LED pilot light

[sbirkenstock]

Hi everybody,

I decided to use an LED pilot light for my amp project.
It is working, but I got a couple of surprising results though.

I got a full rectifier from Radio Shack.
Way too big, it can handle 4 amp.

However, I feed the rectifier from the filament.
Which is at 6,2 volts.
The rectifier puts out 4,3 volts DC.
I thought converting from AC to DC should bring voltage up by the factor of 1.4 and not down?

However, the LED (also from Radio Shack, the LED assemble) wants 1.7 v at 20mA.
To drop 2.6 volts my calculation says 130 Ohms.
Which is close to what you read on the web.
Well, I started a bit higher, but by accident I put in a 2.7 KOhm instead of 270 ohms.
But the led lights up and the voltage drop is perfect.

I actually use 2 LED´s (power, standby).
The second one I put on 470 Ohms. Worked for a while and stopped working.
Both on 2.7KOhm work fine, perfect voltage drop from 4.3 to 1.7.

If I use two 1KOhm it works as well, but voltage goes up from 1.7v to 1.8v.
So it works, but it´s way off my calculation.
Any hint on this?

cheers,

Stephan

[DonMoose]

Hi everybody,

I decided to use an LED pilot light for my amp project.
It is working, but I got a couple of surprising results though.

I got a full rectifier from Radio Shack.
Way too big, it can handle 4 amp.

However, I feed the rectifier from the filament.
Which is at 6,2 volts.
The rectifier puts out 4,3 volts DC.
I thought converting from AC to DC should bring voltage up by the factor of 1.4 and not down?

However, the LED (also from Radio Shack, the LED assemble) wants 1.7 v at 20mA.
To drop 2.6 volts my calculation says 130 Ohms.
Which is close to what you read on the web.
Well, I started a bit higher, but by accident I put in a 2.7 KOhm instead of 270 ohms.
But the led lights up and the voltage drop is perfect.

I actually use 2 LED´s (power, standby).
The second one I put on 470 Ohms. Worked for a while and stopped working.
Both on 2.7KOhm work fine, perfect voltage drop from 4.3 to 1.7.

If I use two 1KOhm it works as well, but voltage goes up from 1.7v to 1.8v.
So it works, but it´s way off my calculation.
Any hint on this?

cheers,

Stephan

A few things come to mind
- you don't mention a filter cap (try a 470uF, 15-25V part), so that voltage you're measuring is some RMS/average of the rectified AC.
- the LED current rating is a MAXimum - a do not exceed value. You get visible illumination at far lower currents.
- the Vf of a diode (LED or not) is fairly constant with current until you burn it up.
-2x 1k in series s going to drop a bit less voltage for a given current (V= I * R)

Add that cap and measure AC and DC voltages across the cap - you should get an AC reading of less than a volt and a DC voltage of Vin * 1.414 - 2*(diode Vf) (2 diodes in the bridge in series with your AC voltage)

Hope that's clear enough to make sense.

[roberto]

Supply them with AC plus a dropping resistor.
It works flawless. And simpler.

[gingertube]

Supply them with AC plus a dropping resistor.
It works flawless. And simpler.

Roberto,
It works up to a point, well up to a maximum voltage anyway.
LEDS have a weakness, very low (typically about 6-7 volts) maximum reverse voltage. So when running a LED on AC it is a good idea to add a series power diode (same orientation as the LED). This (added diode) will handle the reverse voltage and protect the LED.

Cheers,
Ian

[sbirkenstock]

So what would be the right resistor if I use the 6.3 AC of the filament?

Many schematics also show a diode going the other direction parallel.
In this case I should be able to put a cap between them?

thank you for your help,
Stephan

[gingertube]

The power diode "going the other way" across the LED will also protect the LED from reverse voltage as it will be ON when the LED is reverse biased. So the reverse voltage is clamped to one diode drop or about 0.7V

Q & D calc (Quick and Dirty)
Peak voltage for 6.3V RMS is 6.3 x 1.414 = 8.9V
Forward voltage drop for the LED depends on what color the LED is - assume 2.0V for a typical green LED.

That leaves a peak voltage of 6.9 across the dropping resistor.

Aim for say 5mA peak current.
Ohms Law R=V/I so 6.9V/5mA = 1.4 kOhms.

Why so high a peak current? - well the LED will be on for one half cycle ONLY so you want it a bit brighter for that half cycle than if on continuously.

Try 1K5, if its a bit dull try 1K2 or even 1K. If its too bright try 1K8 or 2K2.

Cheers,
Ian

[surfsup]

Wont the rectifier also drop voltage? Since we're in the single digits for voltage i would assume the rectifier drop to be enough to not be ignored?

[sluckey]

Wont the rectifier also drop voltage? Since we're in the single digits for voltage i would assume the rectifier drop to be enough to not be ignored?

If the rectifier diode is across the LED it's not a factor. If you put it in series with the LED and CLR then you would factor in the .7v diode drop.

[ChrisM]

Wont the rectifier also drop voltage? Since we're in the single digits for voltage i would assume the rectifier drop to be enough to not be ignored?

If the rectifier diode is across the LED it's not a factor. If you put it in series with the LED and CLR then you would factor in the .7v diode drop.

Or use a Schotkey diode for a 0.3V drop. Just make sure it has the specs to handle the job.

Must as mentioned the voltage drop is not a factor.

[sbirkenstock]

Hi there,

it works now, but some still surprising results.
Yes, I forgot the cap at the rectifier, I added it (22mf / 16v) and now the voltage DC is higher than the AC voltage. 6.2AC, 7.2DC.

I got two identical LEDs now, one for power and one for standby. Standby I switch on/off with the sandby switch which is DPDT.
On the 1. LED (power) I put a 1.4K resistor. I get a 2 volt drop over this resistor.
If I switch on the 2. LED (in series with a 2.3K, just a different value to try things out) the voltage on the rectifier drop by .4 volts.
The voltage drop over each resistors remains the same: 2 volt.
Despite the different values??
My LED package says 1.7 volt (at 20mA), so I want to get to 1.7 volt, not 2.

I exchanged the first resistor from 1.4K to 4.7K.
The voltage drop now is 1.9, but the LED is kind of dimm now.
Certainly I´d like it brighter, it is fine with 1.4K as well as with 2.2K.
To Ohm´s law the changes should be much bigger???

Well it works, but I still would like to understand what is going on.

I also added an LED directly to the filament just for testing.
I used a 1.4K resistor.
Works fine.
The voltage drop over the resistor is 2.68AC and 2.0 DC.
6.2 - 2.7 = 3.5 volt AC at the LED.
How much AC is good for an LED?

thank you,

Stephan

[DonMoose]

And LED typically has a maximum reverse voltage of about 5V. That's the reason a lot of folks do the thing with a second rectifier diode in series with the LED to protect it. Any handy diode - 1N914, 1N4148, 1N400x - would be more than adequate.

The voltage across the resistor is of interest only to tell you about the current in the LED - and it's linear with the current once you're above about 2mA.