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A 6.3V regulator delivering 2.8A...?
A 6.3V regulator delivering 2.8A...?
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Old 5th February 2008, 06:54 PM   #1
Stixx is offline Stixx  Germany
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Default A 6.3V regulator delivering 2.8A...?

To supply the heaters of my breadboarded Aikido headphone amp (using 6CG7 and 6H30) I have built a regulator according to this schematic (drawn by me but pretty much taken from different sources)

Click the image to open in full size.

Combined current draw will be around 2.8 A which scares me just a little bit...
The smoke test went fine with the regulator putting out a steady 6.4V. This thing generates a lot of heat but the heatsink on the LM 1084 was just warm to the touch after a few minutes. The 50W ballast resistor had to take a real beating and got mighty warm...!
I had to adjust a few things (like changing the potential divider to higher values, now 1.2k / 4.7k) and I will heatsink the R68 dropping resistor.

Apart from the obvious things like providing adequate cooling and such I would like to have the opinion of the experts:

- Anything I missed?
- Anything I could do better?

As always thank you!


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Old 5th February 2008, 07:13 PM   #2
richwalters is offline richwalters
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camera spam ...come on.... get to the point
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Old 5th February 2008, 07:59 PM   #3
zigzagflux is offline zigzagflux  United States
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Your schematic does not quite agree with the picture...

Schematic shows a 0.68 ohm resistor, you have a 2.2 aluminum housed resistor. That would in part explain why the heatsink isn't very hot, but the resistor is.

Also, I'm not a big fan of those 100nF capacitor locations. Consider changing their location. First one goes from regulator input pin to ground, for low impedance power supply to regulator. Second one goes from the top of the 1K resistor to ground. This serves to improve power supply rejection.
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Old 5th February 2008, 08:02 PM   #4
N-Channel is offline N-Channel  United States
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Yes. try a switching regulator! They are so simple to design these days, that they are well worth it! You can do a 5A adjustable regulator using under 10 components (depending on the version used), including the regulator IC.

go here to find out more.....

For an output of 6.3V at 2.8A, at 84% efficiency, the power consumed will be (6.3 * 2.8)/0.84 = 21W, and your heat dissipated will be [(6.3 * 2.8)/0.84] - (6.3 * 2.8) = 21 - 17.64 = 3.36W. At this low level, you could even do a surface-mount version, greatly reducing the size of the circuit and the filter caps. Just a thought.

Steve
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Old 5th February 2008, 08:20 PM   #5
Boris_The_Blade is offline Boris_The_Blade  United States
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Quote:
Originally posted by N-Channel
[B]Yes. try a switching regulator! They are so simple to design these days, that they are well worth it!
Great Idea!

I'm working on an off-line 6.3V@10A + 400V@250mA smps. Just need $ to order parts!
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Old 6th February 2008, 03:17 AM   #6
N-Channel is offline N-Channel  United States
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As for the parts, try "sampling" them from National, ON, and others.
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Old 6th February 2008, 04:16 AM   #7
kevinkr is offline kevinkr  United States
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A 6.3V regulator delivering 2.8A...?
Definitely ditch those 100nF caps - they are in the wrong place and as connected can cause stability issues. Add a 22uF cap across the 1K resistor for improved ripple rejection, and once you know the voltage across the regulator (It should be around 3V at low line to assure good performance) you can shunt the regulator (in to out) with a small power resistor to reduce heating. At 3V if you wanted to reduce the dissipation in the regulator by about half shunt it with a 2.2 ohm 10W resistor. The loop gain is high enough in this part and the output impedance low enough that you won't notice any significant degradation in performance but the regulator will run much cooler.

You do not want to use resistor values any higher than what you now have for the voltage setting resistors, the ADJ current is now about 1mA, the 1084 ADJ pin sources about 100uA worst case over temperature which can cause up to a 10% error in your output voltage. Really this should be more like 120/470 or 243/953, the higher current in the path swamps the adj pin current making it a smaller % of the total.

Add some additional capacitance at the output, say 4,700uF or so..

Correct me if I am wrong, but isn't that 2.2 ohm resistor the load? And it should be on a heat sink so that it doesn't get ruined - you're running it pretty close to its free air dissipation limit which is nowhere close to the heat sink rating of 50W.
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Old 6th February 2008, 07:02 AM   #8
Stixx is offline Stixx  Germany
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@Rich: Danke für den freundlichen Kommentar!

@Kevin: Thanks for your input, exactly what I was hoping for...

Those 100nF's were placed according to a schematic I found on the web...I'll ditch them.

Quote:
Add a 22uF cap across the 1K resistor for improved ripple rejection
Across the resistor means from the resistor to ground (as shown in "Valve Amplifiers")?

Various sources show either small capacitance at the output, others bigger values. I just replaced a 4700uF for the 10uF film
The 2.2 ohm is in fact the load...I switched the regulator off before it started to burn my breadboard...
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