ground loop elimination transformer and phantom power

[tempus]

Hey all;

I have a piezo pickup in my electric guitar which I run to my pedalboard, which then goes out to the PA. To eliminate ground loops, I was thinking about using a transformer (http://www.edcorusa.com/Products/ShowProduct.aspx?ID=304) between the pedalboard out and the PA. I just wanted to make sure there wouldn't be a problem if phantom power got turned on accidentally. Here's my calculations:

phantom power = 48v, typically through 6.81K resistors, so the current hitting the transformer would be 7mA. Now if I use P=VI, I get .37W, so I would need a 0.5W transformer. However if I use P=I²R, I get .007²A*150 ohms = 0.007W.

Which equation is right?

Thanks

[Johan]

the transformer secondary shouldnt be connected to ground( the end seing your mixer).  when the phantom is applied, it reaches both ends of the transformer and they would still have the same potential towards each other( since they are not connected to ground), so nothing hapens...   ..so, no problem...
j

[tempus]

the transformer secondary shouldnt be connected to ground

I thought the CT of the transformer would be connected to Pin 1 of the XLR output, thus connecting it to the mixer's ground. Is this not the case? If not how do I connect the transformer to the mixer's ground?

Thanks

[Johan]

you connect the mixers XLR connector pin-2 hot, pin-3 cold and leave the ground unconnected. if you connect the ground too, the ground loop will remain...
j

[tempus]

So the mixer has no connection to ground (from the balanced out) at all? I thought that it was the transformer itself that was breaking the ground loop, since the grounds actually shared no physical connection.

So my connections should be like this?

[Johan]

So the mixer has no connection to ground (from the balanced out) at all? I thought that it was the transformer itself that was breaking the ground loop, since the grounds actually shared no physical connection.

So my connections should be like this?

..yes..if you cennect this way, it wont matter if your phantom is engaged or not, since both pin 2 and 3 has the same potential.
j

[tempus]

yes..if you cennect this way, it wont matter if your phantom is engaged or not, since both pin 2 and 3 has the same potential.

Right, I understand that, but is this the way it's supposed to be connected? I'm still not getting how it will work at all, since the mixer has no ground reference.

Thanks again

[Johan]

yes..if you cennect this way, it wont matter if your phantom is engaged or not, since both pin 2 and 3 has the same potential.

Right, I understand that, but is this the way it's supposed to be connected? I'm still not getting how it will work at all, since the mixer has no ground reference.

Thanks again

..if you imagene the mixer input looking just like your transformer secondary, only with the ground(pin1) going to the centertap. if signal is fed between pin 2 and pin 3, the electronics will sence what's going on even if you disconnect the centertap from ground.
you could offcourse use the ground and only the hot (pin 2) too, but you would loose the benefit if the ballanced circuit taking care of any hum picked upp by the cable
j

[Mike Burgundy]

Just as a clarification: you're using the transformer as a *balanced* line feed now. A normal signal line carries ground and signal which is moving around it (think sine wave). Anything picked up by a long cable (hum, radio) is present at the end and will be amplified with it.
A balanced line carries ground (pin1), signal (pin2), and *inverted* signal (pin3) - or a 180 degree phase shift. (Nice trick to remember the pinout for these connectors: XLR=Xternal Live Reverse)
Both normal and inverted signal lines will pickup any unwanted disturbances exactly the same, so - neat trick - when you reverse the inverted line it is now *in* phase with the normal line, but all nasties are identical but *out* of phase and will cancel out. This makes for nice long and quiet lines.
Secondly: you don't need ground connected at your end. Connected at the recieving end it will still act as a shield, but the signal is taken as the level *between*pins 2 and 3. The input (probably a regular opamp) doesn't care about anything but the differencence between it's + and - input. Remember feeding a signal into the - input of an opamp will invert it as well as (possibly, depending on circuit) amplify. Feeding into + will just amplify (depending on circuit) - so that's your inversion-and-adding right there. This still leaves you with a strong (twice as strong Vpp as an unbalanced line) signal, done.
Should you feed an unbalanced source into such an input, connecting ground to the reverse line will work just fine - but with more risk of signal degradation.
hih