How to *use* phantom power?

I have a simple balancing circuit using half a LT1365 opamp. as long as it is powered by a 9V battery all goes well. now i want to modify my circuit to get power from phantom (+48).


but i dont know how to get it to work. i connected both lines with a resistor to a cap and divided the resulting line with two resistors to get a virtual ground. when i simulate in ltspice, the amount of the signal on my power line is still to large and the negative signal line now has the positve signal on it, because the virtual ground has a small part of it and that is fed into the non-inverting input of the lower opamp.

so how do i get a stable DC from my signal lines without feeding back my signal into the opamps?:confused:


2010-03-11 10:43 pm
Normally, no power is available from a phantom supply. I'm not sure what you are trying to do. Does the 9V supply float, or is it on ground? I don't see any power line destinations on your diagram.

You can't ground the negative power line on the amplifiers if you expect to ground the input as well. (You should also add an impedance in the input lines to balance the offset current.)
Hi bob91343, thx for helping. I post an updated circuit to show power supply…


On the left side, you see the phantom power from 9V battery. In the middle, you see the power decoupling. Above C2 i have phantom supply of ≈8.946V. The circuit runs well if i use battery V4 in simulation, but when i leave it out and connect to phantom supply (by closing the little gap), all bets are off ;-/
You forgot to ground the signal input.
er, no;-) Its ground virtually(<-correct term?) at half dc phantom power via R25/R26, C1/C3.

And you are indeed trying to draw too much to run the opamps. I think. How much current do they take?
In simulation is about 13mA each.

scotts hint gave me a clue to a now working solution – at least in simulation: i changed R1, R2, R12 & R17 to 1kΩ each. But, as far as i know, phantom is normally driven tru 6k8… so this means i cant drive LT1365s via standard phantom power, right?


2010-04-24 1:52 am
Mic buffers running on phantom power usually use discrete transistors as the active elements so the current draw can be controlled to work well with phantom power. They're usually arranged as a long-tail pair with collectors going directly to the two signal/power lines, and the base(s) as the input. Also, while the standard voltage for phantom power is 48V, cheaper equipment may use 24V or 12V (with the same 6.8k series resistors), so it's prudent to design a commercial product to run on those lower voltages.

A little googling finds this op-amp-based circuit designed to run on phantom power (5th page, page number in text is 100). Like most of these, this is made for an electret element, but it's easy enough to add a capacitor between an unbalanced mic signal and the input, and to change those bias resistors to set the input impedance:
What, exactly, are you trying to accomplish with your circuit? Did you look at the link I posted? It seems to me that Jensen has already done the work for you. Its a hi-fi, 12-volt circuit that provides 10V-p-p, (3.5 Vrms) input/output. Also, it isn't absolutely necessary to use the transformer. You can use balanced circuits like those available from That Corp.

Just a thought. :)