Help with circuit analysis, why is my linear regulator so hot?

Thread Starter

ke5nnt

Joined Mar 1, 2009
384
I'm working on a project and need help figuring a few things out.

My project alternately flashes 2 banks of LEDs, each bank consisting of 30 LEDs, so 60 total.

Both banks are connected from the anodes through resistors to a 9V Linear Regulator, LM7809. The cathodes (collectively 1 from each bank) are connected to a N-channel LL MOSFET, which is driven from an output pin of a uC.

Bank 1:
Diode Vf = 3.2 volts
Diode If = .024A
There are 15 rows, each consisting of 2 LEDs, wired in series.
Resistor value for each series is 120Ω (1/4 watt)

Rich (BB code):
V+----------------------------
      |     |     |     |    |     |...and so on
      R     R     R     R    R     R
      |     |     |     |    |     |
      D     D     D     D    D     D
      D     D     D     D    D     D...D = diode
V- -----------------------------
Bank 2:
Diode Vf = 2.0 volts
Diode If = .024A
This bank is 4 LEDs per series, 7 series plus 1 series consisting of 2 LEDs
Resistor value for 4 series is 47Ω and for 2 series 220Ω (both 1/4 watt)

Rich (BB code):
V+----------------
     |    ___      |
     R   |    |    R
     |   |    |    |
     D   |    D    D
     D   |    D    D
     |___|    |    |
V-____________|____|___
I calculate for bank 1, 9V minus 3.2V drop for each diode, so 9V - 6.4V = 2.6V dropped by resistor
2.6V divided by circuit current (.024A) = 108.3Ω taken to next value = 120Ω
Each resistor dissipates 2.6V times circuit current (.024A) = 62.4mW
Each diode dissipates 3.2V times .024A = 76.8mW
So... for this bank all lit up at once (which it is), 15 series each drawing .024A = .36A
30 diodes dissipating 76.8mW each = 2.304W
15 resistors each dissipating 62.4mW = 936mW
2.304W plus 936mW = 3.24W total power dissipation

Bank 1 summary, .36A draw and dissipating 2.3W

Bank 2
Each LED drops 2V and there are 4 (for 4 series). 9V - 8V = 1V dropped by resistor
1V divided by circuit current (.024A) = 41.7Ω. Next closest 47Ω.
Each resistor dissipates 1V times .024A = 24mW
Each diode dissipates 2.0V times .024A = 48mW
For the 2 series on the end of bank 2
Voltage drop of 5V (9V minus 2.0V x2 LEDs = 5V)
5V divided by .024A = 208.3 up to 220Ω resistor
Resistor dissipates 5V times .024 = 120mW
Each diode dissipates 2.0V times .024 = 48mW

So, each diode dissipates 48mW times 30 diodes = 1.44W
7 resistors dissipating 24mW each = 168mW
1 resistor dissipating 5V times .024A = 120mW
1.44W + 168mW + 120mW = 1.73W

7 series of 4 diodes. Each series draws .024A times 7 = 168mA
1 series of 2 diodes. Add another .024A
Total current draw on bank 2 = 192mA

My reason for explaining all that is, I want to make sure all my math is correct. I've gotten lost along the way trying to do things like this before.

Mainly though, my LM7809 is getting extremely hot. Hot enough to leave a visible mark on your skin when you touch it. It's rated at 1.0A typical (2.2A peak). If it's only connected to the LEDs, even if ALL 60 diodes were lit at once (which they're not), the diodes shouldn't be pulling more than .552A. This is well within the range of the regulator. Does anyone know why it might be getting so hot?
 

steinar96

Joined Apr 18, 2009
239
You need to heat sink your regulator. They are supposed to be heat sinked for most applications. There is a reason for them having the TO-220 packaging (heat sink friendly). The effeciency of linear regulators is quite low so it dissipates quite alot of power.
It getting hot at 0.5A is not a big surprise.
 

studiot

Joined Nov 9, 2007
4,998
Remember also that the total power drawn from the supply is 0.5 amps times the unregulated supply voltage.

Your regulator supplies 0.5 x 9 = 4.5 watts to the circuit. The rest of whatever is drawn from the supply is dissapated by the regulator so if your input is 18 volts the regulator is dissapating another 4.5 watts.

So yes get a heatsink!
 

ELECTRONERD

Joined May 26, 2009
1,147
KE5NNT,

I'm glad you are undertaking the more technical aspect of amateur radio, it is a fascinating field.

Have you ever heard of LED drivers? They are basically miniature power supplies meant to power LEDs. It looks like the LT3598 LED Driver would suit your needs. You can take a look online if you go to the following website:

http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1094,C1766,P86477

It can drive up to 60 LEDs with a maximum defect of 1.5% for current matching.
 

Thread Starter

ke5nnt

Joined Mar 1, 2009
384
Thanks for all the replies. How do you heat sink a surface mount device? I realize the regulator I specified isn't surface mount, but there is a surface mount version and I was just wondering.
 

SgtWookie

Joined Jul 17, 2007
22,210
For an SMD regulator, you would leave a large copper area on the PCB to dissipate the heat. There are formulas for determining the copper thickness and area for heat dissipation. You could also solder a heat sink right next to the pad, bolt it over it, or some other arrangement.

Here's a useful link:
http://www.smps.us/pcb-design.html
Check out the Thermal Design area.
 

bountyhunter

Joined Sep 7, 2009
2,512
Thanks for all the replies. How do you heat sink a surface mount device? I realize the regulator I specified isn't surface mount, but there is a surface mount version and I was just wondering.
You solder it down to copper but the heat transfer capability of PCB copper is VERY limited because of the spreading resistance of thin copper..
 

studiot

Joined Nov 9, 2007
4,998
You solder it down to copper but the heat transfer capability of PCB copper is VERY limited because of the spreading resistance of thin copper..
Not true at all.
Careful when you jump in with a reply to be correct. (I've made that mistake as well)


There are three thermal elements to the functioning of a heatsink.

1) The collection of the heat from a local source such as your regulator

2) The removal of this heat away from the source.

3) The distribution of this heat to the environment.

The thinness of the copper lands on a pc board only affects #2, although this is acceptable for low single digit wattages.

Others have described ways of overcoming #1. However since rate of heat transfer is dependant upon temperature difference this rate will slow if the land is so small that the it achieves an even temperature.



#3 The distribution to the environment is difficult because this is largely effected by convection and the land is mounted on a flat plate which obstructs drawing air over the land and away. Some vertical fins soldered to the land help here.
 
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