I am working in one application where one amplifier class AB will be placed in a specific housing outputs will be 1943 5200 power supply will be 35+35 volts and the duty will be casual low listening probably something like 10-15W at the very maximum
Bias for the amplifier will be no more than 70 ma ( rough calculation i can even go lower )
Problem is that the heatsink will be one aluminum size 35cm x45 cm 2mm thick
do you think that i might get away with it ?
Bias for the amplifier will be no more than 70 ma ( rough calculation i can even go lower )
Problem is that the heatsink will be one aluminum size 35cm x45 cm 2mm thick
do you think that i might get away with it ?
As a rule of thumb that's not going to be enough.
Your application wont need a big heatsink but it will still get quite warm.
You are trying to dissipate 70V x 0.07A under no load = 5W.
For reliability try to keep the output transistors below 65 Degrees C - they can run far hotter.
Lets take a lovely summer day where the air temperature is 35 Degrees (I wish).
The heatsink calculation is Tj = Ta + ( Tjc + Tcs + Tsa)P
P is the power being dissipated - 5W in this case.
Tjc can be gained from the transistor data sheet but is nominally in the order of 1.5
Tcs is down to the mica washer and is generally about 0.3
Tsa is the figure that we are after - Transfer of Heatsink to Air
Tj is the maximum temperature that we want the transistor to reach (as stated 65 Degrees)
Ta is the maximum air temperature (as stated 35 Degrees)
Transposing the formula that gives us ((Tj-Ta)/P)-Tjc-Tjs = Tsa
Or ( ((65-35)/5)-1.5-0.3 = 4.2 Degrees C/Watt
Now if you do a bit of homework you can see what sort of heatsink you would need to meet this requirement.
This is with NO SIGNAL applied.
You will need to increase the figures when the amp is at FULL SWING.
Your application wont need a big heatsink but it will still get quite warm.
You are trying to dissipate 70V x 0.07A under no load = 5W.
For reliability try to keep the output transistors below 65 Degrees C - they can run far hotter.
Lets take a lovely summer day where the air temperature is 35 Degrees (I wish).
The heatsink calculation is Tj = Ta + ( Tjc + Tcs + Tsa)P
P is the power being dissipated - 5W in this case.
Tjc can be gained from the transistor data sheet but is nominally in the order of 1.5
Tcs is down to the mica washer and is generally about 0.3
Tsa is the figure that we are after - Transfer of Heatsink to Air
Tj is the maximum temperature that we want the transistor to reach (as stated 65 Degrees)
Ta is the maximum air temperature (as stated 35 Degrees)
Transposing the formula that gives us ((Tj-Ta)/P)-Tjc-Tjs = Tsa
Or ( ((65-35)/5)-1.5-0.3 = 4.2 Degrees C/Watt
Now if you do a bit of homework you can see what sort of heatsink you would need to meet this requirement.
This is with NO SIGNAL applied.
You will need to increase the figures when the amp is at FULL SWING.
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15W from a 50W amplifier will just about hit the worst case heat dump from the transistors to the heatsink.
If you had said "normal music" levels of 1/2W to 2W, then I would point you to using the Pq as the heatsink load. From there you could estimate the Tc of the devices.
But at 10W to 15W for domestic listening the heat load is going to be far greater than Pq.
If you had said "normal music" levels of 1/2W to 2W, then I would point you to using the Pq as the heatsink load. From there you could estimate the Tc of the devices.
But at 10W to 15W for domestic listening the heat load is going to be far greater than Pq.
ok got it will make some tests and see how far i can go
Due to the nature of the application i have to go like that ...one think i can change though is the thickness of the metal from 2mm to 3mm any comments on that ?
also the shape of the aluminum is not flat ...its something like a W ...with the transistors located in the inner sides of the W ( i am trying to say that obviously a V or W shaped piece of metal will be more effective than a flat one
Will also post pictures as we speak
Due to the nature of the application i have to go like that ...one think i can change though is the thickness of the metal from 2mm to 3mm any comments on that ?
also the shape of the aluminum is not flat ...its something like a W ...with the transistors located in the inner sides of the W ( i am trying to say that obviously a V or W shaped piece of metal will be more effective than a flat one
Will also post pictures as we speak
The thickness of the plate will make little difference.
It is how much alluminium is available to dissipate heat to the air around it.
Hence the design of heatsinks.
If you are limited in space available then you might have to consider forced cooling. (Fans - Water - Peltier )
It is how much alluminium is available to dissipate heat to the air around it.
Hence the design of heatsinks.
If you are limited in space available then you might have to consider forced cooling. (Fans - Water - Peltier )
Here are a few online Heatsink Calculators that I think you may find useful,
Heat Sink Calculator
- heatsink design tools
Heatsink Surface Area Calculator.
enjoy !!!
jer 🙂
Heat Sink Calculator
- heatsink design tools
Heatsink Surface Area Calculator.
enjoy !!!
jer 🙂
One option is to forced fan cool the heatsink. A 40x40mm 12V case fan may fit into your housing.
Another possibility is to use the housing as heatsink, provided it's possible.
hey Michael !!!!
always very good to see you ...i will e mail you pictures ( if you have the time to see )
kind regards
sakis
always very good to see you ...i will e mail you pictures ( if you have the time to see )
kind regards
sakis
A lot depends on the airflow through the enclosure.
I always put two fans on my internal amps.
One on each opposite corner, one to suck in air and the other to blow it out.
PC fans work fine and are cheap. I put the fans in series with a large wattage resistor from B+.
I always put two fans on my internal amps.
One on each opposite corner, one to suck in air and the other to blow it out.
PC fans work fine and are cheap. I put the fans in series with a large wattage resistor from B+.
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