Timeline for Chaining two batteries to achieve a specific voltage range under their maximum
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2017 at 21:44 | comment | added | Holden | By LEDs not powered on, I mean the image displayed is not completely white; some LEDs will be turned off. | |
| Jun 16, 2017 at 21:40 | comment | added | Holden | be powered on. So I'm starting to think that this thing may actually have a somewhat not terrible battery life... Anything wrong with my logic? | |
| Jun 16, 2017 at 21:39 | comment | added | Holden | They should only need to be run in parallel, now that I am going to use a shift register better suited for the application that doesn't need the larger voltage. I'm starting to think of some ways to conserve power. I'll probably be able to dim the LEDs a little bit from 20 milliamps. Also, the CMOS shift registers will not require much current, from what it is sounding like. The Attiny will be sleeping whenever it is not loading a shift register, which is the majority of the time. 12/14 cycles vs. 1 millisecond or 2 milliseconds. Also, at any given time maybe around half of the LEDs won't even | |
| Jun 16, 2017 at 21:27 | comment | added | Redja | @Holden If the batteries are in series then the current will still need to be 100mA through both. If you are really set on using the CR2450 form factor you are going to have a lot of trouble running the device for any reasonable amount of time. | |
| Jun 16, 2017 at 19:00 | comment | added | Holden | Would you recommend any type of special component for letting current through from the cathode rows of the matrix? I was going to have the 8-bit NTE connect to the bases of 8 transistors, but now I could consider a different approach... The only constraint is that the IC needs to be controllable with only 2 lines, (like the shift register is now). | |
| Jun 16, 2017 at 18:31 | comment | added | Holden | I'm set on using the CR2450 form factor. I have already designed this thing to be laser cut and one of its coolest features is the battery compartment. I'm not too worried about battery life, because I don't plan on running this thing continuously for that long, anyways. | |
| Jun 16, 2017 at 18:26 | comment | added | Holden | If I needed to I could I could run the shift registers at really low currents and have their outputs connected to the bases of transistors. That should minimize inefficiency, I would think... I don't think that would be necessary with new CMOS chips, though... | |
| Jun 16, 2017 at 18:25 | comment | added | Holden | Ok, I think you're right about the shift registers... I'm supplying it varying amounts of current and it seems that only a fraction of that is being output on the pins. I'll have to get some new shift registers. That seems crazy to me, though... Where is all the current going? That may be a totally different question entirely... The battery is holding up fine, though. I had it run 4 LEDs at 16mA each, as a test. I didn't test it at 100mA, but once I have all the components sorted out, I don't think each battery should need to supply more than 50mA. (Continued ...) | |
| Jun 16, 2017 at 6:29 | comment | added | Jeroen3 | @Holden Reconsider using AA(A) batteries in combination with a boost converter. (or buck/boost depending on the amount of cells) | |
| Jun 16, 2017 at 6:12 | comment | added | Holden | Lol, that's understandable.. ;) I just really wanted to make an led matrix so I used what I could scavenge at a local makerspace. | |
| Jun 16, 2017 at 6:08 | comment | added | Ale..chenski | From Energizer datasheet, data.energizer.com/pdfs/cr2450.pdf they list 9mA at 2.7 V, making internal resistance about 33 Ohms. So the battery can deliver max 100mA at zero volts, for 2 seconds. Zero volts is not what you want. The 50 mA is "typical" for DM7495. I am really curious, where did you get these ancient parts? | |
| Jun 16, 2017 at 6:03 | comment | added | Holden | Also the LEDs are run in parallel. This thing is charleplexed. So where are you getting a 2-3 volt drop? | |
| Jun 16, 2017 at 5:59 | comment | added | Holden | The recommended max supply current for the DM7495N is 75 mA. I saw the typical 250mW dissapation, but how do you determine 50mA from that? That seems like a lot? Sorry if these are stupid questions. I'm new to electronics... | |
| Jun 16, 2017 at 5:46 | history | answered | Ale..chenski | CC BY-SA 3.0 |