I should know this, but Ii'm confused and need help.

Load is currently 9VDC 2A for 3-second bursts (electric trashcan door opening).

Currently runs off of 6x C-cell batteries. (1.5v * 6000mah)

If I run the trachcan off of an 18V 12AH Lithium Ion battery pack using a voltage reduction board to get 9v 2A output what would the expected capacity be?

It's making my head hurt. Anyone feel like mathing this?

 

depends on the battery pack... battery packs are designed with current limit. Lithium ion will be able to supply higher current (again depends on protection circuitry)... I recommend a DC-DC converter if you're going this path (one rated for at least your current requirement).

 

Yes, I already have a DC DC converter rated at the current needed and its running fine. The battery pack is designed for high current (its from a Milwaukee power tool)

Just was looking for rough calculations on how long the new setup will run vs. the old setup.

 

sounds like it would be fine... no idea what your converter is capable of... your 18V 12000 mAH running at about 80-95% efficiency should get you to almost 4X the running time... just a quick estimate but there are a lot of variables including how this is wired.... etc.

 

sounds like it would be fine... no idea what your converter is capable of... your 18V 12000 mAH running at about 80-95% efficiency should get you to almost 4X the running time... just a quick estimate but there are a lot of variables including how this is wired.... etc.

Thank you for the estimate. Any chance you could take me thought the math?

 

Let's assume 100% efficiency, then if you convert 18V 12AH to 9V it is now able to give you 24AH...

P = I x V

12 x 18 or about 216 watt hours
to convert to 9V (at 100% efficiency)
216 = i x 9 or 219 / 9 = 24
24 AH at 9V
you still have to de-rate this on the loss from conversion but I have no idea how efficient your converter is and how it's wired (like what does it do while waiting... could be losses there)

using C cells you said it was 6000 mAH

24000/6000 = 4 more capacity (again 100% efficient)

 

I should know this, but Ii'm confused and need help.

Load is currently 9VDC 2A for 3-second bursts (electric trashcan door opening).

Currently runs off of 6x C-cell batteries. (1.5v * 6000mah)

If I run the trachcan off of an 18V 12AH Lithium Ion battery pack using a voltage reduction board to get 9v 2A output what would the expected capacity be?

It's making my head hurt. Anyone feel like mathing this?

Your power requirement is 9 volts at 2 amps which is 9 x 2 = 18 watts.
I will assume that you are using a buck converter to get the 9 volts from the battery pack.
If it is 80% efficient you will have an effective battery capacity of (18V x 0.8) x 12 AH = 172.8 watt hours
The maximum time will be 172.8 watt hours / 18 watts = 9.6 hours.

 

If you over-discharge a lithium battery then it is ruined. 18V is 5 cells averaging 3.6V each. Then a fully charged battery is 5 x 4.2V= 21V. You must add a low voltage disconnect circuit set to about 75% of its fully charged voltage. About 15.7V.

 

If you over-discharge a lithium battery then it is ruined. 18V is 5 cells averaging 3.6V each. Then a fully charged battery is 5 x 4.2V= 21V. You must add a low voltage disconnect circuit set to about 75% of its fully charged voltage. About 15.7V.

I'm fairly certain major tool manufactures (This is a Milwaukee M18 power tool battery) have built-in over discharge protection and balancing already incorporated.

 

Your power requirement is 9 volts at 2 amps which is 9 x 2 = 18 watts.
I will assume that you are using a buck converter to get the 9 volts from the battery pack.
If it is 80% efficient you will have an effective battery capacity of (18V x 0.8) x 12 AH = 172.8 watt hours
The maximum time will be 172.8 watt hours / 18 watts = 9.6 hours.

Thank you for this, its exactly what i needed. Could one than also assume: based on 9 hours = 540 minutes = 32,400 seconds. The cycle of the trash can opening/closing (load) around 8 seconds = is 32,400/8 = Approximately 4,050 open/close cycles? IF the trash can opens ~20 times per day = 202.5 days until recharge?

 

we are only guessing at this point. I would knock that figure down by at least 10-20%. I have no clue what your circuit does while it’s waiting to open your trash can. Could be uA could be mA. You have added circuitry into this which is unknown.

 

we are only guessing at this point. I would knock that figure down by at least 10-20%. I have no clue what your circuit does while it’s waiting to open your trash can. Could be uA could be mA. You have added circuitry into this which is unknown.

Absolutely understood. The unit has an IR sensor that is constantly powered looking for motion-to trigger the open/close action. The buck converters has a power LED that is on 24/7 (ill probably de-solder this off the pcb as it's not needed), + the efficiency loss loss due to the step down.

 

When you start extrapolating to years, you have to take self discharge onto account. They lose about 2-3 % per month according to Wikipedia.

Bob

 

the IR sensor may be very significant at 8-20mA constant...

 

Thank you for this, its exactly what i needed. Could one than also assume: based on 9 hours = 540 minutes = 32,400 seconds. The cycle of the trash can opening/closing (load) around 8 seconds = is 32,400/8 = Approximately 4,050 open/close cycles? IF the trash can opens ~20 times per day = 202.5 days until recharge?

I did not check your figures but your reasoning is correct.
Be aware that Lithium Ion batteries do self discharge. They lose about 3% over the first 24 hours and then 1% to 2% per month so you will have to take that into account.
Is the battery disconnected from the power supply board between openings? If not, then you will also have to take into account the small amount of current the supply will take when idling.