Convert 48V Phantom Power to 9V

Hi All,
we want to convert a DC 48V taken from a sound-card Phantom Power.
Currently we are trying to do that using the integrated circuit 7809.

We thought that 48V could be too high as primary input voltage, so we lowered it using a resistive partition. It doesn't work yet and we wouldn't want to use a resistive partition to lower the voltage.

Someone can give as some idea to solve our problem?

Thank in advance!

Onde

The max input voltage of most linear regulators is 27V or lower, and in general you don't want to
use linear regulation of efficiency grounds.

There are DC-DC converters with 48V rated inputs which would be more useful and able to get
more sensible currents out of 48V phantom power (which is limited to a few mA IIRC, something
like a 6k8 series resistor is used).

You want to choose a low quiescent current converter though to avoid wasting any of those mA,
and you will be limited how much power is available to the load circuit.

48V PP is low current (few mAmps max). Moreover 78xx series max input voltage is 35V, afaik.

When I see this question, I suspect that you're trying to use 48VDC phantom power, from a mixer or something like that, to power an electret. I've done that, and it worked quite well. If that's what you're trying to do, or something like it, say so, and I'll try to remember how the gizmo worked.

Yes you can use a regulator to regulate that down to 9V. Just use a big heatsink.
I'd suggesting using a LM317HVT <--- note: HV type

You probably don't want to use DC/DC switching regulator here as any EMI/RF noise may travel back to your mixer and may be heard as noise.
I said MAY because in general, the CMRR of your preamp should take care of that noise and you won't even hear it.
But who knows.

The max. voltage differential (Vout - Vin) of the LM317HV type is 60V. So you're okay*.

• Just use a big heatsink because that will be dissipated as heat.

Thanks all for the answers!

tmd3:
When I see this question, I suspect that you're trying to use 48VDC phantom power, from a mixer or something like that, to power an electret. I've done that, and it worked quite well. If that's what you're trying to do, or something like it, say so, and I'll try to remember how the gizmo worked.

Yes, this is exactly our task If you can remember how the gimzo works, it'll be really good.

Concerning the suggestion of use a LM317HVT it could be a good idea!

vasquo:
Yes you can use a regulator to regulate that down to 9V. Just use a big heatsink.
I'd suggesting using a LM317HVT <--- note: HV type

Is it possible that the sound card phantom power isn't able to sink enough current to make work properly the 7809 or LM317HVT converters?

First you must state how much current you wish to try and draw from the 'phantom power' source. As stated from the below extraction from wikipedia there is only a couple of milliamps available and most all linear regulators require more then that to function let alone supply to the regulated output of the regulator.

So state the voltage and current requirements for the load you are trying to power before asking or offering a possible solution.

Lefty

Phantom powering is not always implemented correctly or adequately, even in professional-quality preamps, mixers, and recorders. In part this is because first-generation (late-1960s through mid-1970s) 48-volt phantom-powered condenser microphones had simple circuitry and required only small amounts of operating current (typically less than 1 mA per microphone), so the phantom supply circuits typically built into recorders, mixers, and preamps of that time were designed on the assumption that this current would be adequate. The original DIN 45596 phantom-power specification called for a maximum of 2 mA. This practice has carried forward to the present; many 48-volt phantom power supply circuits, especially in low-cost and portable equipment, simply cannot supply more than 1 or 2 mA total without breaking down. Some circuits also have significant additional resistance in series with the standard pair of supply resistors for each microphone input; this may not affect low-current microphones much, but it can disable microphones that need more current.

Is it possible that the sound card phantom power isn't able to sink enough current to make work properly the 7809 or LM317HVT converters?

The regulator itself (with no load) doesn't require any significant current. You need to be concerned with the voltage into the regulator, the voltage dropped across the regulator, and the current through the regulator to the load.

And, I would guess that the current required for an electret microphone is no problem either. If you start "pulling" significant current, the regulator can overheat since you are dropping 30V across it. (Heat = Power = Volts x Amps). If you don't know the current required, and if you can't measure it, you'll just have to try it. If the regulator gets too hot to touch, you need a heatsink or some other solution.

I agree with Mark, and I'd stay-away from switching regulators in a microphone preamp circuit (where you want the lowest possible noise).

DVDdoug:

Is it possible that the sound card phantom power isn't able to sink enough current to make work properly the 7809 or LM317HVT converters?

The regulator itself (with no load) doesn't require any significant current. You need to be concerned with the voltage into the regulator, the voltage dropped across the regulator, and the current through the regulator to the load.

And, I would guess that the current required for an electret microphone is no problem either. If you start "pulling" significant current, the regulator can overheat since you are dropping 30V across it. (Heat = Power = Volts x Amps). If you don't know the current required, and if you can't measure it, you'll just have to try it. If the regulator gets too hot to touch, you need a heatsink or some other solution.

I agree with Mark, and I'd stay-away from switching regulators in a microphone preamp circuit (where you want the lowest possible noise).

Well in context of what the OP wants to do (use the 7809 with phantom input power) the statement "The regulator itself (with no load) doesn't require any significant current." is incorrect. A 7809 datasheet I looked at says the Quiescent Current of the regulator is 6ma. Phantom power is limited to one or two ma so there is no way you can use that regulator using phantom power for its input source.

Lefty

OK, I found it. See the schematic.

Phantom power drives the hot and cold leads with +48 VDC, with 0 V on the shield. It's important, form a theoretical viewpoint, to make sure that hot and cold are electrically identical. The intent is that any noise that a long circuit picks up will be impressed equally on the hot and cold leads, since they take the same path. At the mixer, or amplifier, or whatever, only the difference between the hot and cold leads gets amplified, so the common-mode noise is, again theoretically, eliminated. So, the circuit has to be symmetrical with respect to the hot and cold leads.

It also can't draw more than a couple of milliamps without violating the rules.

The electret wants to see a lower voltage - like maybe 2 to 10 VDC - and a not-unreasonable resistance in series.

The resistors in the circuit reduce to an equivalent of a 6.7 VDC source with a series impedance of about 6K1. The voltage is typical for an electret; the impedance is probably a bit higher than is typically used. The voltage divider feeds the electret across the transformer, and the electret delivers AC to the high-impedance side of the transformer, and that puts the low-impedance signal on the hot and cold leads. The matching transformer does a couple of things: it matches the relatively high impedance of the electret circuit to the low-impedance audio system, and it converts the single-ended circuit of the electret to the balanced circuit of the audio system. In the schematic, the electret should be connected to the high-impedance side of the transformer.

As for packaging, the whole thing fit inside one of these:
A95U - Line Matching Transformer (Male XLR to ¼” Male Plug/Female Jack) - Shure USA That was handy, since it had the right connector on each end, and a matching transformer already inside. Three resistors and a little capacitor fit inside with no trouble. I recall that it wasn't particularly easy to get into the enclosure, and it was especially tricky to get in without damaging it too much. I definitely voided the warranty in the process. Later, when that item disappeared, I used a cheaper matching transformer, and I built yet another unit inside an unshielded plastic case - I didn't expect much from that one, but it worked as well as the others. The whole thing didn't seem to be terribly sensitive to the quality of the components.

Purists will hate this circuit, because there's a galvanic connection across the matching transformer. In principle, I don't like that either, but it worked very sweetly for me. I played it for years, in several venues and on several systems, and it always sounded fine and delivered adequate output. Beware, though - the electret, as you might expect, fed back like a banshee.

I recall that I spent a long time figuring all this out - a lot longer than the circuit merits, in retrospect. I recall that I got an unacceptable level of power frequency noise on the first try. I got rid of it by wrapping the electret in aluminum foil, laying the foil across the grounded wire that supported the capsule, and heat-shrinking them together. That eliminated audible noise. I don't recall any other problems in construction.

Your mileage may vary. I might well have gotten lucky with the electret. I'm not sure that I had any right to expect it to perform as well as it did.

Very interesting. I've got a DEITY D-XLR. I was curious so I opened it and made this schematic from it: Schematic on imgur

Phantom power is typically 48V, but I have seen mixers that put out 60V and some condenser microphones with a 1.5V AAA battery that work without Phantom power.

Phantom power is low current, high impedance. You can short out the Phantom power all day with no damage. Because Phantom Power is high impedance, you don't worry about the voltage (you can't) - only that one or two mA is all you are going to get.

You can not beat the "Schoeps circuit" for phantom powering microphones:

There are many minor variations. I prototyped several of them, they solve a lot of issues. I thought I would be really smart with state of the art IC regulators and so on, but in the end they were not better. The Schoeps has the advantage of having a true balanced (differential) output.