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Let's start off by saying I know just about nothing about electronics, thus the reason I am here.

I have a project I have been working on for a while. Although I have jury rigged earlier versions of this project in the past I am looking for the "correct" way to do it now.

If you look at my avatar you can see what I am doing. I am "attempting" to use the phantom power from a mixer to illuminate my custom built microphones.

The power source is my Mackie Mixer (or any other amplifier/mixer that supplies phantom power for use with condenser micorphones.) Since I am using dynamic microphones I do not need this phantom power for the microphone to work thus I am looking to repurpose it for some LEDs.

In the past I have used single LEDS each wired with a resistor and this worked, but that is too much wiring and way too complicated.

I have since found a few 12V LED modules which have 6 LEDs each and are so much easier to install and conceal in my mics.

The phantom power is 48VDC, but the current is only 30-40 milliamps. Because of that, I am able to light a single LED without blowing it even though they are usually only need about 3 VDC.

I have tried a few 12V LED modules. I currently (no pun intended) am trying a 12V 6 LED strip that is meant to be marker lights on a tractor trailer. It lights up and all but it is dimmer than it should be, I assume because of the low current.

Is there a way to up the current so they get brighter?

I hope I explained this right, like I said I really have no clue what I'm talking about.

This yellow one is the old style with 8 individual LEDs with resistors.

The present one is similar but dimmer. I can post a photo if needed but, visualize it as dimmer.

Lighted microphone:

enter image description here

48VDC was not my ideal power source but I've got to use what I've got.

The "old" way I did this I used resistors. When I noticed putting 4 LEDs in series worked all by themselves I stopped putting in resistors. I then made two circuits of four. I then took the XLR cable and connected the negative wires to the negative pin and each positive went to each of the other pins on the plug so as to keep any balance in either leg. No special LEDs, just your generic single yellow LED. The mic still functioned as normal. I couldn't hear any difference as long as the circuits were balanced. I did notice when all the LEDs were on one leg then it got very "tinny" sounding.

It is starting to look like I need to go back to individual LEDs and forget the ease of using a strip, even though it means many wires and trying to route them in the microphone to avoid seeing shadows cast. That was the big issue in my old style of making these.

I'm going to keep trying here. Any more input would be great and a simple solution will be fantastic.

These new LED strips are premade. They have resistors in them but I cannot see them as they are molded into the strip so the values are unknown. They are designed to work on a 12V circuit and there are six in each strip and I am using two strips, connecting them the same way as above so they are balanced. They are not getting as bright as when I test them with 12V.

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    \$\begingroup\$ For this weird power source you should pay attention how the LEDs are wired up. If you put 4 of the same 12v modules in series you should get something reasonable. If you put them in parallel you will get about the same light as 1 module (each module will be dimmer) \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 23, 2023 at 4:27
  • \$\begingroup\$ with individual LED wiring and not modules, you can wire LEDs in series up to 48 volts (or a bit less and then add a resistor for the difference - only one per series string) and you should do this because it gets more power into LEDs and less into resistors. If the mixer is limiting the current then you don't necessarily need a resistor. If some mixers don't limit the current, then you might want a design with resistors so those mixers won't overload the LEDs. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 23, 2023 at 4:32
  • \$\begingroup\$ "this yellow one is the old style with 8 individual LEDs with resistors...." - please show us how you wired this, and tell us exactly which LEDs you used. \$\endgroup\$
    – Bruce Abbott
    Commented Mar 23, 2023 at 4:48

2 Answers 2

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It is not going to work with LEDs that are designed to work at more than a few mA, which your 12V strips probably are.

The 48V from a mixer is only designed to supply very small amounts of current, just enough to bias a condenser mic and power a single FET or valve preamp. The main 48V from the central PSU is distributed to a number of mic channels (anything up to 60 or so) and there are usually distribution resistors, a few kilohms or so, to ensure that a faulty cable or mic won't take out the whole system.

I know that some manufacturers put lights in their mics, but they must be making sure that the current they are drawing is no more than 1 or 2mA. That 30 or 40mA is for the whole desk, if you pull that on a single channel you will likely not get anything like 48V. Possibly the mic may not function correctly. Even if it works on your Mackie, you will have issues plugging into other desks.

EDIT: by request. It's hard to find decent schemas of mic channels (where the 48V is used) but here is one - the 48V switch is SW1, and the 48V bus is the signal coming in from the lower left. The two distribution resistors are 6k8 in this case. So you cannot get more than 7mA. Values from about 2k to 10k are usual.

(FWIW I worked for Soundcraft back in the day, and also serviced and restored others by DDA, Neve and so on, and they all work pretty much this way. The reason is simply simplicity and also keeping the size and heat design of the central PSU sane. Because these supplies have to be ultra-quiet, they are normally linear regulators, so keeping the supply current on the low side makes sense.)

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  • \$\begingroup\$ I downvoted this because you can totally do it if you pay attention to the current and voltage involved. (Dunno how much light you get out of it - probably enough. The yellow one looks pretty cool) \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 23, 2023 at 8:40
  • \$\begingroup\$ that's what I said. Careful design and choice of LEDs, maybe. Off the shelf 12V strips - no. Particularly one meant for a vehicle, where current clearly isn't an issue. Is there a way to up the current? Not really. \$\endgroup\$
    – danmcb
    Commented Mar 23, 2023 at 9:01
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    \$\begingroup\$ Upvoted. This answer doesn't provide a solution for the OP, but it does add valuable info that was not in the OP or previous answer as I'm writing this. From the OP, I assumed "every mic gets 40mA" which is way different than "all mics share 40mA", which is not something I knew despite having a couple of EE degrees and a casual interest in audio equipment. \$\endgroup\$
    – SSilk
    Commented Mar 23, 2023 at 11:28
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    \$\begingroup\$ If you could add a source for the fact that mics share that phantom power rating, that would be good. E.g. an owner's manual for a mixer that shows taht the phantom power rating is shared, not per-device. \$\endgroup\$
    – SSilk
    Commented Mar 23, 2023 at 11:28
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    \$\begingroup\$ @SSilk ok, done. BTW if you are into audio this mic amp circuit is quite a classic design by a guy called Doug Self, and is very common in mixers of this era. It gives a very good combination of being robust, and having low noise, accuracy and headroom, not easy to do. \$\endgroup\$
    – danmcb
    Commented Mar 23, 2023 at 12:16
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I am not sure I understand your case, but let me try to help.

Most LEDs consume somewhere in the range of 5-20mA in order to properly function. What I would do is first of all look at the datasheet of the LEDs you're using. What is the forward voltage drop? How much current do they need for operation?

Take a look at this datasheet as an example. Notice how it says forward voltage (V_f) is 2.5V and that the continuous current is 20mA? If I were to use this LED with your 48V source (which I do admit is a very unusual source for driving LEDs,) I can choose to connect two LEDs in series (for example,) causing a voltage drop of 2.5X2=5V on them, and then add a series resistor to limit current to 40mA:

48-5=43

43/0.02=2150 ohm

Because 2.2kohm is a standard resistor I would choose this value.

In my imagined implementation I would use two LEDs and one 2.2k resistor in series to the power source.

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  • \$\begingroup\$ 20mA is often uncomfortably bright. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 23, 2023 at 9:38
  • \$\begingroup\$ Then start connecting in parallel, I mean, technically you need to recalculate the resistor according to the change in voltage drop, so in my given example, it is now 48-2.5=45.5V-> 45.5/0.02 =~ 2.2k (still) but make sure you connect both leds in parallel and the resistor in series to both. \$\endgroup\$
    – Tom Kagan
    Commented Mar 23, 2023 at 9:39
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    \$\begingroup\$ Your text is incorrect. If you put two LEDs in series then you still only want 20mA not 40mA (your calculations use 0.02 not 0.04). To the human eye for most LEDs 5mA isn't significantly dimmer than 20mA and you can often get away with 1mA or less if it doesn't need to be visible outdoors. Rather than wasting 90% of the power in the resistor you'd probably want to put as many LEDs as you can in series (the number will depend on their voltage) and then drop the current right down, that will cost more and take up more space bit will give the same brightness for less current draw. \$\endgroup\$
    – Andrew
    Commented Mar 23, 2023 at 12:34
  • \$\begingroup\$ You're right, I Should have divided by 0.04. I didn't recommend using more LEDs as it seems that for this application two LEDs should be enough. Stuffing many LEDs into a microphone sounds more difficult than simply adding a larger resistor, don't you think? I also don't believe efficiency is such a concern in this application. \$\endgroup\$
    – Tom Kagan
    Commented Mar 25, 2023 at 15:31

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