Hi,
You have two Zin, the power amplifier Zin=22k and attenuator Zin~=50k.
But it's not that simple.
Firstly the attenuator sees the power amp Zin and this should roughly follow the ten times rule. For a power amp with Zin=22k the source impedance of the attenuator should be less than or equal to (<=) 2k2.
The maximum source impedance of an attenuator is about one quarter of the pot value. This results in a pot value of about 8k8. The nearest value is 10K giving a ratio of 22k/2k5~=8.8, close enough to ten times for our purpose.
The actual source now sees the pot as it's load. BUT, the load varies.
The pot is in parallel to the power amp Zin and as you change the volume setting the impedance of the parallel combination changes.
When the pot is set to minimum, Zin//0ohms and the series resistor is 10k so the load is 10k.
When the pot is at half volume (-6db) there is a series resistor of 5k and then 5k//22k =4k07 the source sees 5k+4k07=9k07. Not much different from 10k so we are alright. The actual attenuation is -7db but this is an adjustable attenuator and an extra db is of no consequence.
When the pot is set to maximum volume the series resistor is 0ohms and the lower leg is 10k//22k=6k88 and the load on the source is 0+6k88=6k88. Again not too much lower than the 10k maximum load. This works because we have stuck to the ten times rule (or close to it).
Your 50k pot and 22k power amp is outside the ten times rule.
Here the max load=50k and min load =15k3 a ratio of 3.2:1 cf previous 1.45:1
The 10k pot with 22k power amp has a minimum load of ~=7k. Using the ten times rule a suitable source impedance for driving this would be <=700r. Many modern sources fall into this category. The few that are 1k or 2k would be less suitable.
Does that clarify things for you?
I see the confusiom.Chivvyp said:
You have two Zin, the power amplifier Zin=22k and attenuator Zin~=50k.
But it's not that simple.
Firstly the attenuator sees the power amp Zin and this should roughly follow the ten times rule. For a power amp with Zin=22k the source impedance of the attenuator should be less than or equal to (<=) 2k2.
The maximum source impedance of an attenuator is about one quarter of the pot value. This results in a pot value of about 8k8. The nearest value is 10K giving a ratio of 22k/2k5~=8.8, close enough to ten times for our purpose.
The actual source now sees the pot as it's load. BUT, the load varies.
The pot is in parallel to the power amp Zin and as you change the volume setting the impedance of the parallel combination changes.
When the pot is set to minimum, Zin//0ohms and the series resistor is 10k so the load is 10k.
When the pot is at half volume (-6db) there is a series resistor of 5k and then 5k//22k =4k07 the source sees 5k+4k07=9k07. Not much different from 10k so we are alright. The actual attenuation is -7db but this is an adjustable attenuator and an extra db is of no consequence.
When the pot is set to maximum volume the series resistor is 0ohms and the lower leg is 10k//22k=6k88 and the load on the source is 0+6k88=6k88. Again not too much lower than the 10k maximum load. This works because we have stuck to the ten times rule (or close to it).
Your 50k pot and 22k power amp is outside the ten times rule.
Here the max load=50k and min load =15k3 a ratio of 3.2:1 cf previous 1.45:1
The 10k pot with 22k power amp has a minimum load of ~=7k. Using the ten times rule a suitable source impedance for driving this would be <=700r. Many modern sources fall into this category. The few that are 1k or 2k would be less suitable.
Does that clarify things for you?
Re: Ultra-Low noise 5V PSU
I was reading LT Application Note 82 (that's one of the places you can find this schematic), and was interested in trying this regulator (and possibly making a negative version of it, too). Pantera (or anyone else, for that matter), did you ever get around to trying this one? I will probably breadboard it, and also try AD587, which is pin compatible.
pantera6 said:
I was reading LT Application Note 82 (that's one of the places you can find this schematic), and was interested in trying this regulator (and possibly making a negative version of it, too). Pantera (or anyone else, for that matter), did you ever get around to trying this one? I will probably breadboard it, and also try AD587, which is pin compatible.
Attachments
If I were to upgrade the power on my DVD player, my goal would be separation more than noise performance. The PCM1738 has 3 analog +5V supplies and 2 analog grounds. Unfortunately, both are unified in the DV47ai, and that's pretty much what's diagrammed in the data sheet! As if such a lay-out can meet the >110db channel separation spec.!
I have some Black Gate 0.47uf NX-HiQ caps. Any point in using them to make at least separate PS bypasses?
I hate data sheets! The mfr gives circuit diagrams that are as simple and cheap as possible to use the part to sell more, but there's no way specs can be met with them. Nine times out of ten, products are then built exactly from the examples. Grrr, we get crap.
I have some Black Gate 0.47uf NX-HiQ caps. Any point in using them to make at least separate PS bypasses?
I hate data sheets! The mfr gives circuit diagrams that are as simple and cheap as possible to use the part to sell more, but there's no way specs can be met with them. Nine times out of ten, products are then built exactly from the examples. Grrr, we get crap.
BillHiFiNutNut said:
See http://home.quicknet.nl/qn/prive/ra.vdsteen/index_en.html for all the schematics you will need for your CD63/67SE. And if you blow it up can I have your 220V 67SE transformer
k.
AndrewT said:
Hi Andrew,
Just trying to reconcile all the different sources of information
I checked the specs for my amp modules, Zin (signal load input), min 14.8, typ 17, max 47.
So I guess I should have a 10K pot rather that the 50K I have currently.
Now, I need 18db attenuation. If I go to http://www.tnt-audio.com/clinica/diyattenuation.xls and put in a load of 10K that gives me values of 8k8 and 1k4. Is this ok or would I be better using something like 1k5, 200 as this would seem to lower the effective resistance to something nearer the original target of 17K/10 (or am I totally misunderstanding the way that this all works?
Regards
Pete
Hi,
you seem to have understood it.
Select available resistor values. From the E24 range 9k1 & 1k3 will give about -18db.
I don't have access to E96 range values.
The effective minimum impedance seen by your source will be [9.1+1.3]//Zin.
I assume your source can drive that.
If you reduce the fixed attenuator values then you improve the power amp end by reducing the source resistance but on the other hand you reduce the load seen by the source.
How low dare you go?
Resistors are so cheap, it may be worth making up pairs of attenuated interconnects with effective impedances of 10k, 5k and 2k and listen to each.
When you have settled on an acceptable load then buying some expensive resistor types may be worthwhile. I tend not to recommend that route to Nirvanna, but others try exotica and report quality improvements.
you seem to have understood it.
Select available resistor values. From the E24 range 9k1 & 1k3 will give about -18db.
I don't have access to E96 range values.
The effective minimum impedance seen by your source will be [9.1+1.3]//Zin.
I assume your source can drive that.
If you reduce the fixed attenuator values then you improve the power amp end by reducing the source resistance but on the other hand you reduce the load seen by the source.
How low dare you go?
Resistors are so cheap, it may be worth making up pairs of attenuated interconnects with effective impedances of 10k, 5k and 2k and listen to each.
When you have settled on an acceptable load then buying some expensive resistor types may be worthwhile. I tend not to recommend that route to Nirvanna, but others try exotica and report quality improvements.
Hi there,
what is this, a CD53 or such?
Maybe this is a good spare player for mine.
http://cgi.ebay.fr/ws/eBayISAPI.dll?ViewItem&item=300058907662&fromMakeTrack=true
what is this, a CD53 or such?
Maybe this is a good spare player for mine.
http://cgi.ebay.fr/ws/eBayISAPI.dll?ViewItem&item=300058907662&fromMakeTrack=true
http://www.d-mpro.com/users/FolderData/{8FFA780C-0318-4D65-80E5-5866936B24C9}/PMD321_front-2.jpg
This is really ours, anyone knows what's inside ?
This is really ours, anyone knows what's inside ?
Malefoda said:
It means the output is split into two parts of opposing phase, which are then recombined in the receiving equipment, and any difference is removed. It cancels some noise pickup on the wiring basically. I'm sure my explanation is incorrect, but that's sort of what it does!
Simon
