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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?

enter image description here

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

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Is the 12V supply regulated already or is it unregulated? The concern behind this question is whether the LED device is subject to overvoltage. – Kaz Jan 16 '13 at 21:38
Actually I don't know. It is one of those switching power supply adapters, I guess they are unregulated transformers(?). Does that mean that if there is no or little load the voltage will rise? – Martin H Jan 16 '13 at 21:43
Switching adapters are almost certainly regulated. They chop line voltage across some inductor to generate a voltage, which is sensed by a feedback loop that controls the chopping. Measure just to be sure, and if you have a scope, it doesn't hurt to look at what the supply is putting out as a function of time. – Kaz Jan 16 '13 at 21:49
I agree with Kaz here - it should be regulated if it is indeed a proper switching regulator module, as opposed to a simple transformer/BR/cap. If you are not totally sure a photo of the device would probably help us let you know. A load test could be discussed if necessary also. Do you have a scope and multimeter? – Oli Glaser Jan 16 '13 at 23:48
Actually I did not buy the supply yet, I thought I seek some advice first. I was looking for a cheap one, the kind you plug in the wall. But as you guys seem to agree that a regulated supply is the way to go, I look for one that is specifically advertised as regulated. I do have a multimeter and a scope to check the actual output. I could even connect a load and measure current but I think this is not necessary – Martin H Jan 17 '13 at 11:17
up vote 3 down vote accepted

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?)

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A 7812 would not do its job, would it? From what I read they need a voltage drop of 3V or so to work. – Martin H Jan 16 '13 at 21:51
It depends on what the actual voltage is. The non-switching type adapters that contain transformers often put out a peak voltage well in excess of what is written on them. – Kaz Jan 16 '13 at 22:00
Isn't this only true as long as there is no load? If I would have a 12V;1A supply and no load, the voltage would be higher then 12V, but with a 1A load it would be very close to 12V, or am I wrong here? I guess a regulated power supply would not be much more expensive if that is preferable. – Martin H Jan 16 '13 at 22:07
That is true. How much the voltage sags for a given load depends on the supply. Anyway, your LED is actually a device that has built-in current regulation, which doesn't need an exact 12V supply. – Kaz Jan 16 '13 at 22:09

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.

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(12v - 9v) * 0.3A = 0.9W. It will get warm, but not too hot. – Passerby Nov 3 '13 at 8:55

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.

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Thanks. Actually the real circuit is a bit more complex with a constant current supply for the LED, I just wanted to keep it simple to illustrate my problem – Martin H Jan 16 '13 at 21:50
I suspect that the "LED" was actually some complex device with a LED embedded in it, simply because "naked" LEDs don't usually run on 12V. – Kaz Jan 16 '13 at 22:07
@Kaz, yeh OP just confirmed this, just making sure ;-) – Oli Glaser Jan 16 '13 at 22:14

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