Power LED from 48 volt phantom power
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 9th August 2011, 04:10 AM #1 TheFixer   diyAudio Member   Join Date: Aug 2011 Power LED from 48 volt phantom power The title says it all. I'd like to power an led using phantom power. A few other threads talked about using phantom power, but I didn't really understand it. Phantom power is generally 48 volts and 15 milliamps, correct? Leds may take up to 20 ma's, but low power versions can run on as little as 2 ma. What components and connections would be necessary to drive a low power led? And, just as important, how can I calculate the values on my own... what is the theory behind it. thanks!
 9th August 2011, 11:36 AM #2 glennb   diyAudio Member   Join Date: May 2004 Location: Melbourne LEDs drop about 1.5 - 3.5 volts, depending on the type, colour, size, brightness etc. Most modern LEDs will glow quite brightly at only about 5 mA. The math is easy to calculate the value of the required series dropper resistor: R dropper = ( Vsupply - Vled ) / Iled eg. 48 volts phantom supply, Led drops 2 volts approx., go for 5 mA ( = 0.005 Amps) Therefore dropping resistor = ( 48 - 2 ) / 0.005 = 9,200 Ohms. Go for the nearest 5% preferred value, 10K Ohms, or 8.2K Ohms The current can be easily calculated for a given resistor: Iled = ( Vsupply - Vled ) / Rdropper eg. ( 48 - 2 ) / 8200 = 0.0056 Amps = 5.6 mA. You should also calculate the power dissipated in the resistor: Power = ( Vsupply - Vled) * Iled = ( 48 - 2 ) / 0.0056 = 0.2576 Watts So you should use a 0.5 W rated resistor for long term reliability. QED. __________________ Glenn.
 9th August 2011, 12:41 PM #3 davidsrsb   diyAudio Member   Join Date: Dec 2005 Location: Kuala Lumpur Even 1mA is enough if you just want a visual indication and not a torch, especially 2mm high efficiency red
 9th August 2011, 10:51 PM #4 Minion   diyAudio Member   Join Date: May 2006 Location: Cowican Bay , vancouver island Also remember that phantom is allready connected to two 6.81k resistors so take that in consideration when calculating your resistor value .....
 10th August 2011, 03:54 AM #5 TheFixer   diyAudio Member   Join Date: Aug 2011 Thank you for the replies, I'm only looking for an indicator light. Just something easily visible. In the calculations, 48v is standard phantom power (although it's wise to check the specs of my particular unit)... I assume the led voltage drop is given in the led advertizing... is "Iled" the current rating for that led? and thus also listed on the packaging? I've read about how touchy leds can be regarding input voltage. Would I need to use any protection to prevent the led burning out? Or is phantom power stable enough to do without? @minion: One 6.81k resistor is attached to pin 2 and one to pin 3, correct? They are then joined to create the "hot" lead (providing 48v). How do these resistors change the calculation for the Rdropper value? (9.2kohm required accroding to calc...subtract resistance already in place = 2.4kohm Rdropper??) A further question: What power is available after the led and it's matching resistor.? That is, we have 48v @ 15ma ---> resistor --> Led --> ?v @ ?ma I'm hoping to have enough voltage and current left after the Led to saturate a transistor or two. Thanks again!
discrete
diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by TheFixer I've read about how touchy leds can be regarding input voltage.
Never feed an LED voltage - feed it (milli)amps.

An LED is still a diode, almost a short circuit once the voltage across it exceeds the inherent forward voltage drop.
If you feed 3.7 V to an LED with a forward voltage of 3.6 V it may pull a few amps (0.1V/very small R) and it will fail.

So use ohm's law and a series resistor: Simplest current source is a resistor in series with a voltage source.

In your case you will get power from either signal line to ground, via a 6.8k resistor. Use as is, or add another ~10k in series for lower current.

 10th August 2011, 10:56 PM #7 TheFixer   diyAudio Member   Join Date: Aug 2011 Ah, that's the best I've heard that explained. So with a supply of, say, 3.4v... the led does nothing. but meet or surpass the forward voltage and suddenly the led allows electricity to pass largely unhindered. If the supply is capable of providing 1 amp, it will draw 1 amp and so on til the led goes "poof". So, that said...would my application require any protection in case the phantom supply varied by some small amount? (say, 47.5 - 48.5 volts?)
discrete
diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by TheFixer So, that said...would my application require any protection in case the phantom supply varied by some small amount? (say, 47.5 - 48.5 volts?)

Nope...the series resistor makes it a crude current source. If the source voltage changes by 1%, the current changes by 1%.
So, for example, instead of 46/6.8 mA, it will have 46.5/6.8 mA....you will not see the difference.
(I used 46V, not 48V, assuming you have a green LED with Vfd =2V)

I would be more concerned for the 6.8k resistor. Older circuits are designed to give maybe 2mA. You will take about 7mA, so 300mW for the 6.8k resistor to handle. Put the 10k in series. So 2.7mA, 74 mW for the 10k, 51mW for the 6.8k, and everone should be happy.

Last edited by discrete; 13th August 2011 at 12:59 AM.

stratus46
diyAudio Member

Join Date: Nov 2009
Location: Los Angeles
Quote:
 Originally Posted by TheFixer The title says it all. I'd like to power an led using phantom power. A few other threads talked about using phantom power, but I didn't really understand it. Phantom power is generally 48 volts and 15 milliamps, correct? Leds may take up to 20 ma's, but low power versions can run on as little as 2 ma. What components and connections would be necessary to drive a low power led? And, just as important, how can I calculate the values on my own... what is the theory behind it. thanks!
Is this strictly for indicating the presence of phantom power or are you going to attempt to indicate AND run a microphone? Will it be on all the time or is this a 'tester' to indicate the port is active? BTW even old low efficiency LEDs will be visible with less than a mA.

TheFixer
diyAudio Member

Join Date: Aug 2011
If this project works as planned the led will only be lit when the mic is muted.
So in the default position, the mic gets all the power, in the mute position pins 2&3 are connected (through a capacitor) muting the mic, and in the process sending some power to the led.

My mixer is a mackie SR24-4. I can't find any indication in the manual of supplied phantom power in milliamps. It was originally bought in 2000, if that helps.

I'm working my way through your statment:
Quote:
 I would be more concerned for the 6.8k resistor. Older circuits are designed to give maybe 2mA. You will take about 7mA, so 300mW for the 6.8k resistor to handle. Put the 10k in series. So 2.7mA, 74 mW for the 10k, 51mW for the 6.8k, and everone should be happy.
Sooo... old phantom power only provided 2mas, while newer version provide up to 15 mas?
My led and it's resistors will require 7mas?

It's late, and I think my brain is already asleep. I'll try to figure this out in the morning.

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