General circuit power question from 12V source

jepherz · Guest

First off, let me start by saying that I am just getting into this whole circuit design, so if it sounds like any of these are nooB questions, well, they are.

Alright, so I made a circuit for automotive use, so my supply is going to be 12V. Currently I am using a 7805 regulator and running it with a 18V wall adaptor. I know when I hook it up to 12-14V, the regulator should be less inneficient, but at this point the regulator gets really hot. I can hold my finger on it for MAYBE 2 seconds. There is also no heat sink on it currently either.

With the regulator, I am powering a PIC, two 7-seg displays and 7447 drivers, a sensor, a 10k potentiometer, and an output LED.

My question is if I am powering this circuit correctly or if I am sucking too much current from a regulator such as the 7805. Should I keep this setup and just add a heat sink no matter how hot it is getting, or do I have another, better option in terms of regulator efficiency?

Thanks for your help,
Jeff

valemike · Guest

The PICDEM-2 uses a wall regulator, but uses a 9V / 500ma (?) rated regulator. Even the PICDEM's 7805 (lm340) gets really hot. That's why they have a heatsink.

In my case, i rectify 12VAC from a transformer, thus giving me 17VDC rectified going into my 7805 (actually LM340). I burn so much current thru LED displays, relay windings, that the VR gets way too hot to touch. Since it's a TO-220, I use my chassis as a heatsink after screwing it onto the chassis wall and using thermal heat transfer pads.

Guest · Posted: Tue Jun 21, 2005 11:51 am

so i'm going about this right and shouldn't change anything?
Thanks

octobersky · Guest

Pin & Size compatible with a TO-220
No Heat: Switch-Mode Power Supply (Eff.>85%)

http://www.datel.com/data/meters/dms-78xxsr.pdf

octobersky · Joined: 21 Jun 2005 Posts: 3

Pin & Size compatible with a TO-220
No Heat: Switch-Mode Power Supply (Eff.>85%)

http://www.datel.com/data/meters/dms-78xxsr.pdf

bfemmel · Joined: 18 Jul 2004 Posts: 40 Location: San Carlos, CA.

jepherz,

The 7805 is certainly going to be the easiest and cheapest solution. If you can't hold your finger on it for more than two seconds, that is probably around a 60C case temperature, not bad, especially without a heat sink. Don't let that amount of heat bother you. When you put your circuit all together in it's final mechanical configuration, add a heat sink by attaching the regulator to some metal if you can. Even the smallest amount of extra surface area will help dissipate the heat. You could even do that by laying it down on the board and providing a little copper pour. That way the heat sill spread into the board. And if you could attach it to the car body that would be great! Very Happy

Bruce

JFK · Guest

If connecting to the car power source then I would suggest additional filtering may be needed otherwiae transients will kill the 7805.

The TO22 package will only dissipate 2Watts @ 25°C if I remember right so doing the math. Max input V lets say 15V - 5V output leaves 10V input to Output, Dissipation 2W/10V = 200mA is the max current you can have @ 25°C.

So if you are drawing close to the 200mA you will need a Heatsink, for any less check the datasheet and it will give you the maximum dissiaption V Temperature. But don't guess how hot it is measure it the ambient inside a car can easily reach 40 - 50° during the day especially in tucked away places where there is no air flow. If it's a permantly installed device attaching the regulator to the chassis will be the best.

JFK

valemike · Guest

SherpaDoug · Joined: 07 Sep 2003 Posts: 1640 Location: Cape Cod Mass USA

It sounds like the 2 seven seg LED displays are drawing most of the current. Could you tone them down a bit? Where will your final product be mounted and how well will it be able to dissipate heat? If it is hot under the hood or getting direct sunshine through glass you may not be able to add too much more heat without problems.
Possibly you could have the circuit monitor temperature and dim the LEDs if things get too bad.
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Guest · Posted: Tue Jun 21, 2005 7:36 pm

newguy · Joined: 24 Jun 2004 Posts: 1907

To dim the LEDs, just drive them with a square wave. By varying the duty cycle, you can vary the brightness. Look into using the PWM on the pic.

asmallri · Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia

Sounds to me that you are not driving the LEDs very efficeiently.

Do you have a schematic?

What current are you supplying to each of the LED segments?

Are you multiplexing the LEDs?
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Stewie82 · Guest

Thomas Blake · Joined: 18 Jan 2004 Posts: 22 Location: Burbank CA

asmallri · Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia

With the values you have specified the worst case scenario is 144mA when all LED segments are active. The "cleanest" solution to your problem is to use a DC-DC converter (switching regulator). This solves the heat dissipation problem. However if you do not want to go down that path another solution is to use two separate power rails.

The 7447 is an open collector device capable of withstanding 15V on the open collector ouput when not being drive - unfortunately not high enough for an automotive application but workable. Use a 5v regulator to drive the PIC and the 74HC47 (this can now be a 78L05 regulator) and use a 10V low dropout regulator to power to the LEDs. Increase the value of the segment resistors to 750R. A low dropout regulator typically requires a minimum of 1.2 volts across the regulator so this will allow the power to drop to 11.2 volts and maintain regulation.

Assume the maximum input voltage you have to allow for is 18V. The maximum current for the LED segments is 144mA. Therefore the maximum power dissipation across the regulator is (18-10)*.144 = 1.15 watts . The result is that you have almost halved the regulators power dissipation and therefore have a much greater chance of avoiding thermal run away when the inside temperature of a car exceeds 60+ deg C. What happened to the rest of the power that was disipated in the original design? It is now distributed across the segment resistors but each segment resistor dissipates less that 100mW.

You could do better with the dual power rail method. How? Use transistors, or FETs to drive the segments. This gets rid of the 15volt cap problem. The LED segments can now be driven directly of the car 12 volt supply. You will need to increase the value of the 750R resistors. The disadvantage with this method is that the intensity of the display will increase with battery voltage (can't have everything). With a little more work (components) you can even solve this problem.
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Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!!