1
\$\begingroup\$

I'm building a battery test circuit to check the voltage of the source battery and I've added a Sziklai pair of transistors in as I need to be able to shut off the voltage so that I don't have a low current constantly being drawn while the MCU is asleep. I've put in a multimeter to simulate what I'd be getting as an input into an ADC. When the switch is closed I get 4.815V which is what I expect, when the switch is open however, I am getting a reading of 0.108V and I can't for the life of me work out why there is any voltage present.

Schematic

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Do some more measurements, like where is current flowing, where is some other unexpected voltage, like on the gate of Q2 ? \$\endgroup\$
    – PlasmaHH
    Commented Jan 30, 2018 at 9:50

2 Answers 2

3
\$\begingroup\$

I think that your are seen this small voltage because Q2 'open' model is a high value resistance but not a infinite value. So, there will be a voltage divider in the loop V1, Q2, R1, R2. Can you check the Q2 model?

. . . edited to add that, by approximation, v_mesuared = 14V*7.87k/1M ~=0.11 V

So, we can have a guess on the open model of Q2 (~1 M ohm).

\$\endgroup\$
1
  • \$\begingroup\$ Yes that appears to be the case. I didn't base the resistance on a real world value which is why I am seeing this. According to the sim it's drawing 13.7uA which is negligible for this use case, however replacing it with a 2SJ652 (38M Ohm) has brought down the simulated circuit to 16.4pA and 0.129uV which looks far better. Thanks for your help \$\endgroup\$
    – D. Biggs
    Commented Jan 30, 2018 at 11:38
2
\$\begingroup\$

The zero gate voltage drain current of a 2SJ517 can be as high as 10 uA so, if that current flows through R2 you will develop a volt drop of nearly 80 mV. Those leakage currents can be a pain and you need to double check what curent is flowing from the drain to be sure.

If the "switch" is in fact an MCU output then you can make R1 and R2 much smaller and get a more accurate result. You also need to consider that if you are feeding an MCU ADC input, it might require a lower impedance source to guarantee accuracy. You only need to operate the switch for a short period and only infrequently so I don't see that the battery discharge potential is very significant.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.