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Old 6th April 2012, 11:32 PM   #401
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Default Matching the NJL transistors

... is not an easy job.

I have etched some small PCBs representing this schematics:

Click the image to open in full size.

and measured a NJL0281DG.

As power supply I use a LAPTOP PSU rated at 15V/4A.

The results: the current that is flowing into the collector of the NJL is 160mA, the current flowing into its base is 2.49mA.
The resulting Hfe is 64.2, which is smaller than the documented minimum of 75 in the OnSemi datasheet.

The reason for this difference is that OnSemi does its measurements with a collector current of at least 500mA
and having a different VCE (5V), while my VCE is about 6V.

I cannot increase the collector current very much without risking the thermal death of transistor T1.

Nevertheless I think that this little PCB will help finding "matched" triples of NJL transistors.

Best regards - Rudi_Ratlos
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Old 7th April 2012, 10:54 AM   #402
AndrewT is offline AndrewT  Scotland
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voltage across T1 should not be a problem.

Try to set up your supply voltage such that T1 is dropping about 1.5V.
This will give a dissipation of 6W @ 4A. A decent heatsink will save this transistor.
If you adjust the base stopper of T2 to get your Vce to 5V then you can check both the base current and the Vbe at near your test values of 5Vce and 4A without risking any components.

If you want you can repeat these measurements at 2A and 1A.
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Old 7th April 2012, 11:21 AM   #403
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I will try, Andrew.

I suppose that OnSemi measures the transistors at a lower supply-voltage: 10-12VDC.

I am lucky to have my LAPTOP power supply and will be able to adjust the supply voltage by means of a 7805.
Maybe I will be able to adjust to IC=500mA @ VCE = 5VDC.

Best regards - Rudi_Ratlos
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Old 7th April 2012, 11:28 AM   #404
AndrewT is offline AndrewT  Scotland
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add a 10r in series with R5. This becomes your base current measurement.
The 2k2 can be a switched resistor for different Ic values.
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Old 1st May 2012, 12:09 PM   #405
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Since we have a public holiday in Germany today, I had some time and soldered the first PCB, a shunt-regulator board.
I am missing a heatsink for the negative rail as you can see.
(I will unsolder my prototype tonight - LoL.)

I connected a 1.8K resistor to the output giving a load of approx. 21 mA.
The transformer used has 2 x 35 VAC secondaries; this time I am using a 36V Zener diode.
The shunt works flawlessly.

Did I ever thank Mihai for developping this wonderful AMP?
If not - thank you, Mihai

Best regards - Rudi_Ratlos
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File Type: jpg Shunt 003.jpg (186.1 KB, 450 views)
File Type: jpg Shunt 004.jpg (103.2 KB, 445 views)
File Type: jpg Shunt 007.jpg (210.0 KB, 137 views)
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Old 1st May 2012, 09:08 PM   #406
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I have been asked if I am sure that the solder pins of the heatsinks that I am using (see attached image) are "electrically coupled"?
I am not sure, but when I measure the resistance between the 2 solder pins, my multimeter returns a value of 0 Ohm.

The datasheet does not reveal, if a connection between a solder pin and the body of the heatsink exists.

Does anybody know for sure?

Best regards - Rudi_Ratlos
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Old 1st May 2012, 09:49 PM   #407
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It all depends of the pins scratchs off some of the black surface but you need to know alu is a bad conductor for electricity
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Old 1st May 2012, 10:01 PM   #408
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Rudi 90% of all these heatsinks are made that way. Evette
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Old 1st May 2012, 10:12 PM   #409
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I believe when these pins are inserted in the heatsink, even if there is some sort of coating (black color), it will be scratched away, these pins are thicker than the actual hole in the heatsink to make the assortment tight. AL is not that real bad conductor any way.
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Old 1st May 2012, 10:23 PM   #410
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Drop a screwdriver on you output transistors and you,ll find out how great a conductor it is. Evette
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