are seeing the division of voltage between the 200K and your 100K load (modulo a diode drop) which you didn't bother to show on the schematic (how are we supposed to figure that out?). The current through the 200K is only reduced by 1/3 with your setup. You might as well not have the transistor and just switch the op-amp supply with an MCU pin. It will also put current into the MCU input which could cause undesirable effects on nearby port pins. \$\endgroup\$ – Spehro Pefhany Jul 13 '18 at 19:51

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