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Old 8th November 2001, 05:57 PM   #11
jam is offline jam  United States
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Andy,

My mistake Andy you are correct. A four transistor cross-coupled buffer would work in the circuit which to me is still a better choice. My favourite is still the complemantary j-fet pair, partly due to simplicity and ease of offset correction.

I find complementary circuits work better than single ended ones when driving loads like filters (equal pull up and pull down), if a single ended buffer is used it has to be idled at higher currents to avoid asymetry.

Jam
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Old 9th November 2001, 01:38 AM   #12
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Cheers everyone this is a good brainstorming session.I am going to have to study my SELF ON AUDIO book for help on the impovments mentioned ie levinson.
Kieran
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Old 13th November 2001, 10:52 PM   #13
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Hello Geoff , Grey , Janne , Jam , Andy ,

Ok I have studied the drawings and can see the complementary feedback pair (ztx384/214)with the other two and the 12k and 68r res as a contant current source.

Just to pause there.....
The CFP I have found notes on but the constant current source in that configuration I can't find anything.Can someone fill in the gap please???

There are three 6dB/Octave filters making 18dB/Octave.

At the input and output there are a 10uF cap and 100k Ohm res in a filter looking arangment. Are they??

At the input the two 330k res are a pot div.
What is the 4.7k Ohm/470pF filter for at the input (after the pot div)?????

The neg feedback is not clear to me because it looks like 3 filters in sucsesion but with the middle one not to 0v but as feedback instead it has lost me.

Last bit now the 2.2k res(input)and 10k res(output) before the ztx384 of the CFP. Why.And can Geoff or someone send me the spiced version of the schematics or EWB version Please.

Kieran


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Old 13th November 2001, 11:24 PM   #14
lohk is offline lohk  Europe
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Ins and outs are coupled with a cap: That is necessary, because the circuit runs with a single supply, you have to prevent a DC offset. 100k is to connect the cap to ground, not to leave it "open" to the outside world for several reasons.
The two 330k resistors set the bias point for the input cap - appr. halve voltage - and the reference voltage point for the input transistor.
All these do not have anything to with filters.
The actual input filter is the 4k7 / 470pf RC network - this is an essential part of naim design, I suppose.

The two constant current sources in the filter circuit are formed with two NPN transistors (ZTX384) and two resistors (47k and 68R).

Neill McBrides schematic drawing is actually a bit ODD and confusing.
Please check out his preamp drawings, than everything will be clear for you immediatly.


best whishes
Klaus
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Old 13th November 2001, 11:32 PM   #15
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Superb. Thank for the notes on the input output parts and the confirmation on the other parts.
But can anyone fill in the gaps with regards the 2.2k res and 10k res?????
DC offset??
The feedback??
Please Kieran
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Old 14th November 2001, 12:24 AM   #16
lohk is offline lohk  Europe
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The 2k2 res. is part of the input circuit and the 10k is the input resistor for the output CFP. It is a part of the network already.
The filters are, as Geoff already mentioned, a passive 6db and an active 12db filter together. The output rail is feed back to the filter.
But I actually just follow the circuit paths with astonishment. These are very simple and very clever engineered circuits.
Neil McBride draw the schematics following the orginal PCB design, which Naim used for all crossover filters.

Klaus
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Old 18th November 2001, 10:38 PM   #17
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Nearly there, can someone explain the 6db passive then 12db active filter. Ican only see it as 3x6db passive!!!!

Help Kieran
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