Hello people!
I'm trying to work out what value of volume pot (well OK, stepped attenuator) to use for my power amp. It will be being driven predominantly by a Cambridge Audio D500SE but I cant seem to find any info on its output impedance or any output cap. The first stage of the amp will be LTP phase splitter of 6H8C (Russian 6SN7 for those unfamiliar) at fairly low idle current (5mA per side)
I know that too high a value will result in treble loss at low volume settings due to miller capacitance, but too low a value could cause bass loss due to source output capacitor if present, and generally be a more difficult load for the source.
Where is the balance?
I'm trying to work out what value of volume pot (well OK, stepped attenuator) to use for my power amp. It will be being driven predominantly by a Cambridge Audio D500SE but I cant seem to find any info on its output impedance or any output cap. The first stage of the amp will be LTP phase splitter of 6H8C (Russian 6SN7 for those unfamiliar) at fairly low idle current (5mA per side)
I know that too high a value will result in treble loss at low volume settings due to miller capacitance, but too low a value could cause bass loss due to source output capacitor if present, and generally be a more difficult load for the source.
Where is the balance?
Hi,
can I assume you have a tube amp after this attenuator?
What recommendations does the manufacturer give?
The Cambridge can be measured.
All you need is a source with a constant fixed frequency 200Hz to 500Hz signal, some resistors for a dummy load and a 2000mVac multimeter.
If you have these come back and we'll explain.
can I assume you have a tube amp after this attenuator?
What recommendations does the manufacturer give?
The Cambridge can be measured.
All you need is a source with a constant fixed frequency 200Hz to 500Hz signal, some resistors for a dummy load and a 2000mVac multimeter.
If you have these come back and we'll explain.
Thanks dshortt9 
That is indeed a useful little sheet, but havent they got the levels mixed up? When the -2dB level is selected (near max volume, atennuated by 2dB) the signal would only be going through a very small value of resistor and so the HF cut off would be very high. Conversely the -60db setting would route the signal through a high resistance (in the case of a CT2 through many resistors) and subsequently have a lower cut off frequency.
Or am I just very confused?
AndrewT,
The manufacturer of the amp? That would be me! the attenuator will be part of the amp, the input stage following the attenuator is as described in the first post.
I can use my computer (with sigjenny) as a signal generator if it doesn't need to be too clean, have everything else.
That is indeed a useful little sheet, but havent they got the levels mixed up? When the -2dB level is selected (near max volume, atennuated by 2dB) the signal would only be going through a very small value of resistor and so the HF cut off would be very high. Conversely the -60db setting would route the signal through a high resistance (in the case of a CT2 through many resistors) and subsequently have a lower cut off frequency.
Or am I just very confused?
AndrewT,
The manufacturer of the amp? That would be me! the attenuator will be part of the amp, the input stage following the attenuator is as described in the first post.
I can use my computer (with sigjenny) as a signal generator if it doesn't need to be too clean, have everything else.
Hi,
Output offset (=DC on output):-
measure the mVdc at the output of the Cambridge.
Do it at switch on, then cold (a few seconds after switch on) and again when warmed up.
If all are zero mVdc then there is effective DC blocking inside.
Output impedance:-
Apply the fixed frequency signal to the Cambridge input (is this a pre or a CDP?).
Measure the AC output voltage into an open circuit load.
Make up a 100k dummy load and measure the output again.
Make up a 10k dummy load and measure again.
Make up a 2k dummy load and measure again.
Give us the readings to play with.
The confusion you're having with volume (attenuation) and bandwidth is down to not understanding the two input filters that operate at the amplifier's input. You should have a low pass filiter to attenuate the Radio Frequency (RF) rubbish that comes in with the signal. You should have a high pass filter to block DC from the source but also to bandlimit the input signal to narrower than the bandwidth of the amplifier. I am wondering how you can design the amp without providing the correct values for these two filters. And an understanding of the operation of the filter is required to enable selection of correct component values. There's something back to front here!
Output offset (=DC on output):-
measure the mVdc at the output of the Cambridge.
Do it at switch on, then cold (a few seconds after switch on) and again when warmed up.
If all are zero mVdc then there is effective DC blocking inside.
Output impedance:-
Apply the fixed frequency signal to the Cambridge input (is this a pre or a CDP?).
Measure the AC output voltage into an open circuit load.
Make up a 100k dummy load and measure the output again.
Make up a 10k dummy load and measure again.
Make up a 2k dummy load and measure again.
Give us the readings to play with.
The confusion you're having with volume (attenuation) and bandwidth is down to not understanding the two input filters that operate at the amplifier's input. You should have a low pass filiter to attenuate the Radio Frequency (RF) rubbish that comes in with the signal. You should have a high pass filter to block DC from the source but also to bandlimit the input signal to narrower than the bandwidth of the amplifier. I am wondering how you can design the amp without providing the correct values for these two filters. And an understanding of the operation of the filter is required to enable selection of correct component values. There's something back to front here!
Yes, like I used to be until someone here put me right. The important thing for HF rolloff due to input capacitance (Miller in this case) is how much resistance is across the grid of the IP tube.
Assuming the source of the signal to have an OP impedance of zero (the best possible case), then the maximum resistance in parallel with the grid will be when the volume control is at 1/2 setting, or 6dB attenuation. At that position, half of the volume control acts in parallel with the other half, giving a total of 1/4 of the volume control across the grid.
Of course, real-world signal sources don't have zero OP impedance, so it's best to err on the side of caution and use a volume control of lower value than you theoretically need. If you're likely to use sources with high OP impedance which can't tolerate a low value of volume control across them, you could play safe by preceding the volume control with a cathode follower.
EC8010 said it, but I'll flesh it out a little. It's true that the filter being discussed is formed by the attenuator and whatever follows it, so it might seem natural when choosing an attenuator to focus on what ever follows it. But, it turns out that the best choice is always the smallest value you can get away with and the limiting factor there is whatever comes *before* the attenuator, so that's what you really want to look at.
Determine the smallest value attenuator that your source can drive and go with that.
Determine the smallest value attenuator that your source can drive and go with that.
Thanks for the replies everyone
AndrewT- I'm sorry if my position seems "back to front". About 80% of my electronics knowlege and understanding has been gained while designing this amp (for about a year now) the rest from my electronics GCSE. I have come to this hobby through a love of music and appreciation of purity, rather than an outlet for my electronics wizardry. So yes, I am learning (isn't everyone?) hence the question.
I understand the principle and calculations of low and high-pass filters and my design has developed with these in mind between each stage, but the series resistor on the grid of the first stage will be fairly small compared to the maximum value of the attenuator.
The D500SE is a CD player.
Dave, Ray and EC8010- I have read this about the volume control before (Morgan Jones) but still don't understand. When the pot/attenuator is at mid point the two halfs are in parallel and so the grid sees 1/4 impedance assuming the source (cd player) has fairly low impedance output. Fine.
But surely this only makes sense if you are feeding the signal through the output (from the grid). The current to charge C(miller) is only flowing through the top part of the attenuator. The ground will not source any current and so, as a low pass filter an equivalent circuit will have the top resistance in series with the signal, the bottom resistanace and C(miller) connected to ground in parallel. Whereas you are saying the two parts of the attenuator would be in parallel in the signal path and only C(miller) shunting to ground.
I don't understand I'm not saying all you guys and MJ are wrong but I just dont see how that model makes sense.
Anyone have any idea what kind of impedance CD players like to see as their load? I guess that should have been my first question really.
AndrewT- I'm sorry if my position seems "back to front". About 80% of my electronics knowlege and understanding has been gained while designing this amp (for about a year now) the rest from my electronics GCSE. I have come to this hobby through a love of music and appreciation of purity, rather than an outlet for my electronics wizardry. So yes, I am learning (isn't everyone?) hence the question.
I understand the principle and calculations of low and high-pass filters and my design has developed with these in mind between each stage, but the series resistor on the grid of the first stage will be fairly small compared to the maximum value of the attenuator.
The D500SE is a CD player.
Dave, Ray and EC8010- I have read this about the volume control before (Morgan Jones) but still don't understand. When the pot/attenuator is at mid point the two halfs are in parallel and so the grid sees 1/4 impedance assuming the source (cd player) has fairly low impedance output. Fine.
But surely this only makes sense if you are feeding the signal through the output (from the grid). The current to charge C(miller) is only flowing through the top part of the attenuator. The ground will not source any current and so, as a low pass filter an equivalent circuit will have the top resistance in series with the signal, the bottom resistanace and C(miller) connected to ground in parallel. Whereas you are saying the two parts of the attenuator would be in parallel in the signal path and only C(miller) shunting to ground.
I don't understand
I'm not saying all you guys and MJ are wrong but I just dont see how that model makes sense.Anyone have any idea what kind of impedance CD players like to see as their load? I guess that should have been my first question really.
That's right, we're saying that both halves of the potentiometer go to ground. One half goes directly, and the other half via the (assumed zero) output resistance of the CD player. It might seem as though that lower resistance doesn't help in charging the Miller capacitance, but it does. Have a re-read of Thevenin and you'll see it.
Most CD players use an op-amp as their output stage. If you read the op-amp data sheets, you'll find that there are graphs of distortion vs output load. Mostly, they don't like to drive less than 2k, so 5k or 10k are fine.
Most CD players use an op-amp as their output stage. If you read the op-amp data sheets, you'll find that there are graphs of distortion vs output load. Mostly, they don't like to drive less than 2k, so 5k or 10k are fine.
.
Sudden flash of understanding!
It's because C(miller) isn't a straightforward capacitance isn't it?!
The source provides current to charge Cag but the votage rise at the anode is multiplied by the gain. So for a gain of 20, it's as if a signal 20 times bigger is being applied from the other side of Cag (Hence C(miller) being Cag X gain). The resulting signal coming out of the grid will source and sink current through the lower leg of the attenuator.
Many thanks EC8010! You have brought me understanding and an answer to my original question.
Sudden flash of understanding!It's because C(miller) isn't a straightforward capacitance isn't it?!
The source provides current to charge Cag but the votage rise at the anode is multiplied by the gain. So for a gain of 20, it's as if a signal 20 times bigger is being applied from the other side of Cag (Hence C(miller) being Cag X gain). The resulting signal coming out of the grid will source and sink current through the lower leg of the attenuator.
Many thanks EC8010! You have brought me understanding and an answer to my original question.
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