Phantom power – it’s spooky!

By | March 8, 2012

I’ve been looking at the design of direct injection (DI) boxes with a view to making a phantom-powered piezo preamp for a friend who has acquired an acoustic bass with a passive piezo pickup but doesn’t have a preamp for it.  He’s tried plugging it straight into the mixer – but has experienced major tone-suckage (not unexpectedly and largely as a result of the very high output impedance of the piezo transducer).

There’s an excellent article on phantom power at Wikipedia and another at, but they don’t quite tell the whole story from a designer’s point of view.   I don’t have a copy of IEC61938 so I am taking the phantom power standards from the above articles:

  • Phantom power voltage (P48) is 48V +/- 4V (i.e. 44~52V)
  • This voltage is presented to the XLR hot and cold (pin 2 and pin 3) by 6.8k resistors
  • The resistors should be matched to within +/- 0.4%
  • The latest spec. allows for 10mA per channel (but see below)

For the remainder of this article we will assume that the phantom voltage is 48V and that the supply meets specification.  Note that if you peruse schematics from commercial designs you may notice that the phantom supply resistors are 6k81 (i.e. 6810 Ohms rather than 6800 Ohms).  This is because surface-mount resistors are commonly supplied as E48 or E96 values.  The extra 10 Ohms is not important.  What matters is that the two resistors have (ideally) exactly the same value – not that they are exactly 6800 Ohms.

Here is a simplified schematic showing the basic idea of phantom power.  Rload represents the power consumed by the DI box.  The choice of Rph1 and Rph2 is up to the designer (but they must be identical values matched closer than 0.4% as for the source resistors).

Here is the d.c. equivalent circuit for the above:

Now effectively Rsource1/Rsource2 are in parallel and Rph1/Rph2 are in parallel. The net of two resistors of the same value in parallel is half that value so the equivalent circuit reduces to the following:

A number of things can be inferred from this schematic.

  • The source resistance limits the maximum current available to the load
  • The load is never going to see the full 48V
  • The lower the value of Rph, the more voltage and current will be available for the load

If we set Rload and Rph to zero we can calculate the short circuit current at 48V / 3400 ohms = 14.1mA.  However, the specification only allows for a maximum of 10mA of load current.  48V / 10mA gives a total circuit resistance of 4800 ohms.  Rsource is “fixed” at 3400 ohms, so Rph + Rload must be 4800 – 3400 = 1400 ohms.

We can’t actually make Rph = zero because this would short out the audio signal.  Also, the output capacitors and the phantom power resistors form a high-pass filter so, the lower the resistance of Rph, the more bass frequencies are rolled off, begging the question: what is the practical minimum for Rph?

Well let’s look at it another way.  What is the minimum useful voltage across Rload?  3V? 6V? 9V?  As stompbox merchants we’re happy with a 9V supply so let’s try that.

If Rph + Rload = 1400 Ohm and current is 10mA then the voltage across Rph and Rload is 14V.  If 9V of that 14V is dropped across Rload then 14 – 9 = 5V must be dropped across Rph.  Therefore, Rph is 5V/10mA = 500 Ohm.

So – neglecting any effect on the audio signal – the lowest practical value for Rph1 and Rph2 is 1k0 Ohm which (when connected in parallel) gives a value of 500 Ohm for Rph.

What about that pesky high-pass filter? Well, the corner frequency of an RC filter is 1/(2. pi. Req. Ceq).  We need to find values for Req and Ceq.

Thevenin analysis leads us to conclude that Req is equivalent to 2x Rsource // 2x Rph and  Ceq = C//C//C//C.

Therefore, for the above example Req = 2 x 6k8 // 2 x 1ko = 13.6k//2k = 1.74k

and (assuming C is 10uF) Ceq = 1/4C or 2.5uF

This gives a corner frequency of 36.5Hz, which is not so bad for a DI box.  Also, it’s worth noting that the balanced input of most mixing desks probably have 22uF or 47uF input capacitors which serve to increase Ceq and lower the corner frequency.  One thing we can see from this is that we need to use large coupling capacitors on the output of our DI box.

So where are we up to? We have a source resistance of 3k4, a phantom resistance of 500 ohms and a load resistance which can go as low as 900 ohms.  What happens as we vary the load resistance?

Load Resistance Graph

Perhaps more importantly, maximum power transfer theorem predicts maximum power is available at a load resistance of 3.9k (i.e. 3.4k Rsource + 500 Ohm Rph).  The graph of load power vs load resistance bears this out:

Load Power Graph

What these graphs tell us is that the sweet spot (with an Rph of 500 ohms) is a load resistance of about 4k ohms.  This allows a nice high load voltage of 24V (giving good headroom) and a generous 5mA of current.  If we manage to design a circuit that requires less than 5mA of current then we can increase the value of Rph to raise the load voltage and thereby increase headroom. 

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