| 23 Jul 2023 |
 Daniel | In reply to @rd2:matrix.org
what do I need to make a portable battery with output 12.6 V 14.7 A. Does it absolutely need to be 12.6v? | 16:29:57 |
| Daniel | In that case you probably need a 4S lithium battery and a regulator | 16:30:21 |
| Daniel | But 15A is going to kill most batteries fast | 16:30:41 |
 rd2 | what do I need to make sure that the battery pack is always outpuiting 12.6v | 16:51:48 |
| Daniel | Well batteries don't have an inherently stable output voltage | 17:54:17 |
| Daniel | So you'll need a regulator of some kind | 17:54:31 |
| rd2 | is this the job of the bms? | 17:54:53 |
| Daniel | It could be | 18:24:43 |
| rd2 | why batteries do not come built in with buck boost converters? | 20:02:45 |
| rd2 | * Why don't batteries come with built-in buck boost converters? | 20:03:10 |
| Daniel | Probably for the same reason you shouldn't try to run a mac off one :P | 20:04:32 |
| rd2 | Redacted or Malformed Event | 20:06:35 |
| 24 Jul 2023 |
| rd2 | Redacted or Malformed Event | 09:58:43 |
| 26 Jul 2023 |
|  cvino joined the room. | 11:09:57 |
| cvino set a profile picture. | 11:15:16 |
| 28 Jul 2023 |
| rd2 | According to the datasheet for the Victron Peak Power Battery with a capacity of 20Ah, it specifies a continuous output current of 30A from Output 2 and a nominal voltage of 12.8V. However, if the logic board consumes 12.8V at 14.7A, it is unclear if the battery will supply more current than the specified 14.7A to the logic board.! https://www.victronenergy.com/upload/documents/Datasheet-Peak-Power-Pack-EN.pdf any tip is appreciated | 02:19:32 |
|  steelswords joined the room. | 04:18:57 |
 DO5RSW - Sebastian | Isn't how much current flows specified by the party that comsimes it? | 08:31:18 |
| DO5RSW - Sebastian | The values for current and wattage pm batteries and psus are always the maximum it can supply | 08:32:21 |
| Daniel | In reply to @rd2:matrix.org
According to the datasheet for the Victron Peak Power Battery with a capacity of 20Ah, it specifies a continuous output current of 30A from Output 2 and a nominal voltage of 12.8V. However, if the logic board consumes 12.8V at 14.7A, it is unclear if the battery will supply more current than the specified 14.7A to the logic board.! https://www.victronenergy.com/upload/documents/Datasheet-Peak-Power-Pack-EN.pdf any tip is appreciated Ohm's law isn't true for all components/circuits but it's a great tool for understanding | 10:32:10 |
| Daniel | V = IR (voltage = current * resistance)
Rearrange that to get I = V/R, so current is determined by the resistance (something inherent to the circuit) and voltage | 10:33:45 |
| Daniel | So the current that a given circuit/component will draw will not change if the voltage does not change (assuming no other changes) | 10:35:00 |
| Daniel | If it helps to use the water analogy, the anount of flow (current) through a pipe is determined only by the size of the pipe (resistance) and pressure (voltage), not by the maximum flow the supply can supply (unless the resistance is too low) | 10:37:19 |
| * Daniel was asleep about 5 minutes ago, hopefully this is somewhat articulate/useful | 10:38:20 |
| rd2 | In other words, the current coming from the battery will step down according to the resistance of the logic board. This resistance determines the maximum allowed current of 14.7A, regardless of the battery's capacity to provide a continuous current of 30A. Have I understood this correctly? | 12:27:50 |
| Daniel | This is not universally true, but as a rule, if voltage is constant current will be too | 12:38:23 |
| Daniel | * The above is not universally true, but as a rule, if voltage is constant current will be too | 12:38:52 |
| rd2 | I have to check the voltage range the bms allows in this batter | 13:00:01 |
| rd2 | 
Download image.png | 13:04:57 |
| rd2 | the maximum power consumption of the logic board states maximum continuous watt consumption 185W and the maximum allowed current is 14.7A so the P= IR = 12.6V*14.7A=185.22W
I calculate the resistance from above R=V/I 12.6/14.7= 0.857142857 Ohm
The allowed voltage range in the battery 14V>=V<=14.2V
Calculate the Currents range for the above voltage I1=V/R = 14/0.857142857=16.33A I2=14.2/ 0.857142857=16.56A
| 13:16:21 |
