Sources for 3 Vdc - Page 1
Forum section: electronics / beginners

author: fixit7
author_rank: Regular Contributor
timestamp: June 08, 2019, 05:11:28 pm
content: I have a laser beam alarm circuit.The detector and piezo speaker uses less than 100 ma.It will be going outside on my porch.It currently runs from 2 aa batteries.I could run a line from outside to supply the voltage but would prefer a battery source.What are my options for the battery source. I would them to last around a year.I already have some 18650s.I would put the detector and laser in some kind of waterproof enclosure.Thanks.

author: sokoloff
author_rank: Super Contributor
timestamp: June 08, 2019, 05:18:12 pm
content: The very first thing I'd do is try them with two D-cell alkaline batteries...

author: DDunfield
author_rank: Regular Contributor
timestamp: June 08, 2019, 07:09:00 pm
content: How much less than 100ma?What is the actual idle (or better yet average) current draw of the device?100ma is a LOT of current to draw from a battery over a year.Very basic calculation with no account for internal leakage, fall-off, self-discharage etc:Year = 365.25 x 24 hours = 8766 hours8766 x 100ma = 876600 mah (milliamp hours)A good alkaline D cell might have 15000 mah.Which means it would run a device drawing 100ma for about 150 hours, or just over 6 days.Dave

author: fixit7
author_rank: Regular Contributor
timestamp: June 08, 2019, 07:47:41 pm
content: I want to make sure I don't ruin my circuit.To find the exact current, do I measure it by putting my meter in between the positive lead and the switch?

author: garethw
author_rank: Regular Contributor
timestamp: June 08, 2019, 08:00:12 pm
content: Put the multimeter in a position so that all of the current passes through it. Right at the battery or in-line with V+ or V-.

author: sokoloff
author_rank: Super Contributor
timestamp: June 08, 2019, 08:01:28 pm
content: Quote from: fixit7 on June 08, 2019, 07:47:41 pmYes, current is the "through variable", so insert your ammeter in series with the power supply.If it's anything off-the-shelf intended to be run on batteries, you're likely (hopefully) going to find that the current is "lower than your ammeter can measure" in steady state. It's going to be difficult to measure average of a widely varying milliamp current with a typical meter.

author: sokoloff
author_rank: Super Contributor
timestamp: June 08, 2019, 08:15:14 pm
content: Quote from: DDunfield on June 08, 2019, 07:09:00 pmThat means it would need battery changes every day running on AA batteries, which leads me to think that if it was designed that way, that the average draw isway, wayunder 100mA.

author: DDunfield
author_rank: Regular Contributor
timestamp: June 08, 2019, 08:17:02 pm
content: Quote from: fixit7 on June 08, 2019, 07:47:41 pmYou need to place your meter (in current mode) between the battery and the device, usually on the non-ground side, which is usually the positive side.Note that in current mode, your multimeter will essentially (**) connect the two leads in a "dead short". Do NOT place them on both terminals of a power source (like the battery).You multimeter will be operating as a "wire" which happens to show the current flowing through it.(**) Worth knowing: There is some "stuff" between the two leads:Any multimeter will have a "current shunt" connecting the two leads. This is a low value resistor over which the meter can measure the voltage drop to determine the current. The size of this shunt will change for various current measurement ranges of your DMM. It can be damaged by too much current!A decent multimeter will also have a fuse in series with the shunt which will blow (hopefully) before damage occurs.In a good multimeter this fuse will be a relatively expensive / harder-to-find high voltage certified fuse.Why am I telling you this:If you try and measure a current on two low a current range, you may blow the fuse and/or damage the shunt.To avoid this, when measuring a current in which you are uncertain of the correct range, start first with the highest range (highest current) that your multimeter has, and then once you know approximately how much current is flowing, you can measure again on a lower range to get a more precise result.eg: If I start on a 10A range, and see that the circuit is drawing < 0.5A, I might then measure in the 500ma range to get a better picture.This a lot better than starting in the 500ma range and discovering that your circuit is drawing 5A (pop goes the fuse).Dave

author: mariush
author_rank: Super Contributor
timestamp: June 08, 2019, 08:21:06 pm
content: You'll definitely need to measure the current.... use a multimeter and measure. But, measure over a period of a let's say 5-10 minutes, and see if the current varies throughout that period.If the amount is much lower, than you may be able to use a tiny solar panel to top up your battery during the day and extend the life of the battery.Digikey stocks some solar cells with very high efficiency (up to 22%), so you could get some amount of power in a small area:https://www.digikey.com/short/pm5j82If you get a panel big enough to output 4.5v or more consistently and at least 20-50mA you could charge up a lithium battery or maybe power your device strictly from solar power during the day.

author: Audioguru
author_rank: Super Contributor
timestamp: June 09, 2019, 01:22:28 am
content: This site is a nightmare to find posts by people. On all other sites I simply click on your profile.I found your Laser Alarm Project but many of your photos were not posted here, instead you posted them over at Drop Box who deleted them.Guess what? Your schematic shows an IC with no part number. It is probably an ordinary old LM555 or NE555 that does not work if its supply is less than 4.5V.Your 3V battery will drop to 2V.

author: fixit7
author_rank: Regular Contributor
timestamp: June 09, 2019, 02:07:19 am
content: Quote from: Audioguru on June 09, 2019, 01:22:28 amIt was a ne555. I abandoned that and found a better design.It uses a S8550 transistor instead of the ne555.See this post.https://www.eevblog.com/forum/beginners/circuit-stopped-working-when-transferred-to-breadboard/

author: Audioguru
author_rank: Super Contributor
timestamp: June 09, 2019, 01:43:55 pm
content: Since you were not using the NE555 timer as a timer I said last month that it was simply a transistor.Your new circuit has no schematic so I do not know if it is correct. If it is correct then its current at 3V will be less than 10mA when it is beeping and less when not beeping.

author: fixit7
author_rank: Regular Contributor
timestamp: June 10, 2019, 02:29:22 am
content: Quote from: Audioguru on June 09, 2019, 01:43:55 pmThanks.I am redoing the circuit as I soldered on the wrong side.When I finish, I will actually measure the actual current used.It currently uses 2 AA batteries. I would like it to last at least a year before changing out the batteries or using some NiMh ones that I already have.Andy

author: ArthurDent
author_rank: Super Contributor
timestamp: June 10, 2019, 02:57:22 am
content: 1) The LMC555 (CMOS) version of the regular 555 timer is designed to work down to 1.5 volts at very low power.2) please reread and understand what many have been telling you about no battery lasting close to the time you say you want.This CMOS 555 circuit may have some useful info. It can flash an LED for up to 6 months on two AAA batteries.https://www.allaboutcircuits.com/textbook/experiments/chpt-8/cmos-555-long-duration-minimum-parts-red-led-flasher/

author: fixit7
author_rank: Regular Contributor
timestamp: June 10, 2019, 03:13:30 am
content: Quote from: ArthurDent on June 10, 2019, 02:57:22 amIt uses a S8550 transistor instead of the ne555.

author: Audioguru
author_rank: Super Contributor
timestamp: June 10, 2019, 03:14:06 am
content: There are good Name-Brand AA alkaline cells available and there are junk ones from China that are labelled Super Heavy Duty. Many of the Chinese ones are dead before they are sold.You showed Radio Shack AA cells. I never tried them.

author: ArthurDent
author_rank: Super Contributor
timestamp: June 10, 2019, 04:04:10 am
content: Quote from: fixit7 on June 10, 2019, 03:13:30 amHow are you powering your source? Whether you can detect light or not is meaningless unless you have a source that will last an equally long time. If you're using a handheld laser it won't last long at all on the small internal batteries. A continuously powered visible LED won't be much better. That post of mine had some useful information that you are completely ignoring. People are trying to help you but you're not making it easy. Scattering your questions over multiple topic is making it nearly impossible. good luck.

author: fixit7
author_rank: Regular Contributor
timestamp: June 10, 2019, 04:36:05 am
content: Quote from: ArthurDent on June 10, 2019, 04:04:10 amI do the best I can.I make mistakes.This is the laser I am using.Depending how fast it depletes batteries will determine if I use batteries or a transformer.Laser diode parameters:Name: Laser small bronze head (red)Transmit power: 150mWStandard size: Φ6 * 10.5Spot mode: point-like spot, continuous outputLaser wavelength: 650nm (red)Optical power: <5mWSupply voltage: 3VDCWorking current: <25mA

author: Audioguru
author_rank: Super Contributor
timestamp: June 10, 2019, 03:05:32 pm
content: You still have not posted a schematic of the transistor Laser Beam-blocked Detector. Since the base-emitter junction of a transistor is affected by temperature then the beeper might not activate when the transistor is warm in summer.

author: Audioguru
author_rank: Super Contributor
timestamp: June 10, 2019, 07:16:21 pm
content: You still did not post a schematic so I made one for you.There is no such thing as a 4400 trimpot. Maybe you adjusted yours to 4400 ohms?The trimpot needs a resistor in series to prevent the transistor from burning out if the trimpot is adjusted too low.

author: fixit7
author_rank: Regular Contributor
timestamp: June 10, 2019, 08:52:08 pm
content: I adjusted the trimpot to get the range.I turned it till I got zero and then turned the other direction to get the 4400 ohms to get the full range.I experimented with 420 -> 800 ohms to see what worked to maintain the speaker at full volume.I adjusted it in testing the circuit, so I may have to buy another one. :-(On the previous circuit that used a 555 chip, I got the circuit to work in ambient daylight by shielding the ldr with a piece of pvc tubing.How can I determine the correct resistor or range I need?Thanks for the circuit diagram.

author: Audioguru
author_rank: Super Contributor
timestamp: June 10, 2019, 09:13:45 pm
content: The trimpot and the LDR form a voltage divider. The resistance of the LDR depends on its sensitivity and on the amount of light it has. The base-emitter junction of the transistor needs about 0.65V for it to turn on and less for it to turn off.If the LDR is 1000 ohms (1k) in bright sunlight then for it to have 0.65V across it, Ohm's Law calculates its current to be 0.65V/1k= 0.65mA.The trimpot voltage will be about 3.0V - 0.65V= 2.35V and its current is about 0.65mA so Ohm's Law calculates it to be 2.35V/0.65mA= about 3615 ohms.The transistor's base current adds to the resistive currents so the value of the trimpot must be less than 3615 ohms, depending on the amount of beeper current.The battery voltage drops as it runs down so you must adjust the trimpot to also allow the circuit to function when the 3V becomes only 2V. Will the beeper work at only 2V?

author: fixit7
author_rank: Regular Contributor
timestamp: June 10, 2019, 10:51:03 pm
content: Quote from: Audioguru on June 10, 2019, 09:13:45 pmOk.I will check if the beeper will work on 2V.I could go the route of using D-cells or 18650s and step down the voltage.

author: fixit7
author_rank: Regular Contributor
timestamp: June 11, 2019, 04:58:32 am
content: Quote from: Audioguru on June 10, 2019, 09:13:45 pmWhile I am waiting on replacement 8550,With a very bright light shining on LDR, ohms is around 30.With laser around 70 ohms.Beeper works at full volume down to 1.7 volts.You mentioned"The trimpot needs a resistor in series to prevent the transistor from burning out if the trimpot is adjusted too low."How can I calculate what resistor I would need?I can do the calculations if you give me the formula.

author: mariush
author_rank: Super Contributor
timestamp: June 11, 2019, 12:39:38 pm
content: The potentiometer is like two resistors in series with values changing as you adjust it.Configure the potentiometer at the minimum where your circuit still works, worst conditions, then measure resistance (between wiper and pin going to transistor).This way you know you can add a resistor smaller than that value and still be able to use the pot. to adjust.For example, let's say you have a 1000 ohm potentiometer and you measure the resistance at 200 ohm. This means you could add a 100 ohm or 150 ohm resistor in series, and lower the potentiometer to 100 ohm or 50 ohm to get same 200 ohm value at the end.The benefit is if you somehow forget and move potentiometer all way to 0, you dontt damage transistor because it still has that 100-150 ohm value

author: fixit7
author_rank: Regular Contributor
timestamp: June 12, 2019, 12:30:09 am
content: Quote from: mariush on June 11, 2019, 12:39:38 pmOk, so with a 150 ohm resistor in series with the pot, the resistance never gets below 150 ohms because that is the minimum resistance of thatpart of circuit (assuming wire and component resistance at zero).

author: MarkF
author_rank: Super Contributor
timestamp: June 12, 2019, 12:49:06 am
content: Quote from: Audioguru on June 09, 2019, 01:22:28 amIt works the same on this forum.-Click on profile-Click on "Show Posts"

author: Audioguru
author_rank: Super Contributor
timestamp: June 12, 2019, 02:55:05 am
content: Quote from: MarkF on June 12, 2019, 12:49:06 amThanks, Mark.