Tag Archives: math

+4 dBu / -10 dBV

+4 dBu / -10 dBV:  These values are typically used to represent ‘Pro’ and ‘Consumer’ audio signal levels - some gear even has a switch to scale the output, or to adjust input gain.  So what do these values mean on a common scale?

First we need to know how to get from dB to volts.  For linear measures like volts, we defined the measure in dB to be 20log10(volts/reference).  To go backwards we’ll need volts = reference10^(dB/20).

You’ll recall from a previous post that the reference for dBV is 1 volt, and the reference for dBu is 0.775 volts.

Let’s convert 4dBu to dBV, via volts:

0.775*10^(4/20) = 1.228V

20*log10(1.228/1) = 1.8dBV

So +4dBu=+1.8dBV, and the difference between +4dBu and -10dBV is really about 12 dB.

dBV, dBu, and dBm

I’ve always have a little trouble remembering the relationship between dBm and the more common signal measurements of dBu and dBV.  The other day I was looking at a transfer function, and I was given results in dBm - the impetus for this post.

First, I should make sure we’re all on the same page.  What is a decibel (dB)?  The decibel is a relative measurement.  For voltage we calculate dB as 20log10(Voltage/Reference), where the reference can be absolute (like with dBu and dBV) or relative (plain old dB).  For an amplifier that outputs a signal twice as large as the input, 20log10(2/1) = 6 .02 dB.  If we ran a signal through this amplifier twice, our output is 4 times as large as the input, or 20log10(4/1) = 12.04dB.  This is the value of a dB scale - we can add gain values in dB instead of multiplying them: 22=4, while 6.02+6.02=12.04.

Before going on I should point out that dBm is not an appropriate measure for a transfer function.  Straight dB is the way to go, as a linear transfer function should produce an output relative to the input - not relative to an absolute value like dBm, dBu, or dBV.  Let’s say I put 0.5 volts into a system, and I get 0.25 volts back out: 20*log10(0.25/0.5) = -6 dB.

So then dBV, dBu, and dBm are all absolute measurements and each has a defined reference value:

  • dBV - 1 Volt
  • dBu - 0.775 Volts
  • dBm - 1 Milliwatt

dBV is straight ahead and simple to work with - this is the measure I use most frequently.  The dBV value of any voltage is 20*log10(V/1), so 1 Volt is 0 dBV, 2 Volts is 6 dBV, and 0.5 Volts is -6 dBV.

Let’s skip to dBm for a second.  This has a reference of 1 milliwatt, a unit of power.  Power is related to voltage by the equation P = VV/R, where R is the resistance the power is dissipated through.  For instance, if I drive 1 Volt into 1000 Ohms of resistance, 1V1V/1000Ohms = 0.001 Watts = 1 mW.  If I increase the voltage by 6 dB (a factor of 2) to 2 Volts, now I’m dissipating 2V2V/1000Ohms = 4 mW.  This is important - doubling the voltage does not lead to a doubling of power, since the power is related to the square of the voltage.  We actually have to use a different formula for dB of power, 10log10(Power/Reference).  With this formula, if we put in our 4mW and 1mW from above, we still get 6 dB of gain.  Math works!

Back to dBu.  dBu is referenced to 0.775 Volts - an odd number to use as a reference.  However, if you calculate the power dissipating by 0dBu through 600Ohms (a long-standing common value for transmission lines), you wind up with exactly 1mW of power, 0dBm.  So 0dBu = 0dBm, if and only if the load is 600 Ohms.  At any other load impedance this relationship changes.

So back to that transfer function in dBm.  In order to get a relative dB measurement I needed two things: the source voltage and the load impedance.  Knowing the load impedance allows you to calculate the voltage output from the system: V = sqrt(PR).  This is still an absolute measurement, so you need to know what the input signal was to calculate the dB out: 20log10(Output/Input).