Powering a 12V and a 9V device from single power supply with LM7809 voltage regulator

Question

I have a little electronics project where it would be nice if I could power two devices from a single power supply (12V switching adaptor).

One of the devices is a high power LED with 350mA, 12V and the other is a 9V lower amperage device.

Using a 12V, 1A power supply, would the circuit below work to regulate down the voltage? Will the LM7809 will have problems with the load if I switch off the LED?

Sorry if the questions seem silly, it is the first time I am dealing with that kind of stuff.

EDIT: I'm aware that the LED is not added correctly. This was more for illustration really. It is a high power 10W RGB LED that I was hoping to connect with this constant current circuit

Answer 1

The LED does not affect the load of the voltage regulator. It affects the supply. The voltage regulator's load is the 9V device. The only way it will cause a problem for the regulator is if its current draw triggers a voltage drop below the point that the regulator is able to maintain 9V.

The main question about your circuit is: is the incoming 12 volts regulated? Or is it just the nominal voltage on a transformer device such as a wall adapter? If 12V is the maximum rating of the LED device, but the supply is unregulated, you may want an additional 7812 regulator there.

(What kind of LED is it; and does it have a built-in resistance for current limiting?)

Answer 2

You should be fine regardless of the state of the LED. As a side note the more current you draw from the 9V regulator though the hotter the regulator will become. At 300mA I would expect it to get warm without a heat sink.

Answer 3

The 7809 will have no problems if you switch off the LED, no. As long as you keep the input voltage above 11V (regulator output voltage + regulator dropout voltage), it will regulate okay.
Also, you will need a heatsink if you want to draw up to 750mA.

The regulators thermal resistance θj-a is 65°C/W, so for every watt dissipated it rises 65°C above ambient temperature.

Your input voltage is 12V, and the output 9V, so at 1A, the dissipation equals:

(Vin - Vout) * Iout -> (12V - 9V) * 750mA = 2.25W

so the temperature rise above ambient without a heatsink would be:

2.25W * 65 = 146.25°C - too much, the ICs thermal protection will kick in and shut it down.

So you need a heatsink, to bring the θj-a down to at least, say 35°C/W, preferably much lower. Here's a reference on Heatsink Basics. This is an example of a heatsink that would help keep your IC cool. It's thermal resistance is 24°C/W, so added to the 5°C/W thermal resistance of the junction to case (which we can't do anything about) we get 24°C + 5°C = 29°C/W. So the temp rise above ambient is now 2.25W * 29 = 65.25.
Not great, but you have some margin there.

For your maximum ambient working temp, subtract this value from 125°C (maximum working temp for the regulator) and you get 125 - 65.25 = 59.75°C. This can easily be reached in an enclosed circuit, so usually you would want a bit more "headroom" than this, e.g. the ability to work up to 70°C ambient. There are loads of heatsinks to choose from out there though, many with very low thermal resistances that will handle this dissipation easily.