Phantom power voltage

What voltage should you see delivered to the mic after the 6.81K resistors. I know that many mics will are not fussy anything between 12 to 52V is fine. But, it seems some of the newer mics are fussy about voltage. The reason I ask is that most phantom power schematics show that you are to trim the voltage regulation stage to 48 to 52V.

I have just built a simple voltage doubler system and have the power/voltage regulator section set to 50V per the diagram. However, the next section is the "Mic input powering" which starts with a series 220 ohm resistor, then has a + to grd. 220uf cap and then a + to grd. LED with a 10K resistor. All of this before it goes to the + splitting 6.81K resistors. At the point just before the 6.81Ks, I now read 34V. Is this reasonable or do I need to trim the voltage up, or, take out that series 220 ohm resistor. Or, is this the attenuation expected if you start at 48V and it really means the mic expects the lower voltage?

I know its a pretty basic question, but, look at the signature below. At least the system works without sparks or anything and will trim solidly between 2 to 60V nice steady DC.

TNX, John

Pretty much by definition you're not going to see 48v with the mic pulling power. 35vdc sounds familiar, though - that's what I've measured a few times.

Mics which pull more current will read lower after the resistor.

A resistor which doesn't drop voltage when it passes current is defying Ohms law, (a fundamental law of physics) and the penalties are stiff for that sort of thing; the courts take a dim view and appeals usually fail. Sentencing is stiff.

The phantom spec says anywhere between 44~52 volts, into a pair of 6.8KΩ resistors. As long as you have that spec, different mics which pull different currents will read different voltages on the MIC side of the resistor.

So there's no answer to your question. Some LDC mics have one FET and not much else, others with multi-patterns have voltage multipliers etc, which might pull more juice in addition to the audio electronic current.

As an example, the Scott Dorsey circuit should pull down to about 36Volts. Other mics will pull down more, and some a little less.

A mic which pulls 48V down to 0 volts has a short in it. A mic which pulls 48V down to 48V is open-circuit. All other mics will do something in between.

Keith

A important thing to know is Ohms law and then DC power

I have 34V going into the 6.81 resistors without a mic connected. This would seem to be out of the 44 to 52V going into those resistors you quote as a spec.

Are you sure the LED is connected before the 6k8 resistors? It's drawing about 5mA and if it were after the 6k8 could account for your drop. Can you take a photo or draw an accurate sketch of that part of your circuit?

A P

The spec (DIN 45 596) quotes 48VDC +-4 V. The DIN spec quotes the maximum current at 2ma per input. The newer IEC spec quotes a maximum current of 10ma. In order to deliver 2ma. through a pair of 6800 Ohm resistors and produce 48V at the connector, you would need to set the supply at 55VDC. I have noticed in the Neumann phantom supplies, the voltage is often set to 52VDC (the value I use also). This will keep you inside the +-4V limit at 2ma.

Here is the schematic I used for the build. I have 50V going to the "mic input powering scheme" section. I have not even hooked up the 2 6.81K resistors. At the point where they would be connected I read 34V to ground. I tried jacking the output to max (about 60V), but can only lift the output from 34V to about 36.5V. Obviously, this is not within the phantom power specs.

http://www.geocities.com/rafafreddy/DIY/pulteq-mb1/phantomps.jpg[/img]

Well, there are a couple of ways to look at this with Ohm's Law.

Case 1:

Let's assume the 220R is correct. You've dropped 50-34.5 = 15.5V across it:

I = 15.5V / 220 ohms = 70.5mA

Which is quite a bit...if that is true, then (assuming a red LED with about a 1.5V forward drop) your LED current limiting resistor is:

R = (34.5V - 1.5V) / 0.0705A = 468 ohms

Is your LED really bright when the phantom is switched on?

Case 2:

What if we assume, instead, that the 200R isn't right, but the LED R is. Then your LED current (the only load you have) is:

I = (34.5V - 1.5V) / 10k = 3.3mA

Which would mean your 220 ohm resistor is actually:

R = (50V - 34.5V) / 0.0033A = ~5k


The only other possibility is that you are drawing current after the 200R from something other than the LED & 10k. Is the 220uF cap known good (not leaky)? Got a photo?

Can I be exceeding the transformers amperage ability and that draws down voltage? Sort of Ohms law driven by current not voltage? First, I measured the volt drop across the 220 ohm resistor, it was .7 volts. Then I shorted across the resistor and I got an additional .7 of output. The interesting thing is that with the switch open I'm at 50V from the power supply. With the switch closed, where I was getting 50V, I now get about 36V. So, it has dragged down the whole voltage output.

The other thing I haven't mentioned is that I am trying to drive 2 mic inputs. So I have 2 of the mic input circuits up to, but not including, the 6.81K resistors. Then, I disabled one of the mic input circuits and the voltage came up to about 41V instead of the 34V I've been seeing.

The transformer I'm using is a 24V, 600ma. I thought 600ma would be enough because the circuit I copied used a radio shack 450ma.

One other question, could the LED be all wrong? There was no value given on the LED so I just tried some I had lying around that were purchased for a stomp box. They do seem to light a nice red. Not overly bright or dim. Somehow, it looks like I'm going to have to reduce the current draw.

TNX, John

Well, now you tell us! So you're pulling about 3mA with the LED load (based on your 0.7V across 220R) which is fine except that causes the LM317 output to go from 50V to 36V. What is the input voltage on the LM317? It sounds like you don't have the required voltage margin for it to regulate properly, so that even a tiny current draw causes it to drop out of regulation. If that's the case you need a transformer with a slightly higher output voltage or you'll need to make a voltage tripler using the existing transformer.

A P

I get 70V at the lm317. What is the purpose of the LED? Could that be taken out of the circuit? Or, are there lower drawing LEDs? Or, could the 10K resistor be increased to reduce current flow?

Correction, the voltage FROM the after the diodes is 77V. HOWEVER, the voltage AT the front end of the LM317 is only 43.5V with the load applied. With the switch open (no load) I get 55V at the LM317. I am dropping 22V across the series resistor just in front of the LM317. That resistor is a 3.3K 2W. The suggested values were between 2.2K and 4.7K. I could reduce the value to raise the voltage at the LM317 but would that be attacking the real problem?

I couldn't open the link to your schematic, but if you are using a doubler circuit to generate your unregulated, are the capacitors sized adequately to provide the current needed? If not it will drop quickly with loading.

In general the the 48V phantom before the 6.8K resistors should be a voltage source and not change appreciably with load, the voltage after the 6.8k will vary depending upon the load and conservative design would hold 48V with the 6.8k resistors shorted. Some mics will drop the phantom down to 10v or less.

JR

Well, I'd really like to see the whole schematic for what you built including the voltage doubler. It is annoying to have to slowly assemble a mental picture of the circuit from piecemeal descriptions over several posts.

John may be right about your doubler caps, but even disregarding that potential problem, it looks like you are simply dropping too much voltage with the 3k3 series resistor feeding your regulator. Do the math. If you intend to run two mics (10mA max each) and the LED (call it 5mA) then you need to have something like 55V at the input of the regulator with a 25mA current draw. If you've got 77V at the doubler prior to that series R, then you want your worst case drop to be 22V at 25mA which is:

R = V/I = 22V/0.025A = 880 ohms

So put a 2W 820 ohm resistor in there. You could go smaller, but be sure to calc the power dissipated in the R and get an appropriately sized one. Or you could eliminate it all together since you're well within the max voltage differential on the 317. You'd just be dissipating the power there instead of splitting it between the reg and the series R. Get a good heatsink and be done with it.

A P

I'm sorry. I thought you would be able to see the whole schematic at the geocites hyperlink. When I tried it off the link above it wouldn't go through because this site added the {img} on the end. I will try to copy the link here again. http://www.geocities.com/rafafreddy/DIY/pulteq-mb1/phantomps.jpg

I assume it will hyperlink this automatically it should end with the .jpg and then it works.

It looks like 470Uf should be adequate for the doubler.. Is it possible there is something weird like the 220uF cap in backwards or broken down?

By all means, Ohm's law will tell you how much current is being drawn.. by the drop across resistors, figure out where it's going.

JR

The diagram given does not have a series dropping R between the diodes and the regulator. That is almost certainly your problem as I said in my previous post (assuming no other differences exist between your build and the schematic you linked).

A P

I guess I linked to the wrong one. He had 2 schemos - the second had the mod adding that series Resistor. If I pull that resistor will I need to worry about a heat sink? If I replaced that resistor with about a 400 ohm. could that be a half watt? I don't have any low value 1 or 2 w resistors on hand. I would have to order just 1 resistor from Mouser. Sioux Falls has only 1 electronic store -- Radio Schack (worthless) and they don't have anything beyond 1/2 w other than the big cements at 10w.

I'm sorry I didn't realize earlier you couldn't get the schemo to come up. You kept asking for a picture and I couldn't understand how that would be any help beyond the schemo. I thought as I described things you were relating that to the schemo.

TNX John

V = IR
P = VI

You can figure out all of the questions you've asked from this basic set of equations...

If it were me, I would jumper the series resistor before the regulator (replace it with a wire) and put a heatsink on the 317. Use the power equation above to figure out how much power the regulator will dissipate:

P = VI = (77-50) * 0.050A = 1.35W

I used 50mA as the current draw through the reg to be somewhat conservative...1.35W would not require a large heatsink, but you probably want something on there. One of those flimsy clip on jobs would probably do OK.

A P