When does using a headphone amp with a DAC/amp lower noise?

[JIW]

When the DAC's line-out is connected to the headphone amplifier, it's signal attenuated by the volume control and raised or lowered by the gain. Thus, using the headphone amplifier can reduce the noise in the signal going to the headphones compared to using the headphone output of the DAC. This of course requires that the noise of the headphone amplifier is lower than that of the DAC's headphone output. Here are some formulas that determine when that is in fact the case. The derivations will be at the bottom.

A given chosen voltage output of the headphone amplifier connected to the DAC's line-out is denoted by W and referred to as the target voltage. Let U_HO be the output voltage of the DAC's headphone output with its volume set to full, let the noise voltage of it be N_HO and let its signal to noise ratio in dB be SNR_HO. Let the signal to noise ratio in dB of the DAC's line-out be SNR_DAC and let the noise voltage of the headphone amplifier be N_HPA. Thus, if the noise from the DAC's line out and the headphone amplifier are uncorrelsted, as long as the target voltage is as below

the noise out of the headphone amplifier will be no higher than that of the DAC's headphone output. If the inequality holds, it is lower. Further, for a given target voltage using the headphone amplifier will have the same or less noise if

Taking the logarithm to express it in dB,

where n_HPA is the headphone amplifiers noise level in dBV, i.e. 20log_10(N_HPA), and n_HO is the noise level of the DAC's headphone output in noise level in dBV, i.e. 20log_10(N_HPA). Of course, the difference between the two is just in dB.


Derivations.

 

[AnalogSteph]

This is the kind of thing that would arguably be most useful in the backend of a little JS calculator that takes the various specs required and lets you play with level while plotting noise floor / SNR for both variants.

It also implies that n_HPA is constant, which is not necessarily the case - even a 10k volume pot is going to have up to 2.5 kOhms of source impedance, so even when paired with an ideal noiseless voltage follower that's up to 0.9 µVrms into 20 kHz at 295 K right there (at V = -6 dB), and probably a bit more when a real opamp is involved... and I'm pretty sure that real amplifiers (that aren't an O2) also tend to have at least some gain after the pot.

 

[JIW]

This is the kind of thing that would arguably be most useful in the backend of a little JS calculator that takes the various specs required and lets you play with level while plotting noise floor / SNR for both variants.

It also implies that n_HPA is constant, which is not necessarily the case - even a 10k volume pot is going to have up to 2.5 kOhms of source impedance, so even when paired with an ideal noiseless voltage follower that's up to 0.9 µVrms into 20 kHz at 295 K right there (at V = -6 dB), and probably a bit more when a real opamp is involved... and I'm pretty sure that real amplifiers (that aren't an O2) also tend to have at least some gain after the pot.

Where is the implication of constant noise from the headphone amp? In the equations, the noise of the headphone amp does not need to be constant and can depend on volume and gain. Admittedly, I did not write anything about it and don't denote the noise of the headphone amp as a function of gain and volume control setting which is suggestive of it being constant.

The main assumption is that the noise floor of the DAC's headphone output is independent of the volume setting for it which allows using that it is proportional to the noise of the DAC's line-out. With a digital volume control acting before conversion, this should be close enough. At least with converters using 24 bit or 32 bits internally.

If anything lowering the volume for the DAC's headphone output would decrease its noise due to lower output from the DAC chip and lower levels along the output path as well as quantisation noise being lower. Whether that is significant or even measurable is another question but it should not increase. Thus, calculating the noise floor from the SNR as I did gives an upper bound for it.

Using the residual noise of the headphone amp of course sets the lower bound for its output noise. Johnson noise from a resistor based volume control may well exceed it. Due to its random nature it can just be added as another squared term in the total noise equation.

How exactly does the volume control add noise?

Following Wikipedia's article on potentiometers, if volume control is done by voltage division with one resistor in series and one parallel with the following load, the series resistor increasing lowers volume while the parallel resistor increasing increases volume. Thus, the volume can be reduced while Johnson noise is also reduced by reducing the parallel resistor. For no attenuation, the series resistor has to be zero while for no signal passing through, the parallel resistor has to be zero. The noise from the series resistor is attenuated by voltage division whereas the noise of the parallel resistor is not. Assuming the following load is much larger than one of the resistors, in approximation any attenuation just depends on the relative resistance between the two meaning the can be arbitrarily high or low as long as they are in the right proportion.


Derivations for the last paragraph.

Circuit and voltage division formula from Wikipedia:

V_L is voltage across following load, V_S is source voltage, R_1 is the series resistor's resistance, R_2 is the parallel resistor's resistance and R_L is the resistance of the following load. Dividing both sides by V_S and dividing both numerator and denominator by R_L gives

V_L/V_S = R_2/(R_1 + R_2 + R_1R_2/R_L).

Thus if at least one of either R_1 or R_2 is significantly smaller than R_L and none is larger, R_1R_2 << R_L and R_1R_2/R_L ≈ 0 and thus

V_L/V_S ≈ R_2/(R_1 + R_2) = 1/(1 + R_1/R_2).

For a given factor of attenuation A = V_L/V_S, thus

1 + R_1/R_2 ≈ 1/A

and thus

R_1/R_2 ≈ 1/A - 1 = (1 - A)/A.

Johnson noise is equivalent to a noiseless resistor in series with a perfect voltage source producing the noise, V_N. Thus, the voltage sources at either end of the parallel part add up to the source voltage V_S and the noise voltage from the series resistor V_N1. Since the perfect voltage source has zero impedance, the resistance of the parallel part is as before, i.e. R_2R_L/(R_2 + R_L). Thus,

V_L = R_2/(R_1 + R_2 + R_1R_2/R_L)*(V_S + V_N1).

However, since V_L = V_2 + V_N2 since it is parallel with the perfect voltage source producing the noise from the parallel resistor V_N2. Thus, with attenuation set to A, the RMS voltage of the total Johnson noise is

V_N = sqrt(A^2(V_N1)^2 + (V_N2)^2).

If temperature and bandwidth are the same, V_N2 = sqrt(R_2/R_1)*V_N1. Thus,

V_N = sqrt(A^2(V_N1)^2 + R_2/R_1*(V_N1)^2) = sqrt(A^2 + R_2/R_1)*V_N1.