I am trying to put together a Thermal Trak NJLxxxx output stage amplifier but I am also using Caddock MP915 emitter resistors.
Should the resistors get their own little heatsinks or is it better to mount them on the main heat sink with the ON Semi Thermal Trak output transistors?
It is a tiny amp, low rail voltage, with only two NJL's for each channel(one pair).
Cheers!
Should the resistors get their own little heatsinks or is it better to mount them on the main heat sink with the ON Semi Thermal Trak output transistors?
It is a tiny amp, low rail voltage, with only two NJL's for each channel(one pair).
Cheers!
This is straightly based on the heat dissipation.
Typically idle dissipation is very low, like 10 mWt (I^2*R, ~0,2A*0,2A*0,2Ohm) and you doesn’t need heatsink at all.
But then check your RMS current output drive. Worst case with 1 pair of output devices, i suggest, could be like 5-7 Amps.
This will result in a 5-10 Wt dissipation and you really need to put them to big common heatsink with good coupling.
So, check your load and signal drive.
So mounting these resistors on the main heatsink is not going to benefit the Thermal track devices?
It would be a 6 ohm pair of speakers coming off a single pair of NJL3281/1302 at 40 V DC max rails. I am not worried about the resistors being OK, I am wondering more about Thermal Trak function. Know what I mean? Is there any benefit to thermally couple the emitter R's to the Heatsink right beside the NJL's?
I am sorry if that sounds nebulous but its the best I can come up with right now.
;-)
I am sorry if that sounds nebulous but its the best I can come up with right now.
;-)
The Caddock resistors have a thermal coefficient that is very small, ranging from -20 to + 80 ppm. As it could be zero, negative or positive coefficient, how can you be sure your'e going to achieve anything useful by tracking (thermally coupling) this minute and uncertain variation?
I wouldn't forget that the internal emitter resistance plays a part in this too and it gets really hot. So, I think an isolated external resistance is easier and prefererable unless the thermal coefficients can be shown to cancel or compensate for each other significantly. It could just as easily perform better - fractionally. If you need cooling, small PCB mounting sinks should suffice.
I wouldn't forget that the internal emitter resistance plays a part in this too and it gets really hot. So, I think an isolated external resistance is easier and prefererable unless the thermal coefficients can be shown to cancel or compensate for each other significantly. It could just as easily perform better - fractionally. If you need cooling, small PCB mounting sinks should suffice.
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I am OK at board layout but its not really my cup of tea. I'd rather pay somebody to do a layout for me. There will be an upfront payment and a final payment. Payment by Email transfer only. It is the layout for this Thermal Trak amp with the Caddocks.
PM me if interested. Preference goes to DIYAUDIO members with online activity in the forums.
PM me if interested. Preference goes to DIYAUDIO members with online activity in the forums.
With +/-40V supply the clipping voltage would be about +/-35V if the power supply is very stiff and doesn't sag at all. This means for a 6 Ohm resisitive load the peak current is 35V/6Ohm=5.83A. But all power supplies sag a few volts at high load so a safe estimate would be 5 Amps. Assuming a class B amp, each emitter resistor sees only half of the current waveform, and the average dissipation is half the peak dissipation, so the average dissipation on one emitter resistor is one quarter of the peak:
I^2*R/4 = 5^2*'x'/4 = 6.25*'x'.
Substitute your emitter resistor value in Ohms in place of 'x', and you'll have the approximate power dissipation on one emitter resistor at full load with +/-40V supply and 6 Ohms load.
For example, if you have 0.22 Ohm emitter resistors, it will be only 1.375 watts. This means that even a 2.5 watt resistor will fill the bill without heatsink.
What must be considered though that someone might use it with a 4 Ohms load (in theory increasing dissipation 2.25 times), or it may be driven to clipping, which increases the average/peak power ratio, or it may be loaded with inductive loads, that increase the instantenous peak currents above resistive loads.
But if you have for example 0.1 Ohm/ 2.5 Watts emitter resistors that need not be heatsinked for a 100W amp.
Besides, if you drive your amp to 100W in your living room that's unbearably loud with the usual speaker sensitivities.
I^2*R/4 = 5^2*'x'/4 = 6.25*'x'.
Substitute your emitter resistor value in Ohms in place of 'x', and you'll have the approximate power dissipation on one emitter resistor at full load with +/-40V supply and 6 Ohms load.
For example, if you have 0.22 Ohm emitter resistors, it will be only 1.375 watts. This means that even a 2.5 watt resistor will fill the bill without heatsink.
What must be considered though that someone might use it with a 4 Ohms load (in theory increasing dissipation 2.25 times), or it may be driven to clipping, which increases the average/peak power ratio, or it may be loaded with inductive loads, that increase the instantenous peak currents above resistive loads.
But if you have for example 0.1 Ohm/ 2.5 Watts emitter resistors that need not be heatsinked for a 100W amp.
Besides, if you drive your amp to 100W in your living room that's unbearably loud with the usual speaker sensitivities.
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