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!
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.
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.
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!
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.
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?)
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.
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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.
G²
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:
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.
Thanks for your replies.
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:
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.
Thanks for your replies.
You won't need to worry about the 6K8 resistors inside the mixer - they will be sized to cope with a short circuit to earth (or ground). The maximum current that can be drawn from each leg of the balanced pair is 7mA - and that is into a short circuit - so there is no voltage left to do anything else with.
Also remember that you need to preserve the balance of the pair - so you must draw power equally from each wire.
I guess that you are trying to do something like this:
Orchid Electronics Balanced Microphone Muting Pedal
Try and work out how they do it - the muting of the microphone is absolutely silent!
Also remember that you need to preserve the balance of the pair - so you must draw power equally from each wire.
I guess that you are trying to do something like this:
Orchid Electronics Balanced Microphone Muting Pedal
Try and work out how they do it - the muting of the microphone is absolutely silent!
I was concerned for the maximum wattage the 6.8k resistor inside the amp/mixer can handle. John Audio cleared that up....makes sense that it would be sized to handle a short circuit (48/6800 = 7mA => 0.339W, so likely a 1W resistor).
I would still add a 10k (or even 22k) resistor in series with the LED. Most LEDs are very visible at 1-3mA.
@ discrete: So the mixing desk should have sufficient protection to guard against a full short.
In your calculation...48volts through a 6k8 resistor would subject that resistor to 7ma of current. 7mas * 48v = 0.336 watts. is that correct?
As an aside, why do they rate resistors in watts rather than mas?
So at this point,
phantom power comes from mixer, is drawn off pins 2 and 3 by standard 6k8 resistors,
hits the rdropper resistor, then the led, then goes to ground on pin 1.
Vled is 2v
Vphantom is 48v
Target Iled is 2mas
So... 46v / .002a = 23000 ohms
with 6k8 provided by the resistors on the supply pins, that leaves me with 16k2 for the value of rdropper. This arrangement would provide 2mas to the led.
Did I do that right?
@davidsrsb: Is dark current also called leakage current? I assume the led powering circuit will be broken when the led is not lit. So no current should flow through the led when it is "off".
In your calculation...48volts through a 6k8 resistor would subject that resistor to 7ma of current. 7mas * 48v = 0.336 watts. is that correct?
As an aside, why do they rate resistors in watts rather than mas?
So at this point,
phantom power comes from mixer, is drawn off pins 2 and 3 by standard 6k8 resistors,
hits the rdropper resistor, then the led, then goes to ground on pin 1.
Vled is 2v
Vphantom is 48v
Target Iled is 2mas
So... 46v / .002a = 23000 ohms
with 6k8 provided by the resistors on the supply pins, that leaves me with 16k2 for the value of rdropper. This arrangement would provide 2mas to the led.
Did I do that right?
@davidsrsb: Is dark current also called leakage current? I assume the led powering circuit will be broken when the led is not lit. So no current should flow through the led when it is "off".
Because it's the watts that determines the physical size and how warm a resistor gets - the current/milliamps for a certain power/watts varies with different resistances.
You got it. Ohm's law will take you far
Closest E12 series resistors are 15k and 18k.
(Preferred number - Wikipedia, the free encyclopedia)
Either will work fine.
Dark current is not leakage current. It is caused by defects in the led material, allowing electrons to pass through without emitting photons. Unfortunately a higher proportion of the current tends to do this at low operating currents, so the light against current curve is not straight and light output drops to zero at a 100s uA current. Dark current is very variable between devices and gets worse with soldering too long. How you solder leds reliably with lead free solder I have no idea
@benb: I'll have to do some research on amps and watts again..It's been a long time since I've been up to speed in all that.
@discrete: Excellent. thanks for the help.
@davidsrsb: Ah, I had googled dark current, but I couldn't tell what it was. So for any given led there is a current level below which no light is emitted? Still, at , say, 2 mas....there should be enough current for visibility?
@stratus: The idea is that in normal operation the phantom power is given to the mic and the led is not lit. When the mute is activated, the led is lit.
I didn't catch that last bit. how does the power/watts vary?
@discrete: Excellent. thanks for the help.
@davidsrsb: Ah, I had googled dark current, but I couldn't tell what it was. So for any given led there is a current level below which no light is emitted? Still, at , say, 2 mas....there should be enough current for visibility?
@stratus: The idea is that in normal operation the phantom power is given to the mic and the led is not lit. When the mute is activated, the led is lit.
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