Blocking phantom

Gentlemen,
My Audix modules share an input for mic and line ( it's switched by a relay at a certain point on the sensitivity switch), but I'd like to have seperate inputs and need to block phantom going to the line input.
I've got this arrangement, but I still have 48V at the line input.
What am I doing wrong?

Ayethengyaw.
peter

Because the negative ends of the caps are just floating in space. Reference them to ground through something less than the 10M input impedance of your voltmeter and that offset should go away. Try, say, a 100K from pin 2 to 1, and another 100K from pin 3 to 1 (on the "line in" jack). This should be a high enough impedance not to load the input, but will still act as a tie-down for your blocking caps.

Thanks Dave.....Do I leave them there, or is this just for testing purposes?

They might not be necessary once that front-end is connected to your preamp--unless the preamp input is AC-coupled--but it won't hurt anything to leave them in place. Bridging 200K across your input will be invisible as far as your input sources and preamp are concerned.

Hi Peter,

How are you measuring the voltage at the Line In socket? Are you sticking the probes of your DMM between Pin 2 and Pin 1, and then Pin 3 and Pin 1? If you're using a DMM and there is no cable connected to the Line In socket, then you will measure +48V. This is because a capacitor has to fully charge up before it can block DC. When you first power up with nothing connected, the caps are empty because there is no path for charging current to flow through them. Try measuring the voltage at pins 1 and 2 again, but this time connect two say 100k resistors from Pin 2 to Pin 1 and from Pin 3 to Pin 1. This will allow a DC path from the ends of the caps to ground, and therefore allow them to cahrge up and block the phantom power.

For a quick test, just get the insert to a male XLR and solder your resistors directly to the solder lugs. Twist together one end of the resistors and solder to Pin 1, and then solder the other ends to Pin 2 and Pin 3. Then plug the XLR into the Line In. If you measure between the pins now you should get no reading (or a small falling reading as the caps charge up)

Give it a go and see if it helps. You can then solder them in permanently. (the 100k resistors will also help discharge the phantom-blocking caps when you switch off phantom power)

:thumb:

Mark

Sorry Dave,

Slow posting- just having a late night cuppa!

:wink:

Mark

Same here... I just finished a cup of Twinings Earl Grey. (I'm all out of PG Tips, and my local store has stopped carrying them).

Yep,

Twinings in the mug here too.

On a Darjeeling tip at the moment!

:grin:

Mark

Marie Brizard in mine..!!
Right, I've put a pair of 100K resistors to ground and have a very slow decline from 48v...currently at 10v. Surely that aint right..??

peter

Yep,

It'll take a while to drop down to 0V due to the Time Constant- the time it takes for the cap to reach 2/3 of the supply is R*C (R in Ohms, C in Farads), or in this case:

100,000 * 0.0001

So 10 seconds- but you need about 4 or 5 time constants, so getting on for a minute to reach "full charge".

Mark

Cheers Mark,
Well it's been 5 minutes now and I'm at 5V. Is there any way to get an instant block...i.e. make it idiot proof?

peter

Well,

You can only reduce the value of the resistors to increase the speed of the "filling up" of the caps- but this will reduce your input impedance. The only dead-cert is to place a mic-line relay in between the two connectors driven off a mic/line switch so that it actually removes the mic connector and phantom-risk completely when a line input is inserted. Can you get switching-XLR sockets I wonder? That would be cool because you could get the relay to switch upon insertion of the Line In XLR

Can't think of any fancy yet simple way off the top of my head.

Night!

Mark

Mark...you're a diamond. :guinness:
Dave....you're a New Yorker. :green: :guinness:

Grassy arse
peter

Peter, you're a gentleman and a scallop. :green: :guinness:

Try using 10K's. The impedance difference between that and the 6.8Ks that are also in the AC signal path will be ignorable. (as opposed to me... just plain ignorant!)

Dave, -No PG? I shall alert the authorities!

Mark;
There was a young girl from Darjeeling,
Who had a peculiar feeling...
So she lay on her back,
And she...

oh, there's the doorbell, I'll just see who it is, shall I?

...

Hello...

Oh, no thanks, we don't want any.

No I'm sure.

Well, we already have one I think.

Okay, I'll ask my wife when she gets home.

Okay, thanks...goodbye.

...

Okay I'm back... where was I? -I forget.

(probably just as well!)

Keef

Pete, Mark, Dave and Keef

:green:

makes me very happy to see some of my bestest friends in the one place.
Still bleeding but perhaps this is home ...

Easy now Kev...sniff.

I havent tried the 10K's yet Keef, but I have another question.

While what is connected to the line input will be OK, how will the mic input be affected with these loading resistors?

pooty

The mic would look into 20K paralleled with the input impedance. At any normal low-Z microphone input, this will do very little difference, if any at all. Only possible drawback is that you use the "floating" property of the input - but using phantom, you've lost it anyway..

Jakob E.

Thanks Jakob.

Marvellous stuff....I now have zero voltage on the line input with phantom engaged, but when I disengage phantom, it fires negative voltage into the line in.
Is there anything I can do about this?

Cheers
peter