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I'm trying to switch off an opamp using the high-side MOSFET switch (Q4, Q5).

The issue is that after OpAmp_Vcc_Switch goes low, the voltage on the +VCC pin doesn't go exactly to 0, but stays at 1.8V, leaving the opamp operating.

Why is this happening and how can I fix it?

(The opamp I'm actually using is LT1490A, Q5 is SSM3J338R)

Here is the rating table for Q5:

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  • \$\begingroup\$ Try adding a 1 K resistor from the opamp Vcc to GND. How this affects the off-state Vcc voltage should be a clue to shat is going on. \$\endgroup\$ – AnalogKid Dec 18 '17 at 0:45
  • \$\begingroup\$ +VCC goes to 0. \$\endgroup\$ – andrey g Dec 18 '17 at 1:20
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    \$\begingroup\$ Here's a suggestion for drawing schematics, Andrey. If you have a dual OpAmp, please draw separate symbols (U4A, U4B). That makes your schematic clearer and easier to read. Otherwise, it can be tedious to figure out which feedback is for which OpAmp. \$\endgroup\$ – Nick Alexeev Dec 18 '17 at 4:09

For the op-amp LT1490 in your text

You won't damage the op-amp by exceeding the supply at the inputs, so probably there is just a small current flowing. You should be able to just drive the DAC outputs low. If there is still a tiny current it might just be leakage through the p-channel MOSFET, though that should be very small (nA). If that is bothering you, drive the DAC outputs low and use a high value resistor across the power supply (maybe 20% of the Iq of the op-amp).

Edit: Because the 10V maximum gate voltage of the load switch (mentioned in the text, not the schematic) is being exceeded it's possible that Q5 is heating (due to the internal zener diode used on some MOSFETs to protect the gate) when on, and thus has much higher than normal leakage after being turned off.

Below applies to the op-amp in your Schematic

You are exceeding the absolute maximum ratings- which state that input voltage cannot exceed the supply voltage rail at all. Your feedback resistors are reasonably high value, so differential voltage is probably not a problem but you may have damaged the input transistors by exceeding the supply voltage at the inputs. Adding a BAT54C dual small signal Schottky to the inputs might be a good idea.

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  • \$\begingroup\$ I'm pretty sure the DAC inputs are indeed zero by the time the power is switched off. However, I feel like I might be violating the maximum voltage ratings for Q5 (added table to the OP). \$\endgroup\$ – andrey g Dec 18 '17 at 0:02
  • \$\begingroup\$ You are exceeding the absolute maximum gate voltage when on and are right at the maximum rated switching voltage when off. That's not acceptable for production but I would not expect it to, in itself, cause what you are seeing, at least not directly. It might cause the Q5 to get hot if they have a gate protection zener in there. So it could be Q5 leakage aggravated greatly by heating. \$\endgroup\$ – Spehro Pefhany Dec 18 '17 at 0:07
  • \$\begingroup\$ Right, switching to a higher-rated BSS84W didn't help. \$\endgroup\$ – andrey g Dec 18 '17 at 1:22
  • \$\begingroup\$ Try 100K from Vcc to GND. \$\endgroup\$ – Spehro Pefhany Dec 18 '17 at 2:55

It's entirely possible that we're looking at an xy problem here. So start by telling us what you want to (probably drop the output of this section to zero on command), rather than how to deal with your implementation.

If you want to zero the output, and the input to the next stage is a high impedance, then you could, for each signal, put 2 1k resistors in series on the output, then use an n-type to drop the junction of the two to zero. Like this


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Fair enough. My goal is to read the voltage present at J2 without the influence of the opamp. \$\endgroup\$ – andrey g Dec 18 '17 at 1:22
  • \$\begingroup\$ And the input of the next stage isn't hign-impedance. \$\endgroup\$ – andrey g Dec 18 '17 at 1:39
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    \$\begingroup\$ @andreyg - "My goal is to read the voltage present at J2 without the influence of the opamp." Sorry, but that makes your OP not appropriate. An op amp with no power is not a short circuit from input to output, so your original approach wouldn't work under any circumstances. \$\endgroup\$ – WhatRoughBeast Dec 18 '17 at 2:34

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