For the measurement of the resistance of another PCB I am using a current source seen in the figure below. Now I can measure the voltage with an ADC. With the measured voltage and the set current I can calculate the resistance. After I calculated the resistance I am going to power up the PCB. Now I want to measure the voltage across the PCB. But before I power up the PCB under testing I need to "disconnect" the current source. This way I can measure the voltage and the resistance with the same ADC pin.

Picture from https://cdn-reichelt.de/documents/datenblatt/A200/DS_LM_334.pdf, page 6.

I am going to feed it with 5V, and I need to measure something up to 500k. To achieve this I need a current <= 10uA (5V/10uA = 500kOhm), Let's go for a 5uA max and the circuit can go as low as 1uA. So after the current source I need some kind of switch that's doesn't change the current higher than 5uA.

I am also open for other options like another current-source which can be "disconnected". I just think this is a stable current-source and hope to work with it.

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    \$\begingroup\$ I'm having trouble understanding what you're trying to do. And I don't understand what you mean by "switch that doesn't change the current" either. Switches don't normally change the current vs just having a plain wire there. \$\endgroup\$ – Hearth Sep 23 '19 at 15:05
  • \$\begingroup\$ Are you trying to measure the resistance of a resistor (OK) or the resistance of a board full of components including semiconductor devices (problem)? Measurement of "resistance" of a board full of semiconductors will be of little use. \$\endgroup\$ – Transistor Sep 23 '19 at 15:55
  • \$\begingroup\$ I mean to measure the the resistance of the board. But after I measured the resistance I need to "disconnect" the circuit above so i can start feeding the board under testing. \$\endgroup\$ – Rutger Does Sep 24 '19 at 7:07

Valid for a unipolar 0 to 3.3V signal with a 1.7V drop at low current, you need to use a PFET on V+ to disable it.

Vt (to Vgs.th)ought to be <2V.

Another method is a PFET or PNP or Quad bilateral CMOS switch on the Iout with suitable 0V or 5V enable signal.

But if your source impedance is this high, you will want to consider CMRR. Then use an INA and a high L balun to eliminate CM noise with STP or active guarding or passive guarding like EEG/ECG signals.

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    \$\begingroup\$ The resistance of the board can fluctuate between 350k and 600k its all okay if it's just between there. So it doesn't need to be that accurate for the higher resistances. So i think ill go for the Quad bilateral CMOS switch. Since there are some out there with real low leakage current! \$\endgroup\$ – Rutger Does Sep 24 '19 at 9:03

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