Making a Joule thief mobile charger  - Page 1
Forum section: electronics / beginners

author: dave_j_fan
author_rank: Regular Contributor
timestamp: October 05, 2017, 06:24:20 am
content: I have been eyeing on few Joule thief designsThis is for MCUand this guy seems made one for cellphonehttps://axomtech101.blogspot.in/2017/10/you-said-joule-thief-you-said-joule.htmlcircuit is there in the linkhow long will it charge a cell phone like MOTOG5 and is there a better circuit ( a simple one with fix resistors )why commencial ones use rare chips to boost voltagePlease comment

author: Codebird
author_rank: Regular Contributor
timestamp: October 05, 2017, 06:32:34 am
content: I see no voltage regulator in that circuit. Without a regulator, you do risk causing damage to the phone. It's probably not a good ides to use it.

author: ataradov
author_rank: Super Contributor
timestamp: October 05, 2017, 06:49:19 am
content: According to google moto G5 has battery capacity of 2.8 Ah. Typical AA battery is 2.5 Ah, So even in a totally ideal world, you would need more than one battery to charge the device.In a real world, you will probably need 4-5 batteries after all the losses are accounted for.And yes, you are feeding switching power supply inside the phone with an unfiltered output of another switching power supply. This is not a good idea, which will eventually kill your phone.

author: dave_j_fan
author_rank: Regular Contributor
timestamp: October 05, 2017, 07:04:24 am
content: what about a zener regulation ?will that not be safe .if yes please suggest a design

author: ataradov
author_rank: Super Contributor
timestamp: October 05, 2017, 07:08:39 am
content: Quote from: dave_j_fan on October 05, 2017, 07:04:24 amIt will be extremely wasteful, probably bringing battery count to 5-10.Quote from: dave_j_fan on October 05, 2017, 07:04:24 amWhat do you want to achieve? A real regulated DC/DC boost converter will probably solve the problem.PS: That first circuit with LM1117 is going to work, but again LDO is wasteful in this situation.

author: Rbastler
author_rank: Frequent Contributor
timestamp: October 05, 2017, 07:09:46 am
content: You can buy little modules from China, that let you power a 5V devices from a 1.5V battery.Sent from my A0001 using Tapatalk

author: dave_j_fan
author_rank: Regular Contributor
timestamp: October 05, 2017, 07:35:27 am
content: wanted to make a useful item from wasted toroids and scrapped electronics .thought 1 celll if it can go 50% its a greatI have seen mintyboost but needs special chipchina!There are always parts in china , shipping is loooooong time so dont like

author: ataradov
author_rank: Super Contributor
timestamp: October 05, 2017, 07:37:21 am
content: Quote from: dave_j_fan on October 05, 2017, 07:35:27 amWith this simple approach - no way.Quote from: dave_j_fan on October 05, 2017, 07:35:27 amWell, yes, you want to get as much efficiency as you can here. And dedicated ICs are the best option.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 05, 2017, 08:15:45 am
content: Here's a possible solution that's regulated:Tim

author: viperidae
author_rank: Frequent Contributor
timestamp: October 06, 2017, 06:39:06 am
content: A 2.5Ah AA is only around 3.25Wh, a 2.8Ah LiIon cell is 10.5Wh. You'd need at least 4 AA's to charge a phone. Probably more due to the losses involved in stepping up to 5V then back down to charge the battery.Most phones these days have a nominal voltage of 3.8VAn AA battery is 1.5V

author: dave_j_fan
author_rank: Regular Contributor
timestamp: November 05, 2017, 10:17:33 am
content: Quote from: T3sl4co1l on October 05, 2017, 08:15:45 amplese suggest partnumbers .. very generic ones i may try