Anyone here technical enough for this...

[tfarney]

Confirm? Deny? Translate into a modern language?


hmm...let's see if I can explain this in the most basic way.

attenuation at input (Vin) reduces the signal level, but does not reduce the detail of the signal (fast forier transform (FFT)will still reveal all the frequency components of the composite sine waves) such that FFT of the source output == FFT after the attenuator, only reduced in amplitude. But having a resistor divider at the output is not like atten at the input, wherein the output is driving a load resistor or load imp. The effect of a high output Z driving a low load Z is that the signal does not transfer well from the output to the load, such that the load is only sensitive to the higher signal sine wave components, whereas the lower signal sine wave components are left out. As a result your FFT at the out put of the driving stage but before the resistor divider != FFT measured after the resistor divider (assuming amp and load have formed a complete circuit)

He's trying to explain how a headphone circuit in an integrated amp, implemented by stepping down the mains through a nest of resistors, kills detail in the signal...

Tim

[lurcher]

Hmm, well, its true that having a impedance mismatch is a bad thing generally, and its likely that using resistive network to reduce the signal level to drive the output possibly will suffer from the frequency dependent impedance of the load causing a non flat response.

But other than that, I think its half confused science half tea pot.

You could always use a transformer to step down from the amp output to the phones, that gives you the correct signal level, and much lower output impedance.

[tfarney]

Hmm, well, its true that having a impedance mismatch is a bad thing generally, and its likely that using resistive network to reduce the signal level to drive the output possibly will suffer from the frequency dependent impedance of the load causing a non flat response.

But other than that, I think its half confused science half tea pot.

You could always use a transformer to step down from the amp output to the phones, that gives you the correct signal level, and much lower outout impedance.

Thanks, lurcher. The debate is over the quality of headphone jacks in integrateds and receivers, many of which simply use the speaker amps, stepped down through a nest of resistors. Theoretically, it seems, this is not the best way to do it. Mysteriously, some of them sound really good.

The guy I quoted above wasn't talking about a simple mis-match of impedance and the resulting change in the frequency response of phones (particularly low-impedance phones, I understand). He was saying, if I understood him, that resistors would remove detail from the signal. Like they were taking information out. Furthermore, he was saying that this is a problem if the resistors are at the output (headphone jack), but not if they are at the input (stepped attenuator). I'm not particularly technical, but I suspect he was blowing a bit of smoke.

Tim

[lurcher]

but I suspect he was blowing a bit of smoke.

So do I, it would require the resistor to be non linear. There could be implications of using resistors with applicable inductance, but thats just using the wrong resistors.

All IMHO of course

[NRG]

Could it be perhaps that a high o/p Z driving some sort of capacitance in cct like the head phone leads forms a low pass filter...

[lurcher]

Yes, you are right Neil, the leads may become important if the output impedance is high, but thinking about it, it would have to be very high to have much effect. Still possible.

For example, if the output imp was 1k, and the lead capacitance was 500pf, the -3dB point would be about 300kHz

[SPS]

I find that quite interesting
i have always found that in inline passive volume control seems to reduce the detail in the signal, compared to have one that replaces the grid resistor in an amp, but that also may be system dependant..

so is he saying that having two sets of resistors in a line doing the same job? is not a good thing?...

steve

[greenhomeelectronics]

There is some fact in what the gentleman says but in reality it would only be true of high impedance headphones and much, much less of an issue on normal 8-32 ohm jobbies. Personally I would always design an amp with a seperate circuit for headphones, there are lots of single chip solutions out there so using a resistive dropper network is a pretty feeble economy.
Just my 2p worth,
Dave.

[lurcher]

There is some fact in what the gentleman says

Let me just check. Are you suggesting that there is some fact in:


The effect of a high output Z driving a low load Z is that the signal does not transfer well from the output to the load, such that the load is only sensitive to the higher signal sine wave components, whereas the lower signal sine wave components are left out.

BTW, I just had a look at your website, I love the idea of saving old kit and getting extra use of it. Great idea.