Is it safe to run 12v DC LED from 8v AC bell transformer?

[haggis999]

I thought that you might like to hear how my project has been progressing over the past couple of days.

An initial attempt to construct the circuit using a 6A terminal block I happened to have in my toolbox was not a success. The screws did not always pin down the narrow diameter component wires when mixed with bell wire, and I doubted if a 3A terminal block would have been much better, so I decided that strip Veroboard would provide a better solution. I decided to start by using no more than the bridge rectifier and a 1.1 kOhm series resistor and took some care to ensure that the Veroboard layout matched my circuit diagram (using a highlighter pen to trace each of the matching connections). A copy of this layout is attached to this post.

My soldering skills are a bit rusty, but I think I managed to complete the Veroboard assembly without overheating the components. As before, I tested it using a direct feed from my bell transformer (i.e. the door bell was not involved at this stage). The rectified no-load supply was 13.6v DC, just as on my initial test assembly, but I was surprised to find that this was reduced to 6.7v DC when my LED was connected (I previously reported a drop to only 12.8v DC). I also now measure an AC voltage of 1.2v (1.0v AC under no load), whereas before the rectifier appeared to deliver no AC at all. Changing the bridge rectifier made no difference.

Do my latest measurements look weird? Have I perhaps damaged the bridge rectifier?

Despite all this, the LED seemed just as bright as before, so I just went ahead and connected my assembly to the bell circuit. There was no obvious drop in the LED brightness, so I guess the solenoid doesn't have a high resistance. The new pushbutton worked fine, with the LED going out when the bell was ringing, as expected. On reflection, I quite like the LED turning off, as it gives a visual indication that the button has been pushed correctly, so I don't now plan to add a capacitor.

 

[Mikefromlondon]

Glad to hear that its all come ding dong! Of course LED's are very efficient, they hardly require full maximum current specified, that is why a 50mA rated LED I will drive it around 10mA, this can prolong their life a hell of a lot.

Don't use cap, as that would increase final voltage. Plus the fact people pressing the button can get feedback

 

[haggis999]

I am planning to try a higher value of series resistor to make the LED less bright and no doubt that will also extend the life of the LED, as you say.

Have you any explanation for my latest DC and AC voltage measurements?

 

[Mikefromlondon]

I can explain, when you connected LED, it started to draw some current so your transformer voltage went down, all transformers have so called regulation, where they churn out more volts than they are rated at, so for example a 12v transformer may actually churn out 14v if it has poor regulation, but a transformer with good regulation will give near enough what is specified as its terminal voltage.

So your 8v transformer has very poor regulation, hence it is churning out 13.6v rectified voltage, it is unsmoothed voltage and you have not yet connected your capacitor and it is recommended that you do not as the voltage may rise and overpower your LED. Capacitor was only suggested in case you get too much shimmering effect and that would be if you were to use a single diode for half wave rectification.


I also have feeling that by just connecting your LED to 13,6v it should not drop down so heavily, so unless the LED is being overdriven by you not having correct value resistor, i.e. you said you used 1.1K, make sure it is 1.1K and not anything less than that, if it is one with colour coded bands, it should have Brown, Brown, and a Red band, with a 4th band with a larger separation which can be any colour like brown, red, orange or gold or silver to denote tolerence in %, brown would be 1% tolerance, red 2%, orange 3% silver 10% and Gold 5%.

Other ac voltage you read and said it was down to 1.2v, may be you did not change the DC range on your meter to AC range.

 

[haggis999]

...Other ac voltage you read and said it was down to 1.2v, may be you did not change the DC range on your meter to AC range.

I don't quite understand that comment. I was making both my measurements across the feed to the LED, and simply switched the multimeter from DC to AC. If I hadn't switched the range I wouldn't have got a different reading.

My resistor carries a 5 band code. It is brown, brown, black, brown, brown. I believe that makes it a 1.1k Ohm, with 1% tolerance.

 

[Mikefromlondon]

Oh right, ok, in this case you are actually measuring the ripple, it would be lower measurement when the LEd is acting as a load, What you are measuring is average voltage between the two lobes of a rectified wave, which with your LED connected is 6.7v, or with LED out it is 13.6v, but this voltage is like an ac ripple, it would be very near zero, hence why 1.2v only, but without load it may go higher as the lobes would also be higher .

Yes your resistor is 1.1k, do me one favour, now measure the voltage across just the resistor and let me know how much you read. this will tell me what current is flowing through the LED

 

[haggis999]

...do me one favour, now measure the voltage across just the resistor and let me know how much you read. this will tell me what current is flowing through the LED

It's 6.1v DC across the resistor, but I also measure 4.0v AC across this resistor.

 

[Mikefromlondon]

Goodness me, 6.1v Dc is your average mean voltage or what we call it RMS, root mean square, you would have known about this, and your peak would be 1.414 times this RMs voltage so in theory it would be 8.6v peak, so if you were truly getting this assuming your multimeter is accurate to around 5%, and no more, this means your LED is drawing 6.1/1100 ohms = 5.5mA, hardly anything, so I don't think you need to lower it any further, worst case scenario if assuming you did not use this resistor and you applied the entire dc 13.6v across that LED, provided it has an internal resistor and switch manufacturer has limited its current to max 10mA, then dropping around 3 volts for a white LED, you would need to drop about 10v from 13.6v across LED's internal resistor at 10mA it would work out about 1K, which is typical industrial standard for such indicator LEds an d most are now HE (high efficiency)

Here is the thing, you can get LEDs for example Red diffused with a viewing angle of 120 degrees, and 20mA forward current giving you 20mcd, and price is 5p, or you can get HE red clear (not diffused) LED with a forward current of 20mA giving you 2000mcd and a viewing angle of 60 degrees, price 8p, what would you go for? I normally go for HE and use a diffuser lense., and run it at less than 10mA, thereby use a 2.2K resistor.

But I am a bit concerned why your 13.6 volts off load voltage collapses to 6.7V dc, for a 5.5mA load it should hardly drop by more than just a few volts. Do me another favour, check your ac voltage at the input to your rectifier when led is connected. Toy should not check ac voltage at the output of the rectifier as that would be your ripple.

 

[haggis999]

The input to the bridge rectifier, with the LED connected, is 15.8 v AC.

 

[Mikefromlondon]

So it seems it is fine, it is not under any heavy load as it is marginally lower than off load voltage , no need to worry, but are you sure your meter read DC rectified voltage 6.7v? because that doesn't sound right. I have some transformers floating around and i will rig a bridge rectifier and check as I believe it should be around about the same level as AC volts minus a few volts dropped through forward conduction of diodes.

 

[haggis999]

So it seems it is fine, it is not under any heavy load as it is marginally lower than off load voltage , no need to worry, but are you sure your meter read DC rectified voltage 6.7v? because that doesn't sound right. I have some transformers floating around and i will rig a bridge rectifier and check as I believe it should be around about the same level as AC volts minus a few volts dropped through forward conduction of diodes.

In my post #69, I reported that the "DC voltage reduces to 12.8v when driving the LED". However, I have just realised that this measurement was taken at the output terminals of the bridge rectifier (when driving the LED). In my post #121, I mistakenly compared this to the voltage measured across the LED, which is 6.7v DC. My initial perception that there had been a change in the DC voltage readings after rebuilding my circuit assembly was therefore incorrect.

Apologies for wasting your time. Senility is clearly taking its toll...

I have now added a couple more series resistors to give a total of 2.76k Ohm (1100 +1100+560 Ohm). This has dimmed the LED to a more acceptable level for my purposes. Replacing this temporary installation with a single resistor of about 3k Ohm will probably provide the optimum solution.

The voltage measured across the LED (and its internal resistor) is now 4.6v DC, while it's 7.9v across my full set of external series resistors, represented by R1 on my circuit diagram.

 

[Mikefromlondon]

That's jolly good, so it is just under 3mA , and like i said if you hadn't used a bridge it would have been all right too with just a single diode and a series resistor

 

[haggis999]

As a postscript to this discussion, future visitors might be interested to read how I modified the wiring for my new illuminated bell push for compatibility with a Grothe Mistral 600D wireless chime kit. I have just installed this to act as an extender for my existing Friedland chimes. The UK supplier of this kit (http://doorchimesuk.co.uk) states that it is unsuitable for use with illuminated bell pushes, as they reduce the life of the batteries in the transmitter unit. The attached diagram shows how I got round this problem.

BTW, this specific Grothe equipment has a very impressive wireless range.