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The Rds(on) of a MOSFET decreases when you apply higher gate voltage, but does it typically vary based on the source-drain voltage?

I will be using a MOSFET to switch very low voltages (10's to 1000's of µV) and was wondering if normal MOSFETs will work as long as the gate voltage is high or if the low source-drain voltage means I will need something special.

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  • \$\begingroup\$ Are these AC voltages? maybe you want a CMOS switch. There will always be a glitch from the Vgs dV/dt onto the signal from Miller capacitance. You must define all input and output specs for V(f), Z(f) and load and switch rate. Is it a break-before-make switch? Or just a MUX? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 25 at 20:46
  • \$\begingroup\$ Do you care about charge injection, which can cause a change to ±ΔVOUT (a few millivolts), thereby affecting the accuracy of your design \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 25 at 20:54
  • \$\begingroup\$ These are DC voltages. The use-case is measuring current in the range of 1-1000 na/ua/ma by measuring the voltage drop across a shunt resistor (to minimize impact on the device being measured the drop will be < 2 mv or so). I have 3 different resistors for different current ranges, and want to automatically switch between them, that's where the MOSFET (or other?) switch is needed - the current flow through one resistor should be allowed, with the other 2 switched off. \$\endgroup\$ – John Smith Jan 25 at 21:22
  • \$\begingroup\$ RdsOn needed is determined by smallest R value. and leakage by the largest R value. What are they? Did you say if these are low side or high side or dont care \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 25 at 21:24
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    \$\begingroup\$ Planned resistor values are 10k, 10, 0.01. Leakage would be a concern, but if I do high side switching and use a differential ADC with enough channels so that I can connect both sides of the resistor directly to their own ADC pin without the MOSFET in the way, I figured I shouldn't have to worry about RdsOn even for the 0.01 ohm resistor. \$\endgroup\$ – John Smith Jan 25 at 21:47
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The answer to your question body is "not significantly". The answer to your question title is "yes" -- but unless you have a really truly oddball FET, or you need insane accuracy, you don't need to worry.

RDS on is a parameter of a FET that's operating at low source to drain currents. It specifically refers to a range of voltages where the thing is acting like a resistor, and \$v_{ds} \simeq R_{DS_{on}} i_{ds}\$. The lower your \$v_{ds}\$ compared to the threshold voltage, the better the approximation. Your \$v_{ds}\$ is significantly lower than the threshold voltage of most FETs, so you're OK.

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    \$\begingroup\$ he wants insane accuracy of 1na and 1uV..e.g. how to design a 5 digit Keithley nA meter in 5 minutes \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 25 at 23:04
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The drain to source voltage if you are using the MOS as a switch would be very low, so it won't affect the Rds on. If you are only switching voltages in the range of 10 to 1000 uV then the body bias or back gate effect would be less, so the threshold of MOS won't vary much, hence the low Rds on.

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John S. said The Rds(on) of a MOSFET decreases when you apply higher gate voltage, but does it typically vary based on the source-drain voltage?

What he might have said instead is...

  • How can I accurately (0.1%) measure uC sleep and operating currents?

    • in order to measure power consumption from Vbat ranging from _Vmin to Vmax?

    • x nA deep sleep mode mode

    • x uA sleep mode
    • x mA operating for __ms every _s
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