5v to 4.1v for camera - Page 1
Forum section: electronics / projects

author: jaipursky
author_rank: Newbie
timestamp: December 19, 2018, 06:10:09 am
content: Hi, My camera doesn't have external power, so I am making an adapter for it which will have an external power source of 5v. My camera runs on Li-Ion battery 4.1v.My question is what is the best way to convert 5v to 4.1. I know I can use a DC to DC converter 2975 however the challenge is the space, original battery is 1100mA pretty small.I tried exploring TP4056 it fits the size but doesn't solve the purpose because it's current dependent. Camera uses 300-800mA.A few diodes in series can reduce the voltage but that's not ideal. I want to avoid an external regulator for sure, so that I can power the camera using any battery bank.What would you use?

author: james_s
author_rank: Super Contributor
timestamp: December 19, 2018, 06:16:15 am
content: I think I would be tempted to try feeding it 5V directly, there's a good chance the voltage is internally regulated down to 3.3V anyway. Other than that a couple of diodes in series is going to be as efficient as any linear regulator, it probably doesn't draw enough power to really matter.

author: mariush
author_rank: Super Contributor
timestamp: December 19, 2018, 06:25:12 am
content: Probably a 3.6v linear regulator in a package like SOT-223 or something similar with a couple of ceramic capacitors on input and outputLithium batteries are only 4.1..4.2v when full, as they discharge the voltage will go down, so the battery should work with as little as 3.7v, maybe even less.3.6v is a very common voltage, so you'll find linear regulators with fixed output voltage at 3.6v if the camera works at this voltage, you're good, otherwise you'd have to use an adjustable regulator and add a couple resistors to the list.Was about to suggest some from Digikey when I spotted this one :0.9$ TLV75740PDRVR IC REG LINEAR 4V 1A 6WSONhttps://www.digikey.com/product-detail/en/texas-instruments/TLV75740PDRVR/296-50009-1-ND/9462849Outputs 4v up to 1A ... It's tiny 2mm x 2mm with 6 contacts but only 4 are actually used, two are NC ... you have input on one side of the IC and output on the other, so it should be easy to solder.I'd add a diode in front of it, as a way to reduce the voltage drop on the chip and spread the heat, if you're gonna use a 5v power supply (usb charger etc)

author: Chriss
author_rank: Frequent Contributor
timestamp: December 19, 2018, 10:00:09 am
content: Hi!"... a few diodes in series..." I'm not sure why you think you would need a few diodes?You would need one diode in series.A diode in series will sink your voltage down to probably 0.7v that would mean around 4.3v from 5v.That voltage would be correct for your camera.The diode should be enough big to deliver around 1A.But, however, I assume this project is not a business project so you can avoid the efficiency of the circuit is not to worry.A simple light bulb is more badly than your circuit with one diode...

author: Siwastaja
author_rank: Super Contributor
timestamp: December 19, 2018, 12:08:40 pm
content: Quote from: Chriss on December 19, 2018, 10:00:09 am... not at all.Contrary to the common misbelief, a diode is not a constant voltage drop. It's a logarithmic function of the current. A digital camera is expected to draw as little as dozens of microamperes when shut down, to several amperes peak (for example, when storing to the memory card and charging the xenon flash capacitor simultaneously). Given a "typical" silicon diode that can handle the peak current, the voltage drop across the diode would vary between around 0.4V and 0.9V, typically. If the camera performs an overvoltage self-checkbeforehaving significant current draw, this could be a problem.Worse, add any temperature variations in the play, and the range gets wider: higher temperature lowers the drop, lower temperature increases the drop. In the end, while it's somewhat common to use diodes in this class of situations ("almost the right voltage; need to drop a bit"), often it doesn't work out when you do the actual numbers over the full actual operating conditions and realize your "0.7V" suddenly becomes something between 0.3 and 1V.In this case, the diodescould perhapswork if you should expect the full working voltage range of the li-ion cell to work; this meansat leastfrom 3.3V to 4.1V (probably from around 3.0V to 4.2V). You just need to aim lower than 4.1V nominal. Assuming a diode drop is min 0.4V, max 0.8V, and assuming the 5V is actually between 4.9V and 5.1V, using two diodes would vary from 3.3V to 4.3V. This already uses slightly optimistic assumptions and still results in illegal output range (to 4.3V), so it has negative margin for error, but yes, two diodes in seriescouldbarely work by luck. One is definitely not going to (or if it does, then the whole premise is meaningless; it's not sensitive to the voltage at all and would probably work with 5V directly as well!)The 3.6V linear regulator suggestion is a good one; just use a beefy one, Iguesssomething specified to 1.5A nominal (therefore current limit way over 2A) could work. The peak current can be more than you think, and the peaks are long enough (~seconds) that a large electrolytic capacitor on the regulator output is not the answer.

author: Chriss
author_rank: Frequent Contributor
timestamp: December 19, 2018, 01:21:07 pm
content: You alright, but I would suggest to give a schematic of the camera electronic circuit so we can see with what we deal.From my practice over almost 20 year of repairing gsm phones, tablet's etc. they all have a voltage regulator between the battery and also between the wall charger connector.I also assume the main voltage what is coming from the battery is regulated 3.3v or 3.7v or something between. All that voltage regulators working from 5v and upward depending on the circuit design.This unit will work also from 5v connected too probably.But all this reading and writing has no sense if we don't see the schematics of the camera.And doing so, we just "assume" does it "can" or "must not" work etc...Maybe to open the camera and check how the circuit is made...Sent from my GT-I8260 using Tapatalk

author: jaipursky
author_rank: Newbie
timestamp: December 19, 2018, 06:19:34 pm
content: Thank you so much for all responses.

author: Richard Crowley
author_rank: Super Contributor
timestamp: December 19, 2018, 06:33:53 pm
content: It is not clear from your question whether your "Li-Ion battery 4.1v." is some sort of replacable primary cell?Or is it a wired-in secondary (rechargable) cell? If it is built-in and rechargable, how do you recharge it?There are dozens of charging boards that take USB 5V and charge up a single Lipo cell.Isn't that the common way of dealing with this?https://www.adafruit.com/product/1904

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 20, 2018, 12:39:34 am
content: The cheap and dirty solution is a series diode, a trick used in Perk mining clusters back in the day. The right way is to use a buck converter module.

author: james_s
author_rank: Super Contributor
timestamp: December 20, 2018, 01:16:55 am
content: If this were something that would draw >2A I'd say go with a buck converter, but it's likely a few dozen mA, use a diode if you're worried, or just feed it straight from 5V. Anything battery powered will have a fairly wide tolerance range. A lithium ion battery is likely to be on fire by the time it hits 5V but at 4.2V nominal designing a circuit that couldn't tolerate 5V to run from it would be cutting things uncomfortably close for me.