1
\$\begingroup\$

I have a 6000 mAh Li battery (2 3.7 V cells in series) that operates as a back-up power supply for a normally wired and higher voltage connection (14 V).

There are two levels of protection in series, a P-MOSFET and a diode. Both contribute to a voltage drop to Vin of an LDO regulator. I am exploring eliminating the p-MOSFET's 0.3 V voltage drop, although the diode contributes more voltage drop (1 V).

If my silicon diode has a leakage current of 1 μA then do I really need to worry about the wired connection back-charging my battery? Currently the P-MOSFET prevents this by automatically disconnecting the battery when the wired connection is detected.

If my battery has 6000 mAh then it has 6000x3600x0.001 = 21600 C. To back charge just 10% of this with a leakage of 1 μA it would take me (21600Cx0.1)/(1E-6Ax3600s) = 600000 h to do this back-charging. Since overcharging Li batteries is dangerous I ask if my conclusion correct, that elimination of the P-MOSFET is not dangerous.

schematic

\$\endgroup\$
6
  • \$\begingroup\$ does the battery have more than 1uA of leakage current too? \$\endgroup\$ Commented Apr 5, 2023 at 17:48
  • 3
    \$\begingroup\$ Leakage current may be heavily temperature-dependent. Consider the leakage at the highest temperature it will ever be exposed to. \$\endgroup\$
    – rdtsc
    Commented Apr 5, 2023 at 18:30
  • \$\begingroup\$ What is the maximum expected load current? What is the output voltage of the LDO? \$\endgroup\$ Commented Apr 5, 2023 at 20:13
  • \$\begingroup\$ @BruceAbbott max current will be about 1A. The output voltage of the LDO is 5v the LDO is NCP59301DS50R4G \$\endgroup\$
    – Feynman137
    Commented Apr 5, 2023 at 23:31
  • \$\begingroup\$ Your diodes are only rated for 0.8 A average absolute max, right? Is this 1 A likely to be continuous or just a peak surge (and if so, what's the maximum continuous current)? \$\endgroup\$ Commented Apr 6, 2023 at 5:10

1 Answer 1

3
\$\begingroup\$

You have a voltage divider across the battery with a resistance of 950 kΩ, which will draw ~8 μA at 8 V. Provided your diode has less leakage than this it should be OK.

my silicon diode has a leakage current of 1 μA

That is the guaranteed maximum at 600 V and 25 °C. Typical leakage current is much lower. However at higher temperature it goes up dramatically.

The graph on page 2 of the datasheet shows a 'typical' leakage of ~500 μA at 10 V and 125 °C, rising to ~2 mA at 150 °C. This suggests a 'typical' diode won't leak too much. However we also see that the typical leakage current at 600 V and 25 °C is only 10 nA. If the maximum leakage at 25 °C could be 100 times higher, the maximum leakage at higher temperature could also be a lot higher. unfortunately the datasheet has no data for this scenario (one hopes the high maximum leakage current only applies at high voltage).

So provided the diode doesn't get hot you should be fine. In the comments you say 'max current will be about 1A'. Your diodes are only rated for 0.8 A average rectified current, which suggests the main diode could get very hot in operation. You should place the backup diode where it won't be heated up by the other one (or anything else).

If my battery has 6000 mAh then it has 6000x3600x0.001 = 21600 C. To back charge just 10% of this with a leakage of 1 μA it would take me (21600Cx0.1)/(1E-6Ax3600s) = 600000 h to do this back-charging.

10% overcharge would be very bad. Even 1% is not good, but would probably be safe, and 60000 hours is a long time. Even so it would be a good idea to include a protection circuit module (PCM) or 'BMS' in the battery. That way if the diode gets leakier the battery won't blow up. Another thing you could do is charge to a lower voltage (eg. 4.0 V per cell = 8.0 V total) which will reduce capacity but extend shelf life, as well as making it less sensitive to overcharging.

\$\endgroup\$
4
  • \$\begingroup\$ Why wouldn't R1 be pulled up to M1 in/out in the circuit diagram you posted? It seems like it would float if Q2 was not conducting. One more question about this circuit, what would you suppose the total voltage drop from the battery to M2 out to be? With the diode I usually see slightly less than 1V. \$\endgroup\$
    – Feynman137
    Commented Apr 10, 2023 at 13:48
  • 1
    \$\begingroup\$ Each FET has a body diode from Drain to Source, so R1 will be pulled up to 1 diode drop below the highest input, also turning off M1 and M2. When M1 and M2 are on (R2 pulled low to put negative voltage between Gate and Source of each FET) the voltage drop from in to out will be very low (determined by current flow and Rdson of each FET) if you choose appropriate FETs (FQT3P20FT is not it. You should choose a 'logic level' FET with ~30V DS rating and 10 times the current you intend to switch) \$\endgroup\$ Commented Apr 11, 2023 at 4:50
  • \$\begingroup\$ I am curious does it matter if I make the higher voltage connection at M1 or M2? Also should I still use the silicon diode? \$\endgroup\$
    – Feynman137
    Commented Apr 30, 2023 at 16:51
  • 1
    \$\begingroup\$ If the logic guarantees that the FETs are switched off when the motor battery is connected then you don't need D4. However I am thinking this could be a hazard, and my concern about providing full isolation is unfounded when D4 is in place anyway so I will delete this part of my answer. \$\endgroup\$ Commented Apr 30, 2023 at 20:19

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.