The Newblifier

[impala454]

So since my speaker project is somewhat on hold waiting for the size of my house to grow, I decided I'd like to build a simple 2-channel amplifier. I read through the stickies but would like some sense of direction on where to start. Is there some kind of simple design already out there I could follow? I figure this will be just for learning experience, and possibly used later on a test bench or as a garage amp, so cheap is the name of the game

[Curt C]

They don't get any cheaper than the chip amp:

http://www.chipamp.com/

Easy build, but should be a good foundation for more extensive projects.

And of course there is the Elliott Sound products site:

http://sound.westhost.com/index2.html

I like the looks of amplifier projects 3A and 101. No, I’ve not built them, but their simple, elegant architecture suggests they could perform quite well when implemented with a good power supply design.

C

[impala454]

Awesome thanks Curt. Just what I was looking for!

[Hdale85]

I have a chipamp and they are actually rather good sounding. Be ready for lots of tweaking to get rid of RF and hum though. Took me quite a while because my speakers were so darn sensitive.

[impala454]

So I got this kit: http://www.chipamp.com/lm1875.shtml

I noticed there's no heatsink, can I just get any old heatsink or is there some specific type or size I need? I know Fry's has a ton of heatsinks and heatsink boxes and such.

[Hdale85]

No they don't generate a ton of heat so something a few inches long and wide should be good. Thats a new kit so I'll be interested to hear how it sounds.

[TacoD]

So I got this kit: http://www.chipamp.com/lm1875.shtml

I noticed there's no heatsink, can I just get any old heatsink or is there some specific type or size I need? I know Fry's has a ton of heatsinks and heatsink boxes and such.

Nice little amplifier, if you build this into a metal case you can easily cool it using the case as heatsink. Do not forget to isolate the chip from the metal.

The 1875 with appropriate speakers (read not a difficult load) sounds very good, if my memory serves me right it is a tat more resolution than the 3875.

[impala454]

cool deal thanks guys. I'll probably build it out here in a few days.

[chasw98]

Look here http://www.apexjr.com/Sinks.htm for cheap, good heatsinks that you can reuse down the road.

Chuck

[impala454]

Cool thanks Chuck. Is there any method to figuring out what size heatsink you need for one of these things?

[dmalphur]

The calculations to determine heat sink size are fairly straightforward. The first thing you need to know is your power supply voltage and the minimum load you will be using.

Look at the datasheet for the LM1875 and you will notice that there are curves which indicate the power dissipation for a given supply voltage, output power and load resistance.

Lets say that you use the recommended 18V transformer. 18V - 1.2V (the forward voltage for the MUR860) = 16.8V. 16.8V*1.414 (capacitor smoothing) gives approximately +/-24V rails. The power dissipation is not proportional to the output power, and at 5W the LM1875 is already dissipating 15W with a 8ohm load. If you want full power you will have to dissipate close to 20W per chip.

Here's the bad news. The LM1875 actually has higher thermal resistance to the heatsink than the LM3886. This means that for a given power dissipation, the LM1875 will actually require a larger heat sink than the LM3886.

Here's the rough math to calculate the thermal resistance of the heat sink required. This is taken from a similar thread that I wrote on diy audio:

Looking at the datasheets for these parts, the thermal resistance of the junction to case for the LM1875 is rated at 3 C/W (though in the application notes it says 2 C/W and they use that number for all of their calculations). The thermal resistance of the junction to case for the LM3886 is 1 C/W. This gives the LM3886 an advantage in that it can dissipate more heat energy for a given rise in temperature. This means that for the same power output, the LM3886 will actually require a smaller heat sink than the LM1875.

While we are calculating thermal resistances, lets go ahead and evaluate the heat sink performance required to output 20W into a 8 ohm load (assuming 20W of heat dissipation to achieve this).

The maximum junction temperature for either chip is 150 C. These amplifiers will be used in a consumer environment, so we will consider the maximum ambient temperature to be 40 C (104 F). This means that the total thermal resistance from the junction to the air has to be (150-40)C / 20W. That's 5.5 C/W total thermal resistance for the system. That number must include the sum of all resistances from the junction all the way to the air. There are three resistances that we will have to consider.

1. The thermal resistance from the junction to the case. This is given in the datasheet and we cannot change this number.

2. The thermal resistance from the case to the heatsink. This will depend on the type of insulator and heat sink paste we use, as well as what type of heat sink material is used and how well it is surface finished.

3. The thermal resistance from the heatsink to the air. This depends mostly on the surface area of the heatsink, though it also is affected by the orientation of the heatsink and the geometry of the heatsink.

If any other junctions are introduced, such as the heat spreader to aluminum housing, they will add to the series thermal resistance (but may still improve the overall thermal resistance).

Starting with our 5.5 C/W requirement we first subtract the junction to case resistance, 2C for the LM1875 and 1C for the LM3886, which leaves us with 3.5 C/W and 4.5 C/W respectively. Second we subtract the case to heatsink thermal resistance, which is estimated by National to be approximately 1.6 C/W for the LM1875 and 0.4 C/W for the LM3886. These estimates are based on a 2 mil mica insulator with heat sink paste on both sides. This leaves us with the required heatsink to air thermal resistance of 1.9 C/W for the LM1875 and 4.1 C/W for the LM3886.

For 30W of heat dissipation (20W power into 4 ohm load with +/- 25V rails), the thermal resistance requirement is .07 C/W for the LM1875 (not attainable) and 2.3 C/W for the LM3886.

Based on those calculations you would need a fairly substantial 1.9 C/W heatsink for each LM1875. Looking at various extrusions on suppliers websites should give you an idea of the size required to get that figure.

EDIT: Looking at the Aavid / Thermalloy webpage, the extrusion selector gives the following size as a 1.9 C/W heatsink profile:



This is based on a length of 3 in, so the above recommendation of a heatsink that is a few inches wide by a few inches tall should work fine. I've always used the old slot style pentium heatsinks with good results. If you want to get full power out of the stereo pair, you probably should go with one heatsink of that size per chip.

Hope this helps,
David Malphurs

[impala454]

Awesome, thanks for that info David.

[dmalphur]

Hey you are welcome! I looked at a couple of surplus pages and found this little number: http://www.allelectronics.com/make-a...-X-37MM/1.html



I would estimate that based on it's profile, length, and fin density it would have around 1 C/W thermal resistance (remember the lower the better). One of these per chip would easily cool your lm1875s even at 20W power output.

An added benefit, they are only $3.00 each!

-David

[impala454]

Oh cool beans. So to attach it do I just use the patch of thermal compound that comes on the bottom of it?

[dmalphur]

If the patch is non-conductive, that will work. The tab on the LM1875 is at negative rail potential, so it should be isolated from the heat sink. I don't see any pictures or hints that tell whether that particular heat sink has a non-conductive pad, but it is not difficult to come across a replacement pad if the one that comes on it is conductive. I can even send you a couple of pads if it comes down to that.

-David