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Schiit Freya S Preamplifier Review

solderdude

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For that you can use something like a passive attenuator.
I recommend something like that because such amplifiers have way too much power for sensitive heaphones (the sensitivity is the culprtit here)

It is quite easy to build such amplifiers. All one needs is stereo potmeters with a reverse log taper or stepped attenuators (relay or switch) and could make the lowest position so that the signal is muted.
In fact my own headphone amplifier design uses an inverting amplifier (for other reasons).

Manufacturers of headphones like to sell more devices so headphones tend to have a high sensitivity and low impedance.
That doesn't mean amplifier manufacturers should not focus on this only and not make amplifiers for higher sensitivity headphones.
There are workarounds to remove the noise issue where it counts.

When you take a look at the Meier Corda Jazz, O.K. maybe not the best example, that uses that topology you can see that noise wise at 50mV the Atom has it beat.
The reason is the topology (non inverting) amp before the volpot, low gain setting and buffer after the volpot.
The only downside is you can clip the input easily on high gain setting but one would not be using this for sensitive headphones.

Also consider that at 80 dB SPL average (which is moderately loud) you can't hear any noise 70dB down let alone 90dB down.

But I do agree that those designing headphone amps and are using stepped or relay volume control could just as easily design it inverting and get slightly better 50mV noise numbers.
 

JohnYang1997

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For that you can use something like a passive attenuator.
I recommend something like that because such amplifiers have way too much power for sensitive heaphones (the sensitivity is the culprtit here)

It is quite easy to build such amplifiers. All one needs is stereo potmeters with a reverse log taper or stepped attenuators (relay or switch) and could make the lowest position so that the signal is muted.
In fact my own headphone amplifier design uses an inverting amplifier (for other reasons).

Manufacturers of headphones like to sell more devices so headphones tend to have a high sensitivity and low impedance.
That doesn't mean amplifier manufacturers should not focus on this only and not make amplifiers for higher sensitivity headphones.
There are workarounds to remove the noise issue where it counts.

When you take a look at the Meier Corda Jazz, O.K. maybe not the best example, that uses that topology you can see that noise wise at 50mV the Atom has it beat.
The reason is the topology (non inverting) amp before the volpot, low gain setting and buffer after the volpot.
The only downside is you can clip the input easily on high gain setting but one would not be using this for sensitive headphones.

Also consider that at 80 dB SPL average (which is moderately loud) you can't hear any noise 70dB down let alone 90dB down.

But I do agree that those designing headphone amps and are using stepped or relay volume control could just as easily design it inverting and get slightly better 50mV noise numbers.
Designing amplifier beating all candidates on the chart here on noise is easy. It doesn't require much to do it. Dx7pro or some others are on the other side of the spectrum, relatively high noise when max volume plus digital volume.
The ap's analyzer's residue noise is also quite high for measuring noise. The residue is comparable with opa1612 with rg of 100ohm and rf of 200ohm. To see most are not even reaching the limit of ap, it's even possible to have an amplifier of 9db gain at output stage and beating all other amps here.
 

solderdude

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What one has to keep in mind is what the practical value is of improving S/N ratio.
Ultimately the limiting factor is the noise on the input of the AP for measurements and the audible noise floor of the ear (becomes less sensitive when we get older).

So driver sensitivity is the key here. When one owns high sensitivity headphones one needs amps with a good S/N ratio at 50mV (which is very loud on such transducers) and limited output power (to protect the ears and driver).
This is where a passive attenuator helps so one can even use 'normal' amps in this case. It is easy to lower noise floors at the output by 20dB.
So one's 80dB S/N ratio amp effectively becomes 100dB.
 

MRC01

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... When you take a look at the Meier Corda Jazz, O.K. maybe not the best example, that uses that topology you can see that noise wise at 50mV the Atom has it beat.
The reason is the topology (non inverting) amp before the volpot, low gain setting and buffer after the volpot.
...
But I do agree that those designing headphone amps and are using stepped or relay volume control could just as easily design it inverting and get slightly better 50mV noise numbers.
What's even more unusual about the Corda Jazz is that has one of the worst SNR at max volume, and one of the best at 50 mV. Its SNR at 50 mV is actually higher than at max volume. It's the only amp measured here whose SNR improves as you turn down the volume.

The main reason for its poor SNR at max was insufficient regulation of its on-board power supply. It suggests that a better implementation of that design could have exemplary performance from full scale to low output.
 

MRC01

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... This is where a passive attenuator helps so one can even use 'normal' amps in this case. It is easy to lower noise floors at the output by 20dB.
So one's 80dB S/N ratio amp effectively becomes 100dB.
This goes back to the original question: you can turn down that volume to 50 mV with a potentiometer. Why would the passive attenuator be quieter? A 10 kOhm resistor has about 1.8 microVolts of noise. That's a 121 dB SNR for a 2 V signal, but it's only 89 dB for a 50 mV signal.
 

JohnYang1997

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This goes back to the original question: you can turn down that volume to 50 mV with a potentiometer. Why would the passive attenuator be quieter? A 10 kOhm resistor has about 1.8 microVolts of noise. That's a 121 dB SNR for a 2 V signal, but it's only 89 dB for a 50 mV signal.
10k pot has maximum of 2.5kohm equivalent resistance.
A pot can turn down incoming signal or previous stages but not the output stage.
 

MRC01

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True, but at low volume settings, a 10k passive attenuator's series resistor is at or near 10k, so that's what the signal passes through.
 

JohnYang1997

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True, but at low volume settings, a 10k passive attenuator's series resistor is at or near 10k, so that's what the signal passes through.
No it's not. It's 10k minus that series resistance and that resistance in parallel. Maximum source resistance is at 50%.
 

MRC01

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Example: at -24 dB, a 10k passive's 2 resistors are 9400 ohms in series and 600 in parallel. The signal that reaches the downstream device passes through the 9400 ohm resistor. Also: at -32 dB (what you need to get 2 Vrms down to 50 mV) the passive resistors are 9750 and 250 ohms.
 

JohnYang1997

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Example: at -24 dB, a 10k passive's 2 resistors are 9400 ohms in series and 600 in parallel. The signal that reaches the downstream device passes through the 9400 ohm resistor.
The noise introduced is 0.4uV.
 

MRC01

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How is that possible? A 9400 Ohm resistor has 1.75 uV of noise at room temperature. The upstream source signal passes through this resistor in order to reach the downstream device.
 

JohnYang1997

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How is that possible? A 9400 Ohm resistor has 1.75 uV of noise at room temperature. The upstream source signal passes through this resistor in order to reach the downstream device.
Like I said, it's the parallel of the two part of resistance. It's 600//9400 in this case. Otherwise it will be most noisy at zero volume, does that make any sense to you?
 

MRC01

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Like I said, it's the parallel of the two part of resistance. It's 600//9400 in this case. Otherwise it will be most noisy at zero volume, does that make any sense to you?
Yes it does make sense, because thermal resistor noise is powered by its temperature energy independent of any signal passing through it. As the signal passes through the resistor, that noise is superimposed on it.
The portion of the signal that passes through the parallel resistor goes to ground and never appears at the downstream device.
I believe that's actually how the attenuator, attenuates.
 

JohnYang1997

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Yes it does make sense, because thermal resistor noise is powered by its temperature energy independent of any signal passing through it.
No it's not. If I ground the input of the following stage, will I still get noise? No. I make amplifiers of noise under 0.3uV and have used apx555 to measure noise(0.8uV of course). 10k pot at 0% doesn't introduce noise. This is basic electronics. Source impedance is all paths seen from the input. Amplifier doesn't know which path is music.
 

MRC01

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Consider that 9400 ohm resistor whose noise is rated at 1.744uV or -115 dB. The first is an absolute voltage, the second is a voltage ratio. It just so happens that this -115 dB ratio is what that voltage would be if 1.0 V were applied across the resistor. Put differently: 20*log(1.0/1.744e-6) = -115.

As you apply different voltages over this resistor, say 2 Vrms versus 50 mV, these figures can't both be the same. One of them must change. Which one changes? If you apply 50 mV across this resistor, does the -115 dB remain the same, in which case the voltage drops to 0.087uV? Or does the 1.744uV remain the same, in which case the ratio changes to -89 dB?
 

JohnYang1997

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Consider that 9400 ohm resistor whose noise is rated at 1.744uV or -115 dB. The first is an absolute voltage, the second is a voltage ratio. It just so happens that this -115 dB ratio is what that voltage would be if 1.0 V were applied across the resistor. Put differently: 20*log(1.0/1.744e-6) = -115.

As you apply different voltages over this resistor, say 2 Vrms versus 50 mV, these figures can't both be the same. One of them must change. Which one changes? If you apply 50 mV across this resistor, does the -115 dB remain the same, in which case the voltage drops to 0.087uV? Or does the 1.744uV remain the same, in which case the ratio changes to -89 dB?
Neither. The SNR will be 50mV over the noise of output stage. The noise of output stage consists of the noise of active components, the feedback resistance, the source resistance. And you are done.
 

MRC01

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Thermal noise is given in as 2 figures: a voltage, and a voltage ratio in dB. The dB is computed assuming a 1.0 reference voltage. But your circuit probably doesn't operate at 1.0 V; it's higher or lower. So when deciding how to compute the noise in your circuit, you'll get different noise figures depending on which you use.
When computing the noise of the feedback and source resistances, which thermal noise figures do you use - the voltage, or the ratio?
 

solderdude

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This goes back to the original question: you can turn down that volume to 50 mV with a potentiometer. Why would the passive attenuator be quieter? A 10 kOhm resistor has about 1.8 microVolts of noise. That's a 121 dB SNR for a 2 V signal, but it's only 89 dB for a 50 mV signal.

The passive attenuator I am talking about is located between the output of the amp and the headphone in question.
It attenuates the signal + noise by 20 dB. Also the resistors are just a few Ohm.
 

MRC01

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Thermal noise is given in as 2 figures: a voltage, and a voltage ratio in dB. The dB is computed assuming a 1.0 reference voltage. But your circuit probably doesn't operate at 1.0 V; it's higher or lower. So when deciding how to compute the noise in your circuit, you'll get different noise figures depending on which you use.
When computing the noise of the feedback and source resistances, which thermal noise figures do you use - the voltage, or the ratio?
I was hoping for answers from those here more knowledgeable than I am. But I also studied this on my own to see if I could find the answer. What I read is that thermal noise doesn't depend on the voltage across or current through the resistor -- other than the indirect effect that higher voltage draws more current which heats it up, which increases thermal noise. This implies that thermal noise is inherently voltage/power. The dB ratio for thermal noise is given relative to an arbitrary 1 V reference. But that is arbitrary. Since it is the voltage & power of thermal noise that is constant depending on temperature but not on voltage, the actual dB ratio of noise depends on the voltage across the resistor.

For example, that 9400 Ohm resistor mentioned above has 1.74 uV of voltage noise at room temperature. If you pass a 1 V signal across it, that noise is 115 dB below the signal. But if you pass a 50 mV signal across it, that noise voltage doesn't change so it's only 89 dB below the signal.

The practical audio take-away is that there's no free lunch when attenuating signals to low levels -- thermal noise becomes a bigger factor for small signals.
 
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