Adapting a microphone to remote phantom power

OK, another potential project: re-jiggering the mike for my Sony Walkman Professional cassette recorder to use phantom power.

I have the WM-D6C Walkman "Professional" recorder and a Sony ECM-737 stereo microphone. The rig still makes very nice recordings. (No, I don't usually use it for music, and I'm not pushing cassette as the ultimate recording medium.)

The annoying thing about it is that while the microphone requires a battery (a single AA cell), the recorder is apparently perfectly capable of supplying power to the mike. The mike jack is marked "PLUG IN POWER", which I suppose is Sony's way of identifying phantom power. (This may not be "phantom power" in the sense of standard supply voltage, but instead a proprietary Sony level.) I can't tell you how many times I've had problems recording which turned out to be a low or dead battery in the microphone. Probably my bad for not buying the next-better mike, which I think did run on the recorder's power.

So I'm thinking I could modify the microphone to accept power from the recorder, without modifying the recorder in any way. There are a few things I don't know, however:

o The voltage of the power supplied by the recorder, which runs on 4 AA cells. (48 volts? I'm guessing not, but who knows?)

o The configuration of the microphone jack. My mike uses a 3-conductor plug, so what happens to the power contact in the jack?

o The wiring inside the mike and the difficulty of rewiring it (haven't opened it up yet to look).

Anyone familiar at all with this equipment? Care to give some advice on how to proceed?

Thanks in advance.

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David Nebenzahl
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On 4/9/2009 4:36 PM David Nebenzahl spake thus:

Further research reveals that the recorder supplies 2.8 volts on both channels of the microphone input.

See above; power carried on the inputs.

There's a small circuit board inside (surface-mount, surprisingly complicated for a microphone preamp). Battery power comes in on two separate wires. If battery - goes to ground, then I might be able just to connect battery + to one of the inputs, perhaps with a 1.5 volt zener diode to limit voltage. If not, I'd be afraid of messing it up. Time to look for a schematic?

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On 4/9/2009 8:27 PM David Nebenzahl spake thus:

Whaddya know? Found a schematic, in this copy of the ECM-737 service manual:

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(from this page:
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So looking at the schematic (page 4 of the manual), I'm wondering if a guy couldn't just connect the battery power input to one of the outputs through a 1.5v voltage regulator. While that would provide power to run the mike, I'm wondering what detrimental effects that would have on the channel it's connected to. Any way to isolate the signal from the DC power?

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Reply to
David Nebenzahl

I doubt it'll be that simple but here's the service manual for you as a starting point with schematics.

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Graham

Reply to
Eeyore

From a quick glance it uses a lot ( relatively for a mic ) of 1.5V power. So the powering scheme from the walkman is unlikely to be compatible and even if it was, you'd run the battery(ies) dry in that in no time.

Graham

Reply to
Eeyore

No, the 2.8V supply is likely VERY current limited. Sounds more like a sound card powering method that's good for about 1 mA.

Graham

Reply to
Eeyore

Phantom power - as regards mics - is only applicable to balanced mics and inputs. A stereo balanced mic has five connections. Mono balanced connectors are usually a 3 pin XLR, stereo 5 pin.

The DC is between both balanced audios and screen - the idea being there is no potential difference across the audio allowing a non phantom powered balance mic to be used on a phantom powered input.

You can feed DC to an unbalanced mic - but my guess is the mic has to be designed for this.

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    Dave Plowman        dave@davenoise.co.uk           London SW
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Dave Plowman (News)

Note that the schematic shows 16mA current draw. I'd invest in some good AA batteries. Wikipedia says you can get up to 3Ah which would be nearly 200 hours of use.

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" Alkaline batteries from 1700 mAh to almost 3000 mAh cost a little more, but last proportionally longer. "

Graham

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Eeyore

On 4/10/2009 2:43 AM Dave Plowman (News) spake thus:

Well, as I pointed out in a follow-up post above (don't know if you saw it), the Sony Walkman in this case applies power (2.8V) across both inputs at the microphone jack. The mike input uses just 3 connections. They call it "Plug-in power", which may be Sony's own version of phantom power. And yes, the voltage is present even if one uses a non-phantom-powered mike; presumably, the voltage is low enough not to damage any unpowered microphones. (In my case, the phantom power is just wasted.)

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On 4/10/2009 2:38 AM Eeyore spake thus:

Hmmm; I could run it through a dummy load (resistor) and measure the current draw and find out. You may well be right. Now I need to get a service manual for the Walkman (WM-D5C); ugh. Finding the microphone manual was easy; I struck out looking for one for the recorder.

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David Nebenzahl

But both inputs in this case are the 'hot' connections of a stereo input?

There are no 'versions' of phantom power - it's a universal standard. Anything else would be a nonsense.

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On 4/10/2009 11:12 AM Dave Plowman (News) spake thus:

Yes. The 2.8V DC is carried on both mike inputs (+2.8V = L/R inputs, with respect to common ground).

Apparently this disproves that theory. (I thought the standard for phantom power was 48 volts.) Talk to Sony. (And remember that this is "Plug-in power"(R)(TM), whatever that is.)

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On 4/10/2009 2:36 AM Eeyore spake thus:

Another unrelated question, since you've obviously looked at the schematic: what do you think the function is of the LEDs in each buffer output stage (D1 & D2)? I'm puzzled since I saw these inside the mike; they're both clear LEDs which are completely hidden once the cover is in place. The one visible LED is the power-good indicator, D5.

Are these used for some kind of voltage regulation or current restriction? Maybe I could just clip them out and reduce that 16mA power draw ...

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David Nebenzahl

Hi!

I was just going to say...

I have what must be a very similar Sony microphone that takes a CR2032 cell in its underside. The battery cover is also the on-off switch.

If I didn't loan it to my brother (which probably means it is now permanently "his"...) I will get a model #.

I've used it with a battletank of a late 70s Panasonic tape recorder and a computer to produce recordings of an acoustic guitar. It works fine, and the audio quality is decent.

William

Reply to
William R. Walsh

Seems to be a problem here with what is phantom power. Perhaps you'd read my earlier post on the subject. What you have is a form of line powering - not the same as phantom powering.

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    Dave Plowman        dave@davenoise.co.uk           London SW
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On Fri, 10 Apr 2009 12:13:16 -0700, David Nebenzahl put finger to keyboard and composed:

They set the bias current for the output transistors, approximately

0.45mA.

- Franc Zabkar

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Reply to
Franc Zabkar

On 4/10/2009 3:31 PM Dave Plowman (News) spake thus:

Aaaaargh.

I've read all the posts here on the subject, as well as other material out there in the greater world, and most people seem to agree that phantom powering = line powering, always. (But not necessarily the reverse, or obverse.)

To use another analogy, if phantom powering is a square and line powering is a rectangle, one can say that all squares are rectangles, but not all rectangles are squares.

Or something like that.

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On Fri, 10 Apr 2009 10:44:30 -0700, David Nebenzahl put finger to keyboard and composed:

Each channel of the mike amp PCB has a 2K2 load resistor. The current draw is therefore ...

2.8V / 2K2 = 1.3mA

I suspect that the Walkman has an internal series resistor of around

1.8K and an O/C supply voltage of 5.0V.

If you could bring the Walkman's 5V (?) supply out to a separate connector, and bypass the preamp's DC-DC converter by connecting the Walkman's 5V supply directly to the preamp's B+, then you would have a mike that drew around 5mA or less from the Walkman, plus the 2.6mA of wasted line power via R12 and R32. The latter could possibly be recouped by removing R12, R32, and the two 1.8K series resistors alluded to above.

FWIW, I notice that in the present configuration both C8 and C23 are slightly reverse biased by 0.4V (= 2.8 - 4.8 / 2).

- Franc Zabkar

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Reply to
Franc Zabkar

It's the standard 'computer mike' way. Not very elegant. And don't twist the connector in use because it's carrying DC and will crackle nicely.

Graham

Reply to
Eeyore

Google computer mike / mic power and you'll see how it works. Very different from your ECM.

Graham

Reply to
Eeyore

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