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)
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)
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?
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- -----------------------------
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-____________|____|___
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?