Help me create Low Voltage switching relay/circuit - Page 1
Forum section: electronics / beginners

author: THATguy
author_rank: Regular Contributor
timestamp: September 09, 2018, 01:23:01 am
content: GOAL:This is for a small handheld torch circuit - all parts should be regular that can be found in any electronics store NOTHING HIGH POWER OR HIGH END - I'm a simpleton trying to make a basic circuit! A similar circuit would be solar powered garden light ( to give an idea of similar componentry, voltage & size of circuit).I have two batteries fully charged. (~1.6v or so, over short distance - torch)Only one will supply the load (single low voltage LED) at a time.I want to keep parts minimal & circuit as simple as possible, I'd prefer not to use resistors, or bulky relay switches especially if that is lossy (no IC's unless one can solve this).When one drops below a suitable voltage I wish for a switch to be activated that lets other battery begin supplying the load.Had trouble finding switches or finding good info on how to make a transistor gate to do this.https://imgur.com/gallery/JTbZ0IMFigure 1. could I wire up a transistor similar to this - in some way that when voltage drops it switches off - but I would still need to link it with the second battery & more transistors somehow??Figure 2. R in circle is a reed switch ( I know it would take more to get them functioning, just would be good if there was a switch that could turn it's self on into a permanent on position until switched off).The battery charges up the cap, when the voltage drops it would go through & turn on the opposite reed switch. (Sorry, there is so much wrong with this circuit lol )Just another brainstorm idea of how it could be done.

author: oPossum
author_rank: Super Contributor
timestamp: September 09, 2018, 02:07:37 am
content: Choose D1 to be about 1 volt less than the low voltage. The relay will stay on until the voltage drops below that point.

author: Brumby
author_rank: Unknown
timestamp: September 09, 2018, 02:53:30 am
content: In reference to the OP's diagram: I can't see how it could switch anything. The diodes are back-to-back, which will effectively remove them and the capacitor from a low voltage DC circuit.As for your inquiry, it lacks some fundamental information.It's like saying "I want to transport some sand" without giving any idea of:- How much? 1kg or 500t?- How far? 5m or 5,000km?- How quickly? 10 seconds or 10 weeks?To even begin to look at this, you need to provide - at the very least - the voltage and current that will be switched. There will likely be other questions as well, but these will emerge along the way.However, even here it is still worth describing theultimate objective, rather than asking for help designing a particular solution that you have already chosen.The reason is simple: there may be better, cheaper, easier and/or more reliable ways of achieving the end result. There may be a design already in existence that does everything you want - and probably more - or one that can be applied to your needs.There is a lot of knowledge on this forum, so you are likely to get some more useful ideas by being more specific in your requirements and less specific in the execution.

author: Brumby
author_rank: Unknown
timestamp: September 09, 2018, 03:04:02 am
content: Quote from: oPossum on September 09, 2018, 02:07:37 amThat is a far more practical circuit - but it still based on an assumption about the voltage available.

author: THATguy
author_rank: Regular Contributor
timestamp: September 09, 2018, 03:07:03 am
content: Just a few volts and very short distance. A switching delay would be fine, keeping the current in the circuit is not critical.Load is just a LED. That is the sole purpose, a torch with a back up battery.I would prefer to not use resistors & keep all voltages just enough to turn the LED on - though if a few more LEDs are required to absorb a bit more excess voltage, or higher power LED that is okay.(I appologise for typos & attrocious spelling - on phone 😭 )

author: JS
author_rank: Frequent Contributor
timestamp: September 09, 2018, 03:23:59 am
content: LED voltage range is quite small, you can take the most out of your battery keeping the losses low using a suitable current source, for small LEDs a J-FET might do, but you need something to keep the LEDs in the happy-working area.To switch between the two batteries a relay could do, but you need some power to keep the relay on, which could be draining your battery noticeable faster than the LEDs. As you will have your LEDs with a CC source you could use them as reference, then switching between two mosfets but that would require a comparator. That would be the only IC in that case, but to ensure start up the preferred state for the mosfet should make them connected and the unconnected should be done by the comparator, for the unused battery. Then you might want to protect the second battery from under-voltage as well, which makes things a bit harder.The LED current would change a lot things for this project. One option might be a miss used CMOS logic IC, using the rails as output and the outputs as rails, check for a controlled rectifier using one chip like that... pretty cool way to solve this sort of problems!JS

author: Brumby
author_rank: Unknown
timestamp: September 09, 2018, 03:47:28 am
content: Quote from: THATguy on September 09, 2018, 03:07:03 amThis is a start. We now know portability will be an issue - and that the power consumption of any solution should be as low as possible.But we still don't know the actual voltage - nor the light output (and, hence, power consumption) of the LED. We also don't have much of a handle on physical size constraints.The phrase "just a LED" is not as helpful as you might think. Is this LED for a pocket sized torch or one of those that can light up someone on the other side of a football field? You haven't made this clear at all.

author: THATguy
author_rank: Regular Contributor
timestamp: September 23, 2018, 06:21:25 pm
content: Okay using an LDR could be a way to switch over to the fresh battery.The second picture shows how the LDR(C) & LED(B) would be setup physically.I've not shown the other side of the battery circuit - but it would be a mirror of this side.The second diode is there because just after this is where the other battery would join.There are a few issues with this:When this battery switches on the LDR would activate again switching it offf - when it switches on there needs to be a way of disabling the LDR... sigh it's already getting messy & too compkex.

author: Zero999
author_rank: Super Contributor
timestamp: September 23, 2018, 06:29:26 pm
content: Quote from: THATguy on September 23, 2018, 06:21:25 pmThe transistor's base is shorted to the collector, which is connected across the battery, causing a large current to flow through the transistor, resulting in its destruction.

author: THATguy
author_rank: Regular Contributor
timestamp: September 23, 2018, 07:31:59 pm
content: Yeah, I might need some resistors, its already getting too big & messy (resistors are wasteful in a low voltage circuit). Don't suppose there are light emitting resistors?Someone has suggested I look at Triac, Thyristor, solid state relay, & analog switches, though analog switches have some specific voltage requirements.Another good suggestion I've had is to look up Low Battery Backup circuitsMost of them seem overly complex.This one looks simple enough, what do you think?strange circuitNot really sure why the battery in this one wouldn't just drain - can't see why there is even a resistor pathway either?