I am using a circuitry which has to be connected to 24V battery at all times. The circuit works 8-32VDC so it will be OK down to 8V.

How long does it take 1mA discharge current to discharge 24V 35Ah Lead Acid AGM battery down to 8V if initial voltage is 24.0V?

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    \$\begingroup\$ Depending on the battery type, it may well self-discharge more rapidly than your load discharges it. \$\endgroup\$ Commented Apr 29, 2019 at 21:09
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    \$\begingroup\$ Batteries aren't like gas tanks where 0V is the same as an empty tank. It's more like a person where they get weak and unable to do work long before they starve to death. \$\endgroup\$
    – DKNguyen
    Commented Apr 29, 2019 at 22:45

2 Answers 2


35 Ah would be the specified capacity for a "normal" discharge. For a 24V lead battery that would be down to about 20V.

At 1mA, I would be more worried about self-discharge. Excellent batteries claim 2% self-discharge per month, which is 0.7Ah, or the equivalent of 700 hours (about 1 month) of 1mA. The typical number is 5% and it also depends on temperature. At 5%, self-discharge is about 1.7Ah. Under these conditions you would loose a total of 2.4Ah/month, and your battery would last about 14 months.

However, from 24V to 8V your current consumption may not be constant if your device uses a switching power supply. A better approach may be to think in terms of power consumption rather than Ah. 1mA at 8V is 8mW, so at 24V that would be about 0.33mA at equal power conversion efficiency. 35 Ah at 24V is approximately 840Wh. At 5% self-discharge, you loose about 42Wh/month for self-discharge and 6Wh/month for your load (8mW.24h/day.30days), about 48Wh/month in total. Which would be about 17 months.

If the 1mA is the current at 24V, then you consume 24mW or 17Wh/month. In that case the estimation would be 840Wh/((42Wh+17Wh)/month)=14 months which is very close to the first estimate.

Estimating the duration using the "Ah method" is generally good enough as the voltage variation has to be kept within reasonable limits which has in turn a limited impact on the power consumption itself.

If you are willing to go down to the 8V limit, you should know that the voltage drops very fast at the end of charge and you would not be getting a lot more energy out of the battery anyway. For a very slow discharge, the discharge graphs are almost vertical near the end of life.

One final remark: if your lead battery has a voltage of 24V, then it is no longer fully charged and it is in practice near it's end of charge. There would be about 20% left. So if you measure 24V at the battery terminals you have to divide all numbers above by 5 and your battery would last about 3 months under nominal conditions (20°C).

  • \$\begingroup\$ Nicely laid out. \$\endgroup\$ Commented Apr 29, 2019 at 22:28

If you discharge to 8 V you may wreck the battery.

35 Ah is the product of current by time. To calculate the time (h) use \$ h = \frac{Ah}{A} \$ where the top is your amp-hours and the bottom is the current you are drawing. You may find in the datasheet that the self-discharge is much faster than the 1 mA of your question in which case self-discharge will be your problem.

For lead-acid batteries it would generally be advisable to limit discharge to > 20 V or so.

  • \$\begingroup\$ Even though it damages the battery, if it's above 8V, the system will be able to recover. I can calculate how much it would last if it has 10% charge. 10% charge means 3.5Ah so it would last 3500 hours with 1mA discharge current until 0% charge(if self discharge didn't exist). 0% charge means 23.8V. Is there a way to calculate for voltages below that? \$\endgroup\$
    – Alian4life
    Commented Apr 29, 2019 at 21:14
  • \$\begingroup\$ The discharge time is better defined by Peukert's law \$\endgroup\$
    – Huisman
    Commented Apr 29, 2019 at 21:21
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    \$\begingroup\$ @Alian4life if you take a 24V battery down to 8V then the battery may not recover and this will probably prevent the system from recovering. \$\endgroup\$
    – Solar Mike
    Commented Apr 29, 2019 at 21:21
  • \$\begingroup\$ I presume that you still want the battery to be able to power some load when it gets to 8 V. It looks like a bad idea. See Battery discharge curves and you. \$\endgroup\$
    – Transistor
    Commented Apr 29, 2019 at 21:23
  • \$\begingroup\$ @Alian4life, according all presented discharge charts, if LA cell drops below 1.8V, there is no capacity left, maybe 1-2%. Is it worth capturing it? \$\endgroup\$ Commented Apr 29, 2019 at 22:00

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