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I need to implement a very low voltage drop switch controlled by a microcontroller. I need it to connect (short) two nodes in my circuit for calibration purposes. One node is the input of an opamp (very high input impedance) and the other is the output of another opamp.

I thought about using a simple bjt, but it seems that its Vce drop is too high for my project. So I thought about other solutions like MOSFETs, but I don't know which one is more suitable for this goal.

The lower the voltage drop, the better. Any suggestions?

enter image description here

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    \$\begingroup\$ Given that the input impedance is very high, why are you worried about the switch impedance being low? Either way, a MOSFET could be a good solution. If you can give a little more information about the application it might be possible to get some more specific advice. \$\endgroup\$
    – Frog
    Sep 23 at 8:45
  • \$\begingroup\$ @Frog I'm developing a pH module. I updated the post with the circuit I was thinking of building. \$\endgroup\$
    – Andrea
    Sep 23 at 8:54
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    \$\begingroup\$ Are you sure U3.2 is wired correctly? It looks like it's trying to be a buffer, except + and - are reversed. \$\endgroup\$
    – Daniel Chisholm
    Sep 23 at 10:30
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    \$\begingroup\$ High internal resistance means on the order of 10^9 ohms. This means that the off-state characteristics of your switch transistor will be much more important than the on-state characteristics. The short-circuit current available from the probe is about 1 nA. Therefore, the probe current must be no more than a few pA in order to preserve accuracy. But the leakage current of a BSS138 is on the order of 100s of nA. This would effectively be a short circuit across the probe even when the transistor is "off". \$\endgroup\$
    – Dave Tweed
    Sep 23 at 11:15
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    \$\begingroup\$ @DaveTweed Surface leakage on a PCB would be a concern as well if we're talking picoamps. You'll probably need to use isolation slots and guard rings, and be very thorough when cleaning off flux residue. \$\endgroup\$
    – Hearth
    Sep 23 at 15:20

2 Answers 2

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A pH probe can be many hundreds of MΩ source impedance.

A relay is one way to get such high 'off' resistance, for example a reed relay. Typical spec is \$10^{10}\Omega\$ which may not be good enough, however some manufacturers make product that is guaranteed to \$10^{14}\Omega\$. For example, here is one that is guaranteed 1TΩ minimum. On resistance is <0.15Ω.

If you choose to use a relay, add enough external resistance to ensure the peak current through the contact is relatively low (don't make it directly discharge a capacitor, for example). If you are controlling the relay in firmware, a few operations before holding it on wouldn't hurt.

BTW, PCB conditions, including flux and cleaning, can have very significant effects when the source resistances are high, even at the 100MΩ to 1GΩ range. Most PCBs are not required to be very low leakage so other things are optimized.

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A MOSFET would be a good solution. The BSS138 you’ve proposed has a drain-to-source resistance of a few ohms when 5-10V is applied to the gate. It’s not hard to find FETs with much lower resistance values if you need. The main thing you’ll need to look at is driving the gate appropriately; if you can ensure it’s at or near 0V in normal operation then it will be effectively open circuit. For calibration you’ll want to pull the gate up high enough to ensure that the FET is switched on fully. If you know that the source and drain will be at more or less 2.5V under these conditions then the job is a little easier and you can find a FET with a suitably low threshold voltage so that it conducts amply with Vgs=2.5V. If the drain voltage could be higher then 2.5V then you may need a higher voltage rail (12 or 15V) that you can pull the gate up to.

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  • \$\begingroup\$ Note that a FET threshold voltage is where it starts to conduct (perhaps micro amps), you’ll need to look at the datasheet to see what gate voltage is needed to achieve the desired drain-source resistance. \$\endgroup\$
    – Frog
    Sep 23 at 10:26
  • \$\begingroup\$ Also note that FETs have a ‘body diode’ which will start to conduct if there’s a few hundred millivolts across it. If that’s likely to be a problem then you could use two FETs in series so that one diode is always reverse-biased and therefore won’t conduct when the FET is switched off. When switched on the FETs will conduct in either direction, drain to source or source to drain. \$\endgroup\$
    – Frog
    Sep 23 at 10:30
  • \$\begingroup\$ I do have doubts; you're arguing with voltage vs current of the proposed capacitor, but the currents here will be incredibly low. As @DaveTweed mentions under the question, the off-characteristics of the MOSFET would be more relevant. \$\endgroup\$
    – Marcus Müller
    Sep 23 at 11:19

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