12V regulator

Go a bit more high tech, maybe the solution, have alook at the applications here

martin

Why not just increase the value of the resistor to run rated current through the LED on the positive half cycle and put something like a 1N4002 across the LED in the reverse direction to keep from destroying the LED by zenering the junction on the reverse half cycle?

Jim

-- "It is the mark of an educated mind to be able to entertain a thought without accepting it." --Aristotle

A .047uF 400Vdc or 250Vac (X2) will limit led current to ~2mA, approximately half of which goes through the led. Placing a 100R 1/2W resistor in series will limit capacitive surge current through the diode required to route reverse polarity current around the indicator terminals.

Due to switch construction, the whole shebang will be live when the switch is on, so suitable precautions are required to isolate this from surrounding circuitry and hardware and to fix the parts securely in position.

Brightness may be an issue - larger capacitors, at this voltage, tend to get physically large.

This is probably only cheaper than obtaining the appropriate switch if all of the parts required are already on hand.

RL

I think you're right, probably best to just go get a neon switch. I was trying to use parts I had on hand, this was a nice looking blue led switch, but it's proving a bit more difficult than I wanted for this particular application. Thanks for the replies, Steve

Simple enough. I don't know if my calculations are correct, but I'm going to assume the switch lamp will draw somewhere around 10mA, & using half wave rectified ac, the resistor would see an average power around .575W, not including the drop across the switch led or resistor. So how about a 1N400X in series with a 5k 1W resistor feeding into the switch, with a reverse-biased 1n400X across the switch terminals (it will be across the internal LED & dropping resistor, I can't access just the led). Should that work, or am I way off base?

Thanks, Steve

You should be able to use about a 10k resistor (about 1-2 watt), and a

1N4004 or higher rated diode, in series with the LED. About 12 mA will flow at a 50% duty cycle, for an average of 6 mA, which might not be bright enough, but should be OK. A 5 k 2W resistor will give twice the brightness with more heat. It should be OK if it is away from any plastic parts, as it will get hot. Capacitors are more efficient, but can produce high surge currents through the LED if there are spikes on the line.

Paul

The switch lamp will draw whatever you allow it to draw with the appropriate dropping resistor. Presume for the moment that you use a 10k resistor. You might be tempted to use the value 110 volts for the source, but the eye is a sort of peak detector, so the led will be drawing (110 * 1.414 / 10E3) about

15 mA on the peak of the cycle.

However, heat goes as RMS, and since you don't have CONSTANT RMS but half-wave RMS, the actual value to use to calculate the power consumed by the resistor is ((110/2)^2)/10E3 or about 300 mW. A half-watt resistor will be adequate, but for a little insurance, I'd probably use a 1 watt device.

A 1N400x isn't going to cut it. The reverse voltage across the diode requires you to use at least a 1N4002 or higher in reverse across the LED. You don't need the series diode. The LED itself is a diode. (LED = light emitting DIODE).

Jim

--
"It is the mark of an educated mind to be able to entertain a thought 
without accepting it."
        --Aristotle


"Steve"  wrote in message 
news:9otnf4d1rcpor5l3paluih6kv0mgikgvik@4ax.com...
> On Sun, 19 Oct 2008 15:32:38 -0700, "RST Engineering \\(jw\\)"
>  wrote:
>
>
> Simple enough.  I don\'t know if my calculations are correct, but I\'m
> going to assume the switch lamp will draw somewhere around 10mA, &
> using half wave rectified ac, the resistor would see an average power
> around .575W, not including the drop across the switch led or
> resistor.  So how about a 1N400X in series with a 5k 1W resistor
> feeding into the switch, with a reverse-biased 1n400X across the
> switch terminals (it will be across the internal LED & dropping
> resistor, I can\'t access just the led).  Should that work, or am I way
> off base?
>
> Thanks,
> Steve

Replace the LED and its resistor with an NE-2 (or any general purpose neon bulb) and about a 220K-470K series resistor, and run it off the

115VAC.

Cheers! Rich

It's not necessary to put a diode in series - the LED is alreaady a diode. But it _is_ necessary to put a diode in antiparallel with the LED so its reverse voltage stays belos its limit.

But the OP has already said that he doesn't have access to the resistor/ LED junction, so he's just going to spring for the neon switch.

Cheers! Rich

Good point, like I said, I am very poor at desiging circuits, even something as simple as this. Thanks for your help,

Steve

So, would putting the diode in reverse across the switch terminals not provide enough protection for the LED? Thanks for all the help, Steve

Now that you mention it, yes - the other diode will clamp the voltage across the resistor/LED series circut, which will serve the same function as only clamping the LED - i.e., very limited reverse voltage.

And don't forget the series resistor! :-)

Have Fun! Rich

So, ditch the lighted switch you have, and get one with a neon lamp and ballast resistor builtin and ready for connection to AC.

d

Oh, no, the worst case involves fire or shock hazard (a shorted cap might explode and send a foil streamer out into the world). Seriously, this switch is a fraction of an inch from someone's finger, and there ARE safety-rated, tested designs out there. Buy one.

But if he does use a proper (say for example a 1N4004) in series, are you saying he still needs an antiparallel diode?

My view is one is always better off using a listed UL product in branch circuit wiring. Modifying stuff and installing it in permanent wiring is not allowed by the electrical code, and can result in safety issues - probably does in many cases. If it is a cord and plug connected device, that's a different story. It may still be unsafe, but the code does not require that you use only commercially produced, UL listed devices to my knowlege.

Ed

Fool.

Jim

I suggested the series diode so that the circuit will only conduct during positive half-cycles, and thus reduce power and heating in the series resistor. The reverse current through the diode should be low enough to avoid damage to the LED when the circuit is reverse biased. A diode across the LED will conduct during the negative half-cycles and just waste power. You could use a bridge rectifier and a higher value resistor, but that adds more complexity.

It's a good idea to use a fusible type resistor that is rated for line voltage use, and the entire circuit should be well insulated and protected from human contact or damage. The neon lamp is the best, but that means getting a new switch or making major changes in the existing one. And you can't get a blue neon lamp. You're stuck with red or yellow (except for special green lamps).

Paul

I'm sure we've all been called worse things by better people - I sure have. :-)

Did you have a technical point in mind that led you to post, or was my omission of the word diode prior to the open paren that prompted it? If that's the reason, then I plead guilty as charged.

On the technical side, do you think that a series diode, such as a 1N4004, would not protect the LED from damage in Paul's circuit?

Ed

[top-post repaired] [someone wrote, and ehdjr snipped the attribution]

Yes, a series diode will not protect a(n) LED from reverse voltage, because there is no guarantee that the 1N4004 will block _all_ of the reverse voltage, because there is no guarantee that its reverse current is LOW enough to prevent the remaining reverse drop from exceeding the LED's reverse voltage rating.

And don't top post; it's uncouth.

Cheers! Rich

I found an app note (for optoisolators) that says a small amount of reverse bias will not harm an LED, and it characterizes the reverse current up to

100 mA at about 5-13 volts breakdown.

The maximum reverse current of the 1N400x series is about 10-50 uA, and typically only 0.05 to 1 uA.

The leakage current will probably be enough to cause the LED to go into its avalanche region, but at 50 uA it will not experience significant heating or other deleterious effects. It is possible that some LEDs, such as the blue one in the OP's switch, may be more sensitive, but I found blue LEDs that allow 10 to 100 uA reverse current:

The Wiki is probably incorrect when they state that any reverse bias on an LED will cause instantaneous failure:

I agree that a reverse protection diode is a good precaution, but I don't think any damage will be done by using only a diode and a current limiting resistor.

Paul