Official M2 schematic

They can certainly take 30V but you may need to reduce Idss values to keep dissipation below 200mW. Cascoding kills a few birds with one stone.

On output 45W to 50W per device is a little high, but probably ok provided they don't get too hot.

Yes! My last jfet pairs had around 6mA idss, so no problem....Need to order again from Spencer.

The output mosfets should take 1/3 of maximal power dissipation! But will have to see in real life...
 
http://www.diyaudio.com/forums/pass-labs/183362-f5x-euvl-approach-20.html#post2575672

Quoting from EUVL:
"If you use the 5R degeneration on the 2SJ74, as I recommended, then you can calculate the Idss of the 2SJ74's using the following equation :

Idss (2SJ74) = Id * (1 + Rs.Yfs)

where Id = Idss of the 2SK170 in this case; Rs =5.1R; Yfs = 0.033 S for 2SJ74 at that bias range.

It works out to roughly 1mA higher than the Idss of the 2SK170s."

Do you think the above is applicable to match the K170/J74 for M2? What will be the Rs for this case?

Thanks
 
For this application I consider 1 mA Idss match plenty good enough,
and the calculation on the Source resistance is correct.

8 ma Idss with a 2SK170 gives 6 mA with a 10 ohm Source resistor.

:cool:

Quoting from EUVL:
"If you use the 5R degeneration on the 2SJ74, as I recommended, then you can calculate the Idss of the 2SJ74's using the following equation :

Idss (2SJ74) = Id * (1 + Rs.Yfs)

where Id = Idss of the 2SK170 in this case; Rs =5.1R; Yfs = 0.033 S for 2SJ74 at that bias range.

It works out to roughly 1mA higher than the Idss of the 2SK170s."

Now I am a bit confused.
looking at the chart and assume Idss of K170 is 7.5ma, Yfs is 0.035, R is 10, Id is 4.7ma and Idss (2SJ74) should be roughly 6.35ma.

So, this is close to what Papa's calculation but conflicting with the statement
"It works out to roughly 1mA higher than the Idss of the 2SK170s."


Now I need to ask Spencer to change the matching parameter if it is not too late.

Thanks
 
Quoting from EUVL:
"If you use the 5R degeneration on the 2SJ74, as I recommended, then you can calculate the Idss of the 2SJ74's using the following equation :

Idss (2SJ74) = Id * (1 + Rs.Yfs)

where Id = Idss of the 2SK170 in this case; Rs =5.1R; Yfs = 0.033 S for 2SJ74 at that bias range.

It works out to roughly 1mA higher than the Idss of the 2SK170s."

Now I am a bit confused.
looking at the chart and assume Idss of K170 is 7.5ma, Yfs is 0.035, R is 10, Id is 4.7ma and Idss (2SJ74) should be roughly 6.35ma.

So, this is close to what Papa's calculation but conflicting with the statement
"It works out to roughly 1mA higher than the Idss of the 2SK170s."


Now I need to ask Spencer to change the matching parameter if it is not too late.

Thanks

7.5 x (1+(10×0.033))
=7.5×1.33 = 10mA for 2sj74 according to the formula provided.

However it's not clear whether Patrick's formula is accurate for all values of source resistance.

I don't think it's worth this amount of headache. I'd just get matched Idss jfets if you are only buying 4.
 
Skylab.
Just so we are all understanding one another.
What Nelson is saying is if you have 2sk170 with an Idss of 10mA then adding 10 Ohms to source will cause Id to drop to 8 mA.
What Patrick is saying is if 2sk170 has an Idss value of a certain value eg 7.5mA, to get perfect matching with 2sj74 once source resistance is added to both devices the actual Idss of 2sj74 needs to be higher.
Nelson has made a general statement that adding source resistance reduces Id through jfet. Patrick is describing
that there is differential reduction in current between the 2 devices with equal amounts of source resistance. Assuming I have understood what he has written.
 
Skylab.
Just so we are all understanding one another.
What Nelson is saying is if you have 2sk170 with an Idss of 10mA then adding 10 Ohms to source will cause Id to drop to 8 mA.
What Patrick is saying is if 2sk170 has an Idss value of a certain value eg 7.5mA, to get perfect matching with 2sj74 once source resistance is added to both devices the actual Idss of 2sj74 needs to be higher.
Nelson has made a general statement that adding source resistance reduces Id through jfet. Patrick is describing
that there is differential reduction in current between the 2 devices with equal amounts of source resistance. Assuming I have understood what he has written.

Hi,
I read it a few times between Nelson and Patrick's statement, just when I put both together then got confused.

Thanks
 
Official Court Jester
Joined 2003
Paid Member
if you keep 1mA difference for Idss , 5R in source difference is OK

in that case, according to EUVL , you have them matched in both Idss (effective) and xconductance

if you choose a couple with greater Idss difference , you can squeeze them together with different source resistors , but xconductance will stay different

if you 're building amps as popcorns , invest in big stash of preputium parts , to have from what to choose

if you're building one , as precious possession , finding two adequate pairs of JFets isn't so hard and well worth greenies paid for