Long skinny builders thread

I will be mounting my devices directly onto the Conrad heatsink, instead of using the L alu bars. I believe this will dissipate the heat away better.

Hi Ivan

What is the size of your Conrad heat sink? Even I am planning to make the second channel with direct mounting on the heat sink. I think there is heat difference between the L type profile and the heat sink for sure during my test but for exact check I am yet to procure a IR heat gun where they usually use for body temperature.

Thanks
 
Second channel is up and running with a different sized heat sink I had couple of them spare - 300/150/83mm. This time as suggested direct mounting of the mosfets and resistors on to the heat sink. Perfect similar bias of 284mV upon 20-30mins of initial power up and settling down with 0.0-0.4mV offset. These heat sinks are not mounted to the cabinet as I need to see the fitment even though the cabinet fits them but need some milling. Now the interesting part of the heat dissipation, seems like these are slightly better compared to the original heat sink but still runs hot after an hour or so. But the entire heat sink is evenly hot unlike the first one. The first one is still running with some fans and undecided to remove the mosfets and resistors remount it to this bigger size heat sinks considering the value of R100 :(




 
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Another J2 build.

I have the boards and the last of parts should all be here this week. I have to say, the parts are a bit of a challenge in todays world. A couple of question I could use some help with.

1. The 2SK170s show Gr grade. Myself and my trusted supplier only have Bl grade. Do I just let it go or make some kind of adjustment. I can keep the Idss down in the 7.5 mA range. Gr ends at 6.5. Not even sure why they are signified as Gr when the 2SJ74s are Bl grade. But, I am a builder not a circuit designer.

2. I found some NOS SJEP120R100 but I don't know how to test them. The standard test with a 9V battery doesn't really work on these (or they are bad). Advice and help would be greatly appreciated. The were not matched by the seller.

Thanks for the help. I'll post pics as I go along but this won't be a fast build.

I can hardly wait to hear the sound everybody raves about.

Regards,

Don
 
@manniraj ,
please don't use plastic housing at this place.
Bildschirm­foto 2022-11-02 um 08.50.27.png
 
The J2 design doesn't use the SJ74/SK170s in a balanced circuit. The SK170's are just implementing a constant current source for the SJ74s (which form a long-tailed pair).

As it turns out, you're actually better off with GR parts for the current source (the SK170s). According to my analysis (with all the usual caveats), you want the total to be 6 - 10mA (so, for instance, a 4mA and a 4.5mA would be ideal).

(I've attached the analysis I did.)
 

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It's not push-pull. The top is a current source for the bottom. (It's not a constant current source, as the optocoupler implements a kind of Aleph current source where it sources what the bottom half needs.)

IIRC, in an active/Aleph/whatever current source the ratio of the source resistors has an effect on the ratio of H2 to H3 (or perhaps even-harmonics to odd-harmonics, or something like that). I can't remember the specifics though....
 
It's not push-pull. The top is a current source for the bottom. (It's not a constant current source, as the optocoupler implements a kind of Aleph current source where it sources what the bottom half needs.)

IIRC, in an active/Aleph/whatever current source the ratio of the source resistors has an effect on the ratio of H2 to H3 (or perhaps even-harmonics to odd-harmonics, or something like that). I can't remember the specifics though....
It would be a current source to the bottom if the output was taken from upper transistor's source. So lower current will be held constant and equal to the upper current plus the load current. And an opposite, if we take the output from the lower transistor's drain, the upper current will be held constant and equal to the lower current plus the load current. And intermediate tap position would give you a current ratio in between. You can write an equation and see it yourself.

IDK how to define a push-pull operation. Is it just a current swing symmetry or both parts must be forward-fed? Taking the last definition it's not a PP anyway because the upper transistor forms a feedback loop around current sense resistors and it's not fed by the input signal.