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chrispol
Joined: 18 Oct 2009
Posts: 14
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 Posted: Thu Nov 05, 2009 4:52 pm |
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Hi,
In general I would say that is not normal. Excessive temperature will shorten the life of the component.
Is the board working at all?
Which component is actually getting hot? The standup part (it looks like a 78XX linear regulator), or the one soldered to the board? Have you got a part number?
Are you using the supplied wall wart power supply, or something else? The power dissipation of the regulator (I don't have a schematic, but I assume it's a linear regulator) is the Voltage in - Voltage out /Current. Excessive dissipation can come from too high an input voltage, or drawing too much current, or a combination of the two.
Joe |
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asmallri
Joined: 12 Aug 2004
Posts: 1635
Location: Perth, Australia
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| Re: Embedded ethernet kit voltage regulator heat |
| Posted: Thu Nov 05, 2009 8:31 pm |
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Well, it won't fry it but it will come close. The problem is the ENC28J60 is a power hog consuming 200mA at 3.3 volts. Lets say you used a standard linear regulator from 12 volts to derive the 3.3 volts. The power dissipated in the regulator = (12 - 3.3)Volts * 0.2Amp = 1.75 Watts. Naturally if the input voltage was higher, say 18 volts then the power dissipated would be (18-3.3)Volts * 0.2Amps = 2.8 Watts.
That is the downside of this Ethernet controller, simple to implement (SPI bus) but power hungry. By was of contrast the RealTek 8019AS consumes around 25mA at 5 volts but is more complex to integrate (parallel address and data bus).
_________________
Regards, Andrew
http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!! |
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chrispol
Joined: 18 Oct 2009
Posts: 14
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| Posted: Thu Nov 05, 2009 10:24 pm |
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So I guess one of the best bet, is to bring the input down to just above 7v the minimum required by the 7805, and may as well heatsink it at the same time.
Gonna have to dig up some power resistors to divide my input voltage. |
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asmallri
Joined: 12 Aug 2004
Posts: 1635
Location: Perth, Australia
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| Posted: Thu Nov 05, 2009 10:56 pm |
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| chrispol wrote: | So I guess one of the best bet, is to bring the input down to just above 7v the minimum required by the 7805, and may as well heatsink it at the same time.
Gonna have to dig up some power resistors to divide my input voltage. |
Well the best thing is to use a switching regulator IC plus an inductor. This will give you high efficiency with minimal power dissipation in the regulator.
_________________
Regards, Andrew
http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!! |
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PICoHolic
Joined: 04 Jan 2005
Posts: 224
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| Posted: Fri Nov 06, 2009 3:44 am |
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I had the same problem.
I have lowered the input voltage to the minimum and installed a heat-sink at the back of the regulator. |
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chrispol
Joined: 18 Oct 2009
Posts: 14
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| Posted: Wed Nov 11, 2009 8:41 am |
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| Yep did the same works now no heat |
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