Burning Amp BA-3

I'm close to fire-up. I could actually get it going and do the biasing here at work (I don't have NEAR enough DMM's to get the job done at home, plus they have the neat little clippy probes), since all the boards are done. I'm waiting for my Alps pot to come in to finish the front panel.

And as I said before some final cosmetic stuff as well. Should look totally killer. I received my straw hat LEDs last night for my back-lit, chrome polished aluminum DIY knobs. Just have to figure out how to glue veneer to aluminum, and I think I have the epoxy that will do the trick.
 
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I'm close to fire-up. I could actually get it going and do the biasing here at work (I don't have NEAR enough DMM's to get the job done at home, plus they have the neat little clippy probes), since all the boards are done. I'm waiting for my Alps pot to come in to finish the front panel.

And as I said before some final cosmetic stuff as well. Should look totally killer. I received my straw hat LEDs last night for my back-lit, chrome polished aluminum DIY knobs. Just have to figure out how to glue veneer to aluminum, and I think I have the epoxy that will do the trick.

I would think epoxy would do it fine. :cheers:

Russellc
 
Hmmm, I did mine exactly like 6L6 did his, 3 on one and three on the other. Of course we both used the predrilled deluxe 5U Chassis, which pretty much forces you to. I dont understand why you put 2 devices on one and 4 on the other...big blank spot on one end. I dont know if this is a risk, but wouldnt the 4 together run hotter than the 2 together, causing the group of 4 to hog more power and possibly "runaway"? I dont know if it would, I really dont know, just wondering.
..............
Precisely.

But I am not going to argue when he has no intention of listening.
 
Try again.

Thermal conductivity in W/m degree K:
Aluminum: 205 (depending on grade and temper)
Aluminum Oxide: 30

Radiation heat transfer is great when the surface is black. If the outside of the heatsink was anodized black, and the inside was masked raw and chemical coated (Iridite) to prevent oxidation, that would be the most ideal setup for cooling.

How thick do you think the anodizing is?
Would it be easy to remove?
Conrad heatsinks polish the mounting surface for the very reason you state.
 
How thick do you think the anodizing is?
Would it be easy to remove?
Conrad heatsinks polish the mounting surface for the very reason you state.

It's usually spec'd at 0.005" or less build-up. Two ways to remove it: grind or milling machine. Anodize will chew up end mills too. Best way is to plunge a couple thousands to break through, and then traverse. People have been building with the 4 and 5U enclosures without any problems, so I don't think you should bother with the anodize. Best case, you'd be able to bias a little bit higher if the heat is tamed a bit more.
 
Hard anodise is more likely to be around 1thou (1mil, or 0.0000394m), ordinary/soft anodize is likely to be <1thou.
The dye in (in rather than on) the anodise may increase the Thermal Resistance, but still bloody brilliant.

Alox thermal spacers are usually 40thou (0.001m) to 160thou thick (0.004m) and some prefer these to using 1 or 2thou mica for best thermal conductivity.
 
0.005'' is 0.000127 meter.
30 W/m·K divided by 0.000127 makes 236,220 W/K for 1 square meter.

For a TO247 (21 x 26 mm), that makes ~129 W/K
aka an additional thermal resistance below 0.01 C/W

W/m^2 degK is area, not thickness. Can you imagine dissipating 236,000W in one square meter? Me neither.

heat transfer (W), q, is (1/((1/h)+(t/k)))*A*deltaT
h=heat xfer coefficient (generally estimated)
t=thickness
k=thermal conductivity (in W/m^2 degK)
A=area

anodize conductivity: 30W/m^2 degK
free air convection on vertical plane, delta T of 30 degC (303 degK): coefficient is 5 W/m^2 degK
Area 1m^2 (imagine a flat sheet, 1m^2 in area)
(1/((1/5)+(.000127/30)))*1*303 = 1515 Watts
 
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W/m^2 degK is area, not thickness.

W/m·K for thermal conductivity is derived with dimension analysis from Fourier's law. (for 1-dimensional heat flow)

Q = - Lambda · A · dT/dx

Q is heat flow (W)
A is area (m2)
dT = temperature difference (K, negative for positive flow)
dx is thickness

W = Lambda · m2 · K/m
=> Thermal conductivity = (W · m) /( m2 · K) aka W/m·K

(Thermo 101, 1st year Delft university, gawd knows how long ago. Which can only imply I'm so insane, I should have become a physics teacher for stupid private school girls with perky breasts)
 
W/m·K for thermal conductivity is derived with dimension analysis from Fourier's law. (for 1-dimensional heat flow)

Q = - Lambda · A · dT/dx

Q is heat flow (W)
A is area (m2)
dT = temperature difference (K, negative for positive flow)
dx is thickness

W = Lambda · m2 · K/m
=> Thermal conductivity = (W · m) /( m2 · K) aka W/m·K

(Thermo 101, 1st year Delft university, gawd knows how long ago. Which can only imply I'm so insane, I should have become a physics teacher for stupid private school girls with perky breasts)

I think I just spit out my coffee! Ah yes, my mistake. I was mixing up the units of thermal conductivity and the heat transfer coefficient.
 
For the P1 and P2 trimmer pots on the gain stage, do I turn them completely CCW to start the biasing procedure? I did some measurements and the pot doesn't go from 0 to 500. It just sorta hangs around the 250 area. I went full CCW AND CW and it only changes a few ohms. Bought Bourne 3296 series, style W. I'm stumped.