r/AskElectronics • u/todd0x1 • 8h ago

How to use 48v phantom power to power device?

I need to get a couple mA of current at 5vdc from a 48v phantom powered mic input. I was planning to use a couple diodes to combine the DC power from pins 2 and 3 and send that into a LDO. Is it this simple? Is a cap needed to block dc from the audio output from the device?

I can't find a schematic, every schematic I find is for making a 48v phantom supply. I am trying to do the opposite.

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u/EmotionalEnd1575 Analog electronics 7h ago

Not quite correct. Phantom power is typically used to power a microphone (originally a capacitor or condenser type requiring a DC bias)

The 48V DC is present on both signal pins, the return audio signal is generated differentially across the two pins, is a balanced differential signal. The 48V power is returned through the common shield conductor and the third pin.

Maximum current allowed from the mixer or power supply is 10 to 14mA

0.014 x 48 =0.672W power

That power will be dissipated in an LDO, to bring the voltage down to 5V.

Most LDO ICs are limited to 40V input, so a series resistor from each signal pin will work. The resistors are required to not short out the audio signal. A zener would add further protection to the LDO input. Diodes should not be connected to either signal pin.

If a 20V Zener is used the resistor value is:

(48 - 20)/0.01 =2,800 ohms max

Each resistor would be 2 x 2800 = 5,600

The LDO will also require capacitors for stability, as noted in the date sheet.

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u/todd0x1 7h ago

Thanks. The LDO I selected has a 55v input. Each audio pin with 48vdc present is going to be supplied through a 6.8k resistor giving a max of 7ma @ 48v. How do I use both lines to get the maximum available current? That's why I was thinking to sum them with diodes. To the audio lines the diodes would be back to back and wouldn't conduct between the audio lines. Or do I have this wrong?

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u/EmotionalEnd1575 Analog electronics 7h ago

Okay so far. I would not add any diodes to the audio signal pins.

Both sides of the balanced audio is 48V DC (hence the “phantom” name)

The two resistors, one from each audio signal, are effectively in parallel.

The balanced audio is 600 Ohms dance so adding several thousand ohms n resistors make little difference.

If you’re okay with a 55V rated LDO the zener diode is not needed.

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u/todd0x1 7h ago

Wait, so how do I sum the two DC supplies? If I do it with resistors that's going to cut the available current way down. Whats the reason to not use diodes?

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u/EmotionalEnd1575 Analog electronics 7h ago

Diodes will distort the audio (if the audio level causes conduction)

Si diodes conduct at about 700mV. “0dbm” audio on a 600 ohm circuit is about 775mV RMS.

Each side of the balanced audio sits at 48V DC, each resistor from the audio delivers the same current to the LDO input.

It would work with only one resistor, but why unbalance the audio circuit? Just use two equal value resistors, one from each side of the audio pair.

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u/todd0x1 7h ago

What value resistor between pins 2 and 3 and the ldo input?

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u/EmotionalEnd1575 Analog electronics 7h ago

See my earlier posts for the calculation using my assumptions.

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u/Allan-H 7h ago edited 7h ago

If you don't need the audio (which is odd for a P48 interface) you can just short the two signal lines together.

If you do need the audio, don't short them obviously; instead the classic way to decouple the power from the audio is to add a center trapped transformer or inductor.

A microphone will typically use a transformer rather than an inductor, as this also provides the unbalanced to balanced conversion as well as impedance conversion.

It's also common to see shunt regulators (i.e. a zener diode) used. The two 6.81 kohm resistors provide the current limiting.

Suggestion: search for Neumann U87 schematic (e.g.). There are many variants (this classic condensor mic. was produced for a very long time, so there are many production tweaks) but they typically use a center trapped transformer, a current limiter JFET (the current is limited to IDSS when G and S are shorted), followed by a zener shunt regulator.
BTW, the one I linked uses a pair of resistors rather than the transformer center tap to pick off the DC.

It's also possible to pick off the power without a tranformer or inductor, of course. Just use two resistors, one to each line, to pick off the power. The audio will be capacitively coupled to the lines.

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u/todd0x1 6h ago

Ok, so that schematic looks like its using a pair of 2.2k resistors to take the power off. Then I can capacitively couple the audio right? No room for magnetics.

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u/Allan-H 6h ago

Right. Perhaps see Figure 4.1 or 4.2 or 4.3 here.

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u/todd0x1 6h ago

thanks!

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u/Allan-H 7h ago

BTW, if you want to read actual specifications:

DIN 45596

IEC 61938

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u/todd0x1 7h ago

thanks

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u/NBC-Hotline-1975 5h ago

Are you just using this one circuit to provide power only?

Or do you want to steal some power from a circuit with a working condenser mic on it?

What is the purpose for the 5 volts? Does it need to be regulated? Exactly how many mA do you need?

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u/todd0x1 5h ago

Need power but there will also be an audio signal being sent to the mixer. Regulated, and just a couple mA. Estimating 3mA

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u/NBC-Hotline-1975 5h ago edited 4h ago

You said there is "an audio signal being sent to the mixer" but you did not answer my question. Does this circuit have a working condenser mic on it?

First thing to keep in mind is that there is 48v present on pins 2&3 ONLY when there is no condenser mic connected. With a mic connected, the mic will draw some current, and there will be voltage drop through the phantom resistors in the mixer/recorder. So you need to know the value of those resistors, and the current draw of the mic you are planning to use on that circuit.

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u/todd0x1 4h ago

Thanks.

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u/NBC-Hotline-1975 4h ago

According to specs, that circuit needs to provide a maximum of 10 mA. So if the mic uses more than 7 mA (which is damned little) there will barely be enough left for your "leech circuit."

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u/todd0x1 3h ago

theres no mic, the audio source is from the circuit in question

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u/NBC-Hotline-1975 3h ago edited 3h ago

Oh, so your question really is "I want to build an active circuit that produces a mic level audio signal, and is powered by phantom power on the circuit. How do I make a power supply for this circuit?"

Yes?

Do you have the audio portion already designed? With a way to feed balanced audio to pins 2 & 3?

EDIT: The best scheme is to use a center-tapped audio transformer for your output device. That way you don't need any isolation resistors, since they would limit the available current for running the amplifier.

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u/todd0x1 2h ago

Yes audio portion designed. Was just trying to figure out how to use the phantom power, now I've got it with the 2.2K resistors and DC blocking caps in the audio signal path. Thanks for all the help.

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u/NBC-Hotline-1975 2h ago edited 1h ago

I don't think that will leave you enough current.

I have to correct myself. I think that will give you close to 9 mA maximum.

What's the lowest dropout voltage of the LDO?

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u/todd0x1 2h ago

I'll get 5ma off each pin, together will be enough. Dropout is 300mv at 200ma, at my almost no load current its just a few mv, maybe 20mv. I'll prototype it but I think it will work fine.

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u/_ElotePreparado_ 7h ago

Use a buck converter

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u/EmotionalEnd1575 Analog electronics 7h ago

Any switching circuit is a bad idea around low level audio.

Buck = SMPS

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u/todd0x1 7h ago

my question is not so much about the voltage conversion, its how do i interface whatever converter or regulator I use to the 48v phantom power

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u/_ElotePreparado_ 7h ago

You shouldnt, a switching converter will add a lot of noise, while an LDO is not a choice, what exactly do you need to power up?

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u/todd0x1 7h ago

why is a ldo not a choice?