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#1 |
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diyAudio Member
Join Date: May 2006
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Hi,
can somebody help me how to find the heatsink for more than one bjt. ex if I have two bjt that will need to be fixed to a heatsink, what type of heatsing will I need to use please ? are there any calculators, not to find the heat dissipation capability of the heatsink but how much heat dissipate the bjt so I can find a heatsink for them. thanks ron |
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#2 |
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diyAudio Member
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Well, we must of course know what those BJT's do. Are they in an amp? If so, supply voltage? Bias current? Load impedance?
Jan Didden
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#3 |
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diyAudio Member
Join Date: May 2006
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is there any calculators that I insert the values and gives me an output ?
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#4 |
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diyAudio Member
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Yes, of course. You need to find the Vce times Ic at no signal, to find the idle disspation. Do this per transistor on the heatsink, and add it up. That is the MINIMUM you need to be able to dissipate.
Then, you need to decide what the average signal-related dissipation will be. That's a bit of guess work. Generally, music requires a long-term average of anything between 1 Watt and 15 Watt, depending on the speakers. Pick a number, say 10W to be safe, and calculate the Vce and IC resulting from an output power of 10W in your load impedance (8 or 4 ohms or whatever). Calculate the total disspation of the transistors on the heatsink. Add the two numbers together. This is not strictly correct but errs on the plus side for some more safety margin. Does this help? Jan Didden
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#5 | |
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diyAudio Member
Join Date: Nov 2006
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Quote:
In the case of two devices: you halve the junction-case thermal resistance and the insulator thermal resistance and add this "equivalent thermal resistance" to the heatsink thermal resistance (which is the same as before) and, after subtracting from max junction temperature allowed (usually 200 °C for metal packages or 150 °C for plastic packages) the maximum ambient temperature expected (wich is not so easy to define: the "ambient" temperature is not your room temperature but the temperature inside or near the enclosure of your equipment!) and calculate (divide the temperature span by the the thermal resistance obtained before). Piercarlo |
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| Thread | Thread Starter | Forum | Replies | Last Post |
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