More newbie questions

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Hi Everyone,

Just found this board the other day and have read everything with great interest. I am relatively new to the DIY area, so far the most complex thing I've designed and built is my subwoofer (based on a MASS 2012 driver).

I've seen the amp design by Pass Labs, Leech, ESP, etc. and am wondering just what I might be getting myself into. So I have a few questions:

1) How much money (on average) are we talking about to assemble something like the Pass a40 or Leech amps? Can someone ball park this for me $100-$200?, $400-$500? more?
It seems that the most expensive parts are the chasis, transformer, and heat sinks... Is this correct?

2) How much experience do I need to be able to successfully build one of these projects? I am pretty handy with my soldering iron and volt/ohm meter. Do I need any other type of test eqipment?

3) Are these projects that can be assembled in a weekend, or are we talking much longer time frames here (assuming all parts are already on hand)?

Thank for your feedback, I do appreciate it!
1) Figure $300-500 depending on which design, how much power, etc. A 60w amp will cost more than a 30w one, because of the bigger power supply. A Class A amp will cost more than a class AB one because of the larger heatsinks and larger numbers of output transistors. Heatsinks, chassis, transformers, and power supply capacitors will probably be over 80% of the total cost.

2) You probably want a function generator (or at least a CD of test tones). A scope would be handy, too, and can be picked up for about $100-200 at ham fests or $400-500 new. You don't need more than a 20MHz scope for most audio work.

3) Depends on how much concern you have for the final result. If you just want to slap together something that makes a sound, you can probably do it in a couple of days, especially if you use premade boards. If you want to do a quality job and make something that looks and sounds excellent, it will take some time.

Thanks, Jon! Just looking to gather some basic information and see if this is something I can reasonably accomplish. I am really interested by the Pass a40 design. I figure I'll spend most of my time sourcing the parts for it.

I do have a test tone CD, but don't have a scope - nor do I know how to use one. Guess it's time to start learning!

I'd really like to take my time on this one and build it right. Sound quality is most important to me, but I also want the final product to be something visually appealing, too.

Guess I was hoping for a little less cost, but if I take my time and purchase parts little by little (more like 'as I find them') it won't be any big deal...
If you buy stuff surplus, you can usually get it a lot cheaper, especially heatsinks, powersupply capacitors, and transformers.

$500 might sound like a lot, but a design like the Pass A40 is probably in the same league as commercial amps costing 3 to 4 times that.

Here are a couple of links to surplus stores online:

The only one I've dealt with is the first one, which is actually here in my neck of the woods (SF Bay Area), so I'll usually just go down there and poke around.

Most of the places have stock that they don't list on their webpages (i.e. Halted often has small quantities of great heatsinks), so you might try calling them, too.

Also, most reasonably sized metro areas will have a couple of electronic surplus dealers, more if your in an area with a lot of manufactures, military bases, etc. Try looking in your local yellow pages.

Toroid of Maryland has all kinds of toroidal transformers available. I think Plitron may sell some direct, as well:

Heatsinks, you're probably better off getting surplus. Newark electronics sells a variety of the Wakefield heatsinks, including the popular 423, which is .67 C/W, for about $20. Unfortunately, it's double sided, which means you can only really use it with TO3 cased devices.

A site that someone posted in another thread here, , is offering two varieties of flat-back heatsinks, one at .80 and the other at .35 C/W. The .35 C/W one is only $35 which is a steal for something that size. Never dealt with them, though, so I don't know.

Finally, if you're good with metalwork, this guy has some fantastic looking DIY chassis/heatsink designs:

A little out of my league, though.

Heat sink calculations

Jon: Thanks for the toroid links, I think I've found what I need in that dept! I have a question for you about calculating heat sink values, though. Can you tell me if I have calculated this properly?

For the Pass Labs a40 design (100 wpc draw, 40 wpc output to speaker) we have a 40% efficiency. This means at idle, I need to dissipate 100wpc heat and at full load, I need to dissipate 60wpc heat. Assuming we have 25c ambient temp and wish to limit thermal rise to another 25c, we have the following formulas:

Idle: c/w = 25c/100w = .25c/w maximum dissipation
Full output: c/w = 25c/60w = .42c/w minimum dissipation

So, I should look for a heat sink with dissipation values somewhere between .25c/w and .42c/w - the closer to .25c/w the better, right?

Have I done this correctly?

For heatsinks, you always design for the worst case, so if the A40 is dissipating 100w at idle, that's what we need to design the heatsinks for.

If you want to keep the temperature rise to 25C, you'll need a junction/ambient resistance of 25/100 = 0.25. But it is not just the heatsink that factors into this figure. You also have to worry about junction/case resistance and case/heatsink resistance. The junction/case resistance is on the order of .5C/W (less if you use TO-3 transistors) and the case/heatsink resistance is about .1C/W. Since the A40 uses 6 output devices, you divide these figures by six.

So you can find the necessary heatsink by solving:

0.08 + 0.02 + X = 0.25

Which means you'd need a .15C/W heatsink, which is very big indeed. You could probably do this easily in a monoblock configuration, but it'd be a mighty big amp in stereo. Now you know why the big class A Krells and Levinsons are 10 inches tall and weigh 200lbs...

In reality, you'd probably just have to accept a 40-50C junction temperature rise, which would require a total junction/ambient resistance of 0.4-0.5, or a heatsink/ambient resistance of 0.3-0.35. That's a big heatsink, but in the realm of the possible. You could use one of those big .35 C/W heatsinks from Seal electronics, or a 2-3 Wakefield 423Ks (.67C/W each) and be ok.

Using this approach is probably overdesigning a bit but it's better than frying expensive output transistors.

BTW, if you are going to buy the heatsinks surplus, the ESP site ( has a nifty little heatsink thermal resistance spreadsheet you can use to figure out how efficient your latest surplus find is.

Heat sink options

Jon: OK, so I need more heat dissipation. Does all of the heat dissipation need to be on a single heat sink for some reason? Or can I use multiple smaller sinks?

Assuming I use eight TO-3s and put each one on its own heatsink, that 0.15c/w total figure gets divided by the 8 output divices (not 6 as you mentioned above). So that reduces the thermal requirements for each transistor on its own heatsink to something more like 1.2c/w. Is this correct?

Assuming the above is valid, I have found a few alternative configurations at
(one of the original sites you provided above).

1) On page 1 they have the (HSK)NC-421A double surface heat sink already drilled for a single TO-3. Its rating is 1.1c/w - just about right.

2) On page 2, they have the (HSK)641A, which is rated to dissipate 35 watts. If we have a total of 100 watts to dissipate, spread over 8 output devices, shouldn't each device only need to dissipate 12.5 watts (eventhough its smaller than the one listed above)?

What's your take on these? Each alternative is cheaper than one or two larger sinks, and I can put them side by side on the sides and back of the chasis to keep a "smaller" overall appearance of the final stereo amp.
If there are 8 devices, the junction/case and case/heatsink resistance drops to .075 C/W, so the total heatsink/ambient resistance you need to shoot for is .175 C/W.

There's no reason you can't use multiple heatsinks. In fact, in this application, I can't see any other way. You simply divide the resistance of each heatsink by the number you are using. The formulas for series and parallel resistance is the same as for electrical resistance.

If you mount each transistor on its own heatsink, you'll only need about 1.4 C/W heatsinks which should be relatively easy to find.

The NC-421A you mentioned is exactly the same as the Wakefield 423, but half as high. The 423 is rated at .67C/w.

I can't seem to get to the site right now, but if I remember correctly, the 641 is a little less efficient 1.3 or something in that ballpark. Their "wattage" figure is a little high -- they assume a greater than 25 degree temperature rise.

I would go for either 8 of the 421's or 4 of the 423's. If you're building a stereo amp, remember, you'll need twice this many. I personally would think the 4 (x2) 423s would be better. Mounting 16 621s on a normal sized chassis would be hard.

Actually, if you used 6 621s or 3 623s per channel, the total heatsink resistance would be .183 and .223 respectively. These would give a 26 and 29 degree rise over ambient, which would probably be fine.

One final thing to remember, the resistance rating on these are for them mounted with the fins vertical and with adequate airflow through the sink. If you mount the fins horizontally, they'll be less efficient, but I don't know by what factor.

output devices & heatsinks

Jon: I think we have a little misunderstanding: for the a40 amp, there are 4 output devices per channel, for a total of 8 devices for a stereo amp. These 8 total devices need to dissipate a total of about 100w (50w per channel) at idle.

Does this then raise the thermal resistance requirement for each heatsink to 0.7c/w? (half of the 1.4 you mentioned above)
I'm afraid the dissipation is 100w in each channel in the A40. In a class A push/pull design, you dissipate at least twice as much power as you can deliver into the load.

In general, the formula for calculating the total thermal resistance of the whole mess is:

R = (Rj + Rc)/n + Ra/m

Rj is the junction/case resistance of the transistor
Rc is the case to heatsink resistance of whatever insulator you're using
Ra is the heatsink/ambient resistance of the heatsink (the normally quoted efficiency)
n is the number of output devices
m is the number of heatsinks

Typical values for Rj and Rc are 0.5 and 0.1 respectively, but look on the data sheets for your transistors.

If you find the contribution of Rj and Rc too high, you can always add additional output devices in parallel. Four output devices for 40w class A seems really low to me. Pass's own A75 design uses 24 devices per channel for 75w.

Also, you can probably run the junctions higher than 25C above ambient. Most power transistors can be run with junction temps in the 100-200C range. Just make sure that the power dissipation at that temperature doesn't exceed the safe operating area given on the spec sheet.

oops! Your're right!

Ooops, I had calculated the power dissipation accross each of the push/pull output stages for each channel. According to the calculations in the paper on the A75, I calculated the dissipation for each bank at 54w (54w push, 54w pull), so there are a total of 108w per channel, or 216w for the stereo pair - duh! Got lost in my own calculations...

Thanks for pointing out my error - could have lead to a big problem!
One more try

OK, here goes another try putting together what we have gone over:

Following Pass's calculations on dissipated power (A75 amp article), the a40 will need to dissipate 216w in total for stereo operation. We wish to keep thermal rise to 25c. Therefore, the target thermal resistance = 25/216 = 0.115c/w. OK so far?

In the article on the A75, this figure is simply divided by the number of heat sinks to use, and no mention it made of junction/case and case/heatsink loss. If we ignore these losses, I could use 8 heat sinks each with a 0.92w/c thermal resistance (8 * 0.115 = 0.92). This seems pretty easy to accomplish...

However, if we factor in the junction/case and case/heatsink loss for a total of 8 output devices (stereo) and plan on using 8 individual heat sinks (onee per output device), we then get the formula:

(0.5+0.1)/8 devices + cw/8 heatsinks = 0.115 total thermal resistance
cw = (0.115 - 0.075) * 8 or <b>0.32c/w for EACH</b> of the 8 heatsinks?!?!

8 sinks with thermal resistance of 0.32c/w seems REALLY excessive... Am I wrong here? This is like having 8 of those Seal heatsinks!

Any guess why Pass's article ignores losses in thermal resistance? In the last paragraph of the heatsink section, he simply indicates that the case temp rises 10c over the 50c temp of the sink - is this because no additional thermal losses are accounted for?

[Edited by Eric on 12-08-2000 at 03:43 PM]
How to recognize a thermal feedback loop

I'll go back to the plans and re-read them to see if he mentions such a loop. On the plans, the wiring guide for the transistors simply indicates 4 output devices and four resistors that are not mounted on the main amplifier circuit board, so these are the components located near/on the heatsinks. I'll go back over the article to see what I can find. The two pictures of this project on the pass web site show 1 design using 2 heat sinks, and a second design using 4 heat sinks...
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