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Sender

Well-known member
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May 17, 2005
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242
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Is it really as simple as W=V*A???


The issue is that I want to implement a "Phantom On" LED on a Pre. I understand the current limiting side to arrive at the value,

R=(V1-V2)/A1. [V1 being supply voltage, V2 being voltage drop across LED, and A1 being draw]

R =(48-2)/0.015
R = 3,066 (nearest size being 3.09k)

So once you plug the numbers into the watt equation....

W = 48 (or 46 to accommodate the voltage drop?)*0.015
W = 0.72


So do I need a 3.09k 3/4W resistor for this application?
 
[quote author="Sender"]R =(48-2)/0.015[/quote]
The maximum amount of current that a mic can draw from P48, with both lines shorted to ground, is 14mA, so this LED would double the amount of power drawn from your +48V supply. You may want to look for a high-efficiency (HE) or high-brightness LED; these only need a mA or two to be bright enough to be seen in full daylight.

JDB.
[you could almost run a discrete op-amp from 15mA@48V]
 
[quote author="jdbakker"][quote author="Sender"]R =(48-2)/0.015[/quote]
The maximum amount of current that a mic can draw from P48, with both lines shorted to ground, is 14mA, so this LED would double the amount of power drawn from your +48V supply. You may want to look for a high-efficiency (HE) or high-brightness LED; these only need a mA or two to be bright enough to be seen in full daylight.

JDB.
[you could almost run a discrete op-amp from 15mA@48V][/quote]

I'm not concerned too much with the draw on the power supply. I have about 1.25 amps available, but efficiency is noted for further design notes.

But I do suppose if I use a LED with less draw I won't have to worry about having a crazy high value of resistor needed.

At 48v, with a 2mA draw, thats roughly 0.1W.

I guess this is the advantage of having a low draw LED, yes?
 
Thats a lot of global warming for one LED.
Think if everybody did that.
If you are stuck knocking down 48, then maybe consider incandescent bulbs, the come in all kinds of voltages, I bet even 48.

Then get a cool Fender blue lens from Angela or Antique, and you do not need a resistor.

Can you tell I have been where you are before?
Thats becuase I'm old.
But not crusty.
Yet.
 
[quote author="mshilarious"]Why do you want to drop the (nearly) full phantom voltage across the LED resistor? Pull it from another power rail.[/quote]

The only reason I am doing this is that this LED is a 'phantom on' indicator, so when you turn the phantom on for that channel, the LED illuminates, hence trying to make it run off of 48V. I suppose I could use a DPDT switch and have the LED running off of something like 15V.


Lets take a look at Joe Malone's 'Go Between' circuit, for educational purposes.

In this example, Joe does the current limiting 1/4W resistors.

phantom.jpg


Does the 470 & 6.8k (in series with the LED) resistor act as a voltage divider, in order to reduce the voltage going into the 6.8 resistor? [/img]
 
No, it's just another load on the rail. But 6k8 gets you 7mA current, for about 0.3W in the resistor. The 470 is not in series with the LED, so no divider there. Actually I am not crazy about the 470, if you stack up a few mics at 10mA draw, you're dropping a fair amount across that resistor. Even one channel, you are right at the edge of a noncompliant supply. If it means a bigger cap, so be it, but the 48V rail should be happily filtered and regulated and a low enough impedance to supply all inputs without dropping below 44V.

Anyway the question is how bright do you need the LED to be? It's not hard to try it and see, maybe a 10k will do. Then you don't have to worry about 2W resistors, 1/2W will do.
 
IMO you might not even need a hi-eff LED. its often amazing to me how little current it takes to lite up a standard part. especially the smaller packages. if you went with the original plan, be prepared for a blinding, annoyingly bright LED.
 
[quote author="mshilarious"] Actually I am not crazy about the 470[/quote]

I thought the 470 ohms was chosen for filtration, as part of the RC filter formed with the capacitor, which is 10uF in the above schem. I usually use something larger, up to 470uF. Perhaps it is overkill, but it makes for a nice smooth phantom supply.

Check out the way the LED is implemented in Eddie Cilletti's schematic:
http://www.tangible-technology.com/power/Phantom_frying.html

I actually built that a few years back, and it works perfectly. It just occurred to be that it is odd to have the LED anode at the full rail voltage and the resistor from cathode to ground.
 
It makes no difference if the resistor is in front of or behind the LED.

Yes, the 470 is there to filter the phantom supply along with cap. But again, I don't see the point of creating a low-impedance phantom supply like in the last schemo you linked and then limiting it so (it also seems pointless to design a circuit that has to shed 30V unless that rail is being used for something else). That schemo says it is for several mics, but there is no way it would be compliant with a full 10mA draw on several mics. Instead, I would filter the zener diode more thoroughly in front of the transistor (much larger C4, and it could stand a resistor too), and use something like 100 ohms for R3 and 1000uF for C3. That would get you four full 10mA mics, or many less thirsty ones.

There are better phantom supply schemos out there, probably somewhere in the metas, I would encourage you to look for them.
 
Yeah, building that design was a good experience because it made me identify all of its shortcomings. It was very worthwhile for the depth of understanding it offered, not its praticality. Since then I've built numerous supplies using standard regulators.

The difference I was alluding to was in the value of the current limiting resistor, and how it might be affected by whether the LED is hung directly on the rail or in series with the resistor to the the rail. With the LED attached to the rail, the low side of the resistor is grounded, so the resistor must dissipate any power remaining after the LED. If the resistor and LED are swapped, as I usually see done, then the LED has to dissipate any extra volts that remain after the drop across the resistor.

Getting back to the JLM design, take a look at his discrete regulator for the phantom supply on the BabyAnimal boards. It's so wonderfully simple.

jlm_ba_reg.PNG
 
[quote author="skipwave"]The difference I was alluding to was in the value of the current limiting resistor, and how it might be affected by whether the LED is hung directly on the rail or in series with the resistor to the the rail. With the LED attached to the rail, the low side of the resistor is grounded, so the resistor must dissipate any power remaining after the LED. If the resistor and LED are swapped, as I usually see done, then the LED has to dissipate any extra volts that remain after the drop across the resistor.[/quote]
No.

In virtually all practical series circuits[*] it makes no difference whatsoever in which order the individual elements are hooked up. There's nothing magical about the example you mention: no matter which is connected to ground, the resistor and the LED get to divide the voltage across the both of them. The resistor follows Ohm's law, and the LED's I-to-V curve shows that the voltage across it is 'close enough to constant' for a wide range of currents. It does not matter which component is connected to ground. The electrons neither know nor care.

JDB.
[*] The only situations where you care which part is grounded is when you are dealing with high-impedance circuits where you run a risk of capacitive coupling with unwanted signals, or in RF designs. There it might make sense to ground the most susceptible part, like the outer layers of a big foil capacitor or a particular end of a transformer.
 
Ahhhh, thanks for the correction, JD. I was mistakenly thinking that the LED had a constant voltage drop and the remainder was dropped across the resistor to ground, or in the other scenario the current draw of the LED determined the drop across the resistor, but I see how that doesn't make sense.
 
IIRC from a class at college, although LED's use less power than incandescent they are less efficient wrt power in to power out. I'll have to see if I still have that textbook. When I am connecting an LED, I find some random used LED, and connect it to whatever supply with a 1k resistor, and adjust the resistor value as needed. You could use the Phantom power to energize the Base of a transistor to power the LED from another supply. Just a thought.
 

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