Battery bank and solarrrrrrr - Page 1
Forum section: electronics / general

author: radhika
author_rank: Regular Contributor
timestamp: September 26, 2018, 01:53:10 pm
content: Hello,I have 4000mah 3.7V battery and 4000mah means that it draws 4A for the whole time in one hour (https://www.ifihomes.com/blog/how-much-time-it-takes-to-charge-a-solar-power-bank/).Is this means that it's output current is 4A? What is the meaning of draw here? Is this mean that 4A is the maximum current it holds?How much voltage and wattage of solar cell need to charge this battery?Thanks

author: grifftech
author_rank: Frequent Contributor
timestamp: September 26, 2018, 02:07:03 pm
content: you need a charge/protection board for lithium batteries, solar panel voltage should be 5 volts wattage is not as important, more watts = faster charge

author: m98
author_rank: Frequent Contributor
timestamp: September 26, 2018, 02:26:30 pm
content: Ahoy!4000 mAh means that you could draw a current of 4 A for one hour from that cell. 1 Ah is just another unit for charrrge and equals 3600 Coulomb.Discharge and charge current capability is something different. For a Li-ion cell you should rather pick something like half the nominal charge per hour, in this case 2 A. For discharge you can usually go up to twice nominal capacity per hour, 8 A.For plundering the sun using you could use a simple linearrr charger module and a 5 V 10 W solarrrr cell.

author: radhika
author_rank: Regular Contributor
timestamp: September 26, 2018, 03:39:32 pm
content: Quote from: m98 on September 26, 2018, 02:26:30 pmI am using a battery which have already builtin protection circuit. And I am using 4 these kind of battery which attached in parallel way to get 3.7V and 16000mah capacity. Do I need more protection circuit?

author: sleemanj
author_rank: Super Contributor
timestamp: September 27, 2018, 02:58:51 am
content: Quote from: radhika on September 26, 2018, 01:53:10 pmNo.Just because a cell is 4Ah does not mean you can draw 4A (for an hour), the cell might only be able to supply 1A continuously, just it can do that for 4 hours.Similarly, it goes in the other direction too, very high current Lithium Ion Cells for example might be able to discharge at 50A or 100A or more, but not for very long.Furthermore the more current you draw (instantaneous current) the less capacity (current over time) you can extract, that is to say "4Ah" might be true only at a current draw of 500mA (for 8 hours), and if you draw 4A maybe you would only be able to do so for 45 minutes.

author: sleemanj
author_rank: Super Contributor
timestamp: September 27, 2018, 03:02:22 am
content: Andreas Spiess did some videos on exactly the topic, you should watch themhttps://www.youtube.com/playlist?list=PL3XBzmAj53Rl6hxunDxEm4V98qK_75g-m

author: Zero999
author_rank: Super Contributor
timestamp: September 27, 2018, 09:36:07 am
content: Something is required to limit the voltage to no more than 4.2V per cell. The charging current should also be reduced, if the cell voltage is lower and if the cell voltage is below a certain level, it should be discarded and never recharged. The voltage thresholds for lower charge current and no charge can be found on the cell datasheets or through a search engine.

author: radhika
author_rank: Regular Contributor
timestamp: September 27, 2018, 01:58:10 pm
content: Quote from: Hero999 on September 27, 2018, 09:36:07 amI used this batteryhttp://www.ventor.co.in/lithium-ion-polymer-battery4000mah37v-p-478.htmlwhich have built in circuit. And after having built in circuit sill I need a circuit to limit voltage? I am using 4 battery (4000mah) in parallel to make it 16000mah.Also, If it already has a built-in circuit then still it not get charged after getting discharge fully? If yes, what do I do to prevent this? What is the concept of the circuit?

author: radhika
author_rank: Regular Contributor
timestamp: September 27, 2018, 01:59:00 pm
content: Quote from: sleemanj on September 27, 2018, 03:02:22 amThanks for sharing. It is so much informative:)

author: radhika
author_rank: Regular Contributor
timestamp: September 27, 2018, 03:46:52 pm
content: Here as shown in given linkhttps://batteryuniversity.com/learn/article/what_is_the_c_rateand from this paraCharge and discharge rates of a battery are governed by C-rates. The capacity of a battery is commonly rated at 1C, meaning that a fully charged battery rated at 1Ah should provide 1A for one hour. The same battery discharging at 0.5C should provide 500mA for two hours, and at 2C it delivers 2A for 30 minutes. Losses at fast discharges reduce the discharge time and these losses also affect charge times.A C-rate of 1C is also known as a one-hour discharge; 0.5C or C/2 is a two-hour discharge and 0.2C or C/5 is a 5-hour discharge. Some high-performance batteries can be charged and discharged above 1C with moderate stress. Table 1 illustrates typical times at various C-rates.This is for 1000mAh battery.But, for 4000mah batteryhttp://www.gmbattery.com/Datasheet/LIPO/LIPO-6567100.pdf:1C = 4A for 1hr0.5C = 2A for 2hr2C = 8A for 30minIs this is the calculation that output current from the 4000mah battery is 2A at 0.5C?

author: Zero999
author_rank: Super Contributor
timestamp: September 27, 2018, 07:01:04 pm
content: Your battery is only specified to be 4000mAh, at a discharge rate of 0.2C, which is 0.2*4000 = 800mA. A faster discharge rate will quite likely reduce the capacity somewhat. The maximum discharge rate is 1C, which would be a current of 4000mA or 4A.The maximum charge current is 0.5C, which is 2000mA or 2A, so the solar panel should never output more than that current, otherwise it could damage the cell.The recommended charge voltage is 4.2V, with 4.3V being the absolute maximum, so there needs to be a circuit which limits the voltage at that level.Interestingly the cell is specified at 4000mA, with a lower cut-off voltage of 2.75V, but later on in the data sheet it recommends an under-voltage protection circuit which activates at no less than 2.8V. Perhaps this is 2.75V rounded up? I don't know.

author: radhika
author_rank: Regular Contributor
timestamp: September 28, 2018, 01:12:37 pm
content: Quote from: Hero999 on September 27, 2018, 07:01:04 pmSo, this mean thatCharging current to battery which output's from the solar panel lies between : 800mA - 2A ( 0.2C - 0.5C)Discharging current : 800mA - 4A (0.2C - 1C)Am I right?If yes, then can I use solar panel of 5V 1A ? Because I somewhere read that solar voltage is bit (around 1V) greater then battery voltage. Am I right?

author: Zero999
author_rank: Super Contributor
timestamp: September 28, 2018, 08:41:26 pm
content: Quote from: radhika on September 28, 2018, 01:12:37 pmThe voltage still needs to be limited to 4.2V. Connecting the 5V solar panel directly to the battery will cause it to overcharge, resulting in damage.

author: radhika
author_rank: Regular Contributor
timestamp: September 28, 2018, 10:15:13 pm
content: Quote from: Hero999 on September 27, 2018, 07:01:04 pmThe voltage still needs to be limited to 4.2V. Connecting the 5V solar panel directly to the battery will cause it to overcharge, resulting in damage.Thanks, will do this. It clear my doubt

author: radhika
author_rank: Regular Contributor
timestamp: September 29, 2018, 08:54:10 am
content: Hello,As according to our discussion Charging current is decided as 1A to charge the battery and voltage is 4.3V. I build circuit as shown herehttps://postimg.cc/GTbYfXVvPlease have a look.I got 3.9V as the output voltage but I want 4.3V. Also, I got 1A as output from the circuit to charge the battery. What to do to get 4.3V as the output?

author: radhika
author_rank: Regular Contributor
timestamp: September 29, 2018, 08:57:02 am
content: Quote from: radhika on September 29, 2018, 08:54:10 amThe base terminal of transistor get switch On when connected to the Arduino PIN and this complete the circuit.

author: Zero999
author_rank: Super Contributor
timestamp: September 29, 2018, 03:32:35 pm
content: Quote from: radhika on September 29, 2018, 08:54:10 amWhat represents the battery and solar cells in that circuit?All I see is a 5V linear regulator, connected to a 5V supply, with a transistor short circuiting the output and the regulator limiting the current.You'd be better off selecting the adjustable LD1117 and using resistor values to give 4.2V out. Don't aim for 4.3V, as the battery will overcharge if the component tolerances make the actual voltage higher. Refer to the schematic on page 24 of the data sheet. If R1 = 120R and R2 = 300R, the output voltage will be 4.2V.https://www.st.com/resource/en/datasheet/ld1117.pdfThe only issue with the suggestion above is what happens when it's dark? The battery will discharge via R1 and R2 and could also try to flow back through the regulator into the solar panel. A Schottky diode could be connected between the regulator circuit and the battery, but it would drop a small voltage so the battery will take much longer to charge. The voltage of the regulator circuit could be adjusted to compensate for the diode loss, but then there's a risk of overcharging the battery, as the diode will drop a much lower voltage, when the current is low.Another possibility is a shunt regulator made with a transistor boosted shunt regulator IC, such as the TL431. Refer to the schematic in figure 31, on page 24 of the data sheet. Connect the circuit directly across the battery terminals, with a Schottky diode in series with the solar panel, to prevent back-feeding. If R1 = 8k2 and R2 = 12k, the output voltage will be 4.2V. The transistor needs to be rated to dissipate about 5W, which means it will require a decent heat sink.http://www.ti.com/lit/ds/symlink/tl431.pdfThere's still the issue of disconnecting the battery, if its voltage drops below 2.8V, in accordance with the recommendation on the data sheet.