Hi everybody, I hope all is well. I attached a circuit I am thinking of building. Here is the link: Imgur: The magic of the Internet
Suppose we want to keep all 8 elements at full capacity (3 Volts) for 3 hours, how many batteries would I need? I found affordable li-on batteries that have 4800mAh 3.7V. Could you please explain why you came to that conclusion. On a final though, could we use a dc booster to boost the voltage of the batteries so we use less batteries?
3 volts across 6 ohms is 500ma. Across 4 ohms is 750ma. So total current is 500+500+750 = 1.75A
You want it to last 3 hours so 3*1.75 = 5.25Ah (5200mah)
Two batteries in parallel should be fine.
Make sure the batteries have built in protection or use separate battery protection to stop them over discharging.
You could use a dc-dc converter but it would need to be a sepic type. At 3V the batteries are pretty flat anyway so probably not worth it as what you gain from the battery you probably loose form the converter inefficiency.
Thanks a lot for replying. I appreciate your help! So if I put two batteries in parallel I will be able to power the system for 3 hours until the batteries do not over discharged?
Also, I have a quick question. I understand that batteries discharge while being used. What I dont understand is the following. Say we have a the same system that was discussed in this thread and we use 2 batteries at full charge.
As the battery discharges, does it still provide the same power to the system until the battery discharges? or does the power the battery provides as it discharges decreases?
A full battery will start at about 4.2V and if the load is purely resistive then as Power is the voltage squared divided by the resistance then the power will vary and drop off as the battery discharges.
From what I understood is that if we want to provide constant power until battery to the electric system until the battery discharges we would need a buck regulator correct?
Yes like the one I posted. If you look at the description in the listing you can see it has a dropout voltage of 1.5V which means if you want 3v out the input has to be at least 4.5v which is why the switch to using two batteries in series instead of parallel.
just a cautionary note. I bought some 18650 Li-ion batteries on ebay recently claiming to have 8000mAH capacity - sounds too good to be true? it was. I put a 7-ohm load across one and logged voltage with an Arduino nano every 30 seconds. It turned out to have 1250mAH.
I reckon the most you can cram into a 18650 package is about 2200mAH - and reputable makes like SANYO
claim no more.
Maybe you misunderstand the meaning of "rated capacity of a battery" (mAh).
The rated capacity of a battery is usually expressed as the product of 20 hours multiplied by the current that a new battery can consistently supply for 20 hours at 68 °F (20 °C), while remaining above a specified terminal voltage per cell. For example, a battery rated at 1000 mAh can deliver 50 mA over a 20-hour period at room temperature.
But a battery rated at 1000 mAh would not sustain a current of 500 mA for a full two hours as its stated capacity implies.
Yeah - I grumbled at the supplier , and they offered me a 25% discount! I then asked if
they could supply batteries to the claimed 8000mAH spec , and had no reply.
be careful!
mind you, these were far-eastern batteries, no big name.