Led form single 8v ac wire - Page 1
Forum section: electronics / beginners

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 02:15:05 pm
content: It's missing the current limiting resistor but is this circuit good enough to power an led of a single 8v ac tap from a transformer?

author: ledtester
author_rank: Super Contributor
timestamp: February 01, 2023, 02:31:48 pm
content: Here's a Falstad simulation with the resistor included:(Falstad simulation)The LED will flicker in intensity at the mains frequency. You can play around with the component values to see how they affect the LED output.

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 02:35:31 pm
content: OK, thanks for that. Anyway to stop the led flashing? Attached is the original circuit

author: CountChocula
author_rank: Unknown
timestamp: February 01, 2023, 03:01:49 pm
content: You should test this with actual components and make sure it actually flashes first… simulations aren't always reliable. If you do observe flashing, you can increase the size of your cap… 220µF will probably do it.

author: shapirus
author_rank: Super Contributor
timestamp: February 01, 2023, 03:03:22 pm
content: Quote from: reddish75 on February 01, 2023, 02:35:31 pm1) bigger cap2) full-wave rectifier instead of that single diode

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 03:39:10 pm
content: Can I use full bridge with this circuit?

author: wasedadoc
author_rank: Super Contributor
timestamp: February 01, 2023, 03:41:49 pm
content: Quote from: reddish75 on February 01, 2023, 03:39:10 pmYes.

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 03:49:59 pm
content: So forgive the stupid question but how to connect the full wave rectifier? Ac from transformer to positive side, ground to negative side?

author: wasedadoc
author_rank: Super Contributor
timestamp: February 01, 2023, 03:58:12 pm
content: Quote from: reddish75 on February 01, 2023, 03:49:59 pmNo.https://www.circuits-diy.com/full-wave-bridge-rectifier-circuit/

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 04:02:31 pm
content: Only one output from the transformer for this lamp

author: wasedadoc
author_rank: Super Contributor
timestamp: February 01, 2023, 04:08:49 pm
content: Quote from: reddish75 on February 01, 2023, 04:02:31 pmWhat point are you trying to make by that sentence?

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 04:45:26 pm
content: With my understanding, probably wrong but don't you need two AC inputs to use a full bridge rectifier? Point I was making that the lamp circuit that the led is replacing only has one AC wire coming from the transformer? How would I connect this to utilise a full bridge rectifier?

author: wasedadoc
author_rank: Super Contributor
timestamp: February 01, 2023, 04:50:42 pm
content: Quote from: reddish75 on February 01, 2023, 04:45:26 pmAll transformer windings have two ends. Would not work otherwise.

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 05:06:31 pm
content: Ah right OK. So the other transformer wire goes to a fuse shown circled in red in this attachment, any issues connecting this up to the led circuit?

author: floobydust
author_rank: Super Contributor
timestamp: February 01, 2023, 06:04:17 pm
content: For OP's Kenwood power amplifier, it's just a pilot light but on receivers the transformer lamp winding is for the entire front panel and would power many lamps.For Marantz fuse-lamp replacements, a bridge rectifier, filter cap 47uF then resistor then LED works well with low flicker.You could also just use two LED's in parallel, one for each 1/2 wave as 120Hz ripple you can't really see. Guessing this is a grain of wheat lamp so (only one?) 3mm white LED fits.Still need the dropping resistor though.

author: reddish75
author_rank: Regular Contributor
timestamp: February 01, 2023, 07:15:56 pm
content: Thanks for reply, the original was an 8v ~ 0.15A lamp

author: RJSV
author_rank: Super Contributor
timestamp: February 03, 2023, 03:53:26 am
content: Your bridge will need TWO ac inputs, connected to opposite corners of the bridge. Then the other two opposing corners are the bridge output, rectified to DC + and DC -. You can easily develop the skill to, stare at the thing for a second, and mentally 'trace' the current flow. Only 2 of those 4 diodes, in the bridge, are active, so, mentally you can 'erase' or ignore the other 2. Try drawing that way, and you'll get it. You will have it drawn twice, once for each of the AC input directions.Do that and soon you can visualize how the bridge will always direct an input to the DC output, in correct direction. (That way later you don't need to have memorized it, you can drive it in a couple seconds).I wouldn't trust any circuit like that unless it has a series resistor, to drop a couple volts off the top.A 1 kohm at 10 mA will drop 1 volt off of your average, maybe taking 5 V down to 4V.Led will 'drop' a characteristic voltage AFTER you've put the right amount of current through it. That means the LED runs on current, with voltage being a response, not an initial goal or setting.For example, you might go with 2 k ohms if LED current is too low when a 1 k ohm was used.Voltmeter = real, actual voltsSimulator = ...(sometimes matches reality)Confusing, I know !

author: wasedadoc
author_rank: Super Contributor
timestamp: February 03, 2023, 10:21:58 am
content: Quote from: RJHayward on February 03, 2023, 03:53:26 amThat is back to front. Who wrote "confusing"?

author: mariush
author_rank: Super Contributor
timestamp: February 03, 2023, 11:23:11 am
content: If you have to replace multiple bulbs with leds, it may make more sense to make the circuit a bit more complex than just a bridge rectifier and a resistor for each led to limit current.For example, let's say there's a bunch leds each with a forward voltage of 3.2v (white / blue) and you want 100mA in total ( 10 leds x 10mA each, or 8 leds x 12.5 mA , whatever)The schematic shows 7.5v AC winding, but you can't rely on that, depending on input AC voltage the output AC voltage may vary, and also the AC output may be 10-20% higher with no load. So probably that's why you're measuring 8.1v AC, because it's almost no load.So let's say a minimum of 7v AC is guaranteed, but you could have up to 8.5v AC.When you rectify the AC voltage to DC, you're gonna have a peak DC voltage equal to sqrt(2) x Vac - 2 x voltage drop on diodes inside rectifier.A typical voltage drop on diodes inside rectifier is around 0.7v but may be smaller at low currents like those 50-100mA you'd use for those 4 leds.Let's calculate the lowest and highest peak voltage based on the above :7v AC : Vdc peak = 1.414 x 7 - 2 x 0.7 = 8.5v DC8.5v AC : Vdc peak = 1.414 x 8.5 - 2 x 0.7 = 10.6v DCThis tells you that it's not safe to use a 10v rated capacitor after the bridge rectifier, because it's possible to get more than 10v and damage the capacitor. So you'd go for 16v or 25v rated capacitor.You add a capacitor after the bridge rectifier to smooth the output voltage and get the minimum voltage above some threshold.A formula that approximates how much capacitance you need is this :C = Current / ( 2 x AC frequency x [ Vdc peak - Vdc min desired ) ]In this scenario, our current is 100mA (0.1A) , the Vdc peak is 8.5v DC (minimum we think it's gonna be), and Vdc min desired is whatever minimum voltage we want.For example, if you just plan to connect the leds with current limiting resistors, then you'd want at least 3.2v + a few .10v , let's say 3.5vBut, let's say we introduce a linear regulator to provide a constant voltage getting rid of fluctuations this way. We can use a voltage regulator that outputs a constant 3.3v, then use a resistor for each led to limit the current.In order for the voltage regulator to output 3.3v at 100mA, it will need its input voltage to always be above 3.3v by some amount which varies from regulator to regulator, but a typical amount is 1vSo let's say we want at least 4.5v DC (because 4.5v - 3.3v > 1v) all the time and put the number in the formula - gonna use 60 Hz because schematics show 120v ACC = 0.1 A / [ 2 x 60 x ( 8.5v - 4.5v) ] = 0.1 / 120x4 = 0.1/480 = 2.083 x 10-4= 208 uFSo, a 220uF or a bit higher will be enough to guarantee there's always gonna be at least 4.5v after the capacitor.Now you can use a basic 3.3v linear regulator to get 3.3v and then you can use a resistor to limit the current to 10mAVin - V led = Current x Resistor => 3.3v - 3.2v = 0.01A x Resistor => Resistor = 0.1v / 0.01A = 10 ohmA simple 3.3v fixed voltage regulator like AP2120 would work :https://www.digikey.com/en/products/detail/diodes-incorporated/AP2120N-3-3TRG1/4470762orhttps://www.digikey.com/en/products/detail/diodes-incorporated/AP2139AK-3-3TRG1/4470806orhttps://www.digikey.com/en/products/detail/microchip-technology/MCP1700T-3302E-TT/652676etc etc

author: shapirus
author_rank: Super Contributor
timestamp: February 03, 2023, 11:37:54 am
content: Quote from: mariush on February 03, 2023, 11:23:11 amOnce you reach the point of considering linear regulators, you may find it more reasonable to use an adjustable linear regulator like LM317 in current limiting mode (one of the typical circuits from the datasheet). No dropper resistor, no guesswork or careful calculations, just set the desired max current and it's good to go for any number of LEDs as long as input voltage is high enough to turn them on and the thermal constraints for the regulator are met.

author: Kim Christensen
author_rank: Super Contributor
timestamp: February 03, 2023, 06:55:30 pm
content: I'd just keep it simple. A resistor (820 ohms will give 10ma peak current for 3V white LEDs) and two back to back LEDs. You could even build this into the old bulb's base and pot it with epoxy or something so you don't even have to modify the original equipment.Yes, it'll flicker at a 100Hz or 120Hz rate but your eye won't really notice much. I'd try it on a breadboard and see if flicker bothers you or not.

author: floobydust
author_rank: Super Contributor
timestamp: February 03, 2023, 07:35:08 pm
content: Not sure why Kenwood used such a big lamp. There's lots of room, I'd put a couple 3 or 5mm LED in there and a dropping resistor I'd try 10k because it doesn't need to light up the planet. Or one LED and the bridge rectifier+cap.The original lamp seems to have two wires, so the ground is not in the way. But pulling off the front panel I think is not good, risk mangling things.pic fromhttps://audiokarma.org/forums/index.php?threads/set-to-restore-a-kenwood-model-500.711239/

author: donlisms
author_rank: Frequent Contributor
timestamp: February 04, 2023, 08:43:46 am
content: I think I'd try one resistor and one led. It's a diode, so it will take care of its own polarity issues - it doesn't really need another one. Vision sorta blurs at 20 or 30 Hz, depending on circumstances, so maybe it's fine like that. Just a thought.