I did anticipate using a very old mains powered Klaxon ET-H as a door bell, but it's so frighteningly loud I decided against it. The neighbours probably wouldn't have liked it much either...
Is it safe to run 12v DC LED from 8v AC bell transformer?
- Thread starter haggis999
- Start date
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Did this once when kids used to play knock knock ginger with my door bell, so at the time I worked in Fire Alarm Industry, and rigged up a loud claxton sounder to my bell circuit that worked on 24v, they soon stopped playing! I suspended the sounder facing down from a window upstairs directing sound straight down at anyone about to ring the bell. Sadly my post man wasn't amused, he complained that he nearly had a heart attack!
As it happens, our house is fitted with a Friedland chime. Ding dong, indeed.
I have summarised most of the advice provided so far in a set of three wiring options, shown in the attached diagram, and I would very be grateful if you good folks would check it carefully for any mistakes. At present, I am tending towards Option C.
I have summarised most of the advice provided so far in a set of three wiring options, shown in the attached diagram, and I would very be grateful if you good folks would check it carefully for any mistakes. At present, I am tending towards Option C.
Seems all OK, perhaps you may want to show the capacitor symbol with one side not infilled, just the outline to represent positive plate of the capacitor, for an LED that does not draw more than about 20- 50mA current a value of 22uF would suffice, rated at approx 16 to 25v , avoid using too high a value, as this may produce some arcing across bell contacts and this could result in switch contacts burning eventually causing early failure, a 22uf would stop simmering effect of half wave DC . On a full wave you may not even need it and avoid it at all cost because if your ac off load voltage is 16v then the final voltage after rectifying and adding that smoothing capacitor will raise it by 1.414, so if you multiply 16 x 1.414, you will get approx 23Volts! with LED connected across it may drop a bit to like 18V, it is still way above the rated value of the LED at 12v. With half wave your average value of DC voltage will be a lot less than 23v it will probably be around 12 - 14v dc. (all depends of course what current the LED is pulling through. It can only be determined by actual circuit and measuring it. Theory is one thing, practice is another.
Theoretically I could send a rocket to moon, but practically I know it will probably land somewhere in Timbuktoo. This is why most missions to Moon and Mars fail the first few times because even the most advanced experts get it wrong as calculations and formulas rarely reflect reality. Effects of solar flares, which no one can predict can deflect intended route.
I do not think you need any smoothing cap on a full wave rectification for an LED.
Theoretically I could send a rocket to moon, but practically I know it will probably land somewhere in Timbuktoo. This is why most missions to Moon and Mars fail the first few times because even the most advanced experts get it wrong as calculations and formulas rarely reflect reality. Effects of solar flares, which no one can predict can deflect intended route.
I do not think you need any smoothing cap on a full wave rectification for an LED.
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Post deleted. sorry i quoted my own post and so deleted now, I was meant to edit it.
Trying to bump up your post count?! 
Probably not, since that would reduce the percentage of posts that had received Thanks or LikesTrying to bump up your post count?!
Kind Regards, John
Capacitor symbols now corrected. However, I'm not sure I know the circuit symbol for solar flares...
Would it not be a good idea to start with an additional series resistor to limit the current through the LED?
yes one can incorporate an extra series resistor if the voltage gets too high, another 1k ohm 1/4 watt resistor, would not harm, but value really depends how bright you need your led to be, if gets too dim, reduce the value. As for solar flare, there are no symbols, as far as I know, what we know is that they even disrupt our major power supplies and causes havoc on earth as well. Engineers didn't consider solar flares into their equations when designing power stations and transmission lines. Satellites can be destroyed or damaged, or provide false data.
There's a bit of a chicken and egg situation here, as I won't be able to measure the real world voltage until I actually connect the circuit. However, I don't want to find out the hard way that a DC voltage of around 23v causes too much current to flow through the LED. It seems safer to deliberately start with an extra series resistor. If the LED is too dim then I could simply try a lower value of resistor, as you said.
Modified diagram attached.
Modified diagram attached.
Yes you could do that, or wire up but don't yet connect to LED, then first measure voltage and if the voltage is way above 12v, then choose a resistor.
Tip on how to work out value of a resistor,
Let us assume on average most LEDs are quoted for their maximum mcd (light intensity measured in micro candle power) at a maximum quoted current, on average this usually around 50mA maximum, (typically 20mA) I never drive LEDs at their maximum ratings, I usually use High brightness LEDs and use around 10mA,
so say your LED was designed to use 50mA, at 12v DC supply, a red LED drops around 2.2v so you are looking to dissipate 10volts at 50mA across the current limit resistor , this works out 10/0.05 =200 ohms, (nearest value 220 ohms) that the switch maker may have used
or at say 25mA, dropping 10volts across the resistor, we would get 10/0.025 =400 ohms or nearest value 390R (R means ohms)
but let us now say you were getting 16volts and not 12v, so you are going to have to dissipate an additional 4 volts on top of your 12volts already catered for by internal resistor within the switch, so at 50mA then the resistor would be 4/0.05=80R or nearest prefered value would be 100R
or at 25mA it would be double that or 160R, nearest preferred value would be 150R or 180R
I stated 1K earlier but this works perfect for most of my applications as I really hate driving LEDs to their maximum current, a;lso helps me keep my power consumption to minimum, however you could find out what your switch maker has used for your LED resistor by measuring current at 12v, so by applying 12v and measuring current through the circuit using an accurate multimeter you could work out your resistor value by dividing 12v minus th3e forward voltage drop of a Red LED at 2.2v so you can take 10v and divide that by current you measured.
Voltage is in volts, and current is in Amps, so 50mA equates to 0.05A.
another example, say your 16v ac after rectifying and smoothed by a capacitor peaks at 23volts, so now you will need to drop nearly 23-12= 11volts at 50mA would need a 220R, at 25mA you will need 440R or nearest value would be 470R.
Tip on how to work out value of a resistor,
Let us assume on average most LEDs are quoted for their maximum mcd (light intensity measured in micro candle power) at a maximum quoted current, on average this usually around 50mA maximum, (typically 20mA) I never drive LEDs at their maximum ratings, I usually use High brightness LEDs and use around 10mA,
so say your LED was designed to use 50mA, at 12v DC supply, a red LED drops around 2.2v so you are looking to dissipate 10volts at 50mA across the current limit resistor , this works out 10/0.05 =200 ohms, (nearest value 220 ohms) that the switch maker may have used
or at say 25mA, dropping 10volts across the resistor, we would get 10/0.025 =400 ohms or nearest value 390R (R means ohms)
but let us now say you were getting 16volts and not 12v, so you are going to have to dissipate an additional 4 volts on top of your 12volts already catered for by internal resistor within the switch, so at 50mA then the resistor would be 4/0.05=80R or nearest prefered value would be 100R
or at 25mA it would be double that or 160R, nearest preferred value would be 150R or 180R
I stated 1K earlier but this works perfect for most of my applications as I really hate driving LEDs to their maximum current, a;lso helps me keep my power consumption to minimum, however you could find out what your switch maker has used for your LED resistor by measuring current at 12v, so by applying 12v and measuring current through the circuit using an accurate multimeter you could work out your resistor value by dividing 12v minus th3e forward voltage drop of a Red LED at 2.2v so you can take 10v and divide that by current you measured.
Voltage is in volts, and current is in Amps, so 50mA equates to 0.05A.
another example, say your 16v ac after rectifying and smoothed by a capacitor peaks at 23volts, so now you will need to drop nearly 23-12= 11volts at 50mA would need a 220R, at 25mA you will need 440R or nearest value would be 470R.
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Indeed, you see now, how hard it is to even earn a single "Thanks" I still haven't earned one yet on this thread, yet you see others how they give their mates fake thanks by making sarcastic remarks and insults. but there you go the world is not perfectly round, there are some flat spots.Probably not, since that would reduce the percentage of posts that had received Thanks or Likes
Kind Regards, John
Half the time thanks and likes are not given as the buttons are hidden!
LOL! Thanks everyone! feel good!
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