Calculating the proper OHM rating for a project

[pierhogunn]

Gent's I have found an electronic component that I want to use in a project.

it's an array of 940nm LED's, it draws 12v - 13.8v with optimum output at 13.2V, and it pulls right under 300ma

I went rummaging through my drawer at work and found an old 13V, 300ma power supply.

I put a meter on it, and low and behold it's a 18.16V power supply, while this does not surprise me, it does lead me to ask the question what sort of resister should I put into this project, and on which lead of the LED array should I put the resister on.

A helpful fellow told me that the formula Voltage Drop ( Supply Voltage - Desired Output Voltage) divided by .3 should give you how many ohms the resister should be

okay, great, 18.16 - 13.2 / .3 = 16.53 ohm

so is it that simple? do I have to worry about watts, the resisters I have found are listed as X ohms Y watts...

being electronically naive, I am begging for help picking the right resister

another thing, there is a little diagram on the brick <->--(*--<+> , does this mean that the little pin in the center of the plug is the + lead?

[Slik Geek]

I'm guessing that the 18.16V measurment was "no load". Based upon your measurements, it appears that the power supply is unregulated. When a 300 ma load is placed on its output, the voltage will probably drop to somewhat near 13V. That output voltage will vary due to tolearances in the transformer and the AC line voltage.

The important parameter for your load (the LEDs) is the current. Monitor the current flow and keep it within the ratings of the LEDs and you'll be fine (assuming that the polarity of your supply to the LEDs is correct!).

If you have some resistors available, you can experiment with various loads on the power supply and confirm that it matches its ratings. (Using just a resistor for the load: 13V divided by 300ma = 43.3 ohms load). That resistor will dissipate about 4W, so if you resistor isn't rated that high, work very quickly in taking a measurement!

It's likely that IF you need a resistor in series with your LEDs, it will not have to be as high of a value as you have calculated. The power dissipated in the resistor is equal to the voltage drop across the resistor multiplied by the current flowing through the resistor.

You can easily determine which pin is the + lead with your voltage meter. I don't always trust the diagram.

[scorrpio]

Lets see... Series connection: I = constant, V = V1 + V2, R = R1 + R2. V = I*R. Voltage drop on the resistor needs to be ~5V, so at 0.3A current it needs to be 16-17 ohm, so you appear correct.

I suggest getting a 20ohm trimmer and finding the best result. The one I linked, however, is only 0.5W rated, while you need approx. 1.5W. Locating a 2W, 20ohm pot is going to be tricky. What you can do is get three 100 ohm trimmers, put them in parallel for a 0-33 ohm range at a total of 2 1/4 Watts.

[pierhogunn]

ah questions from questions, what is the difference between regulated and unregulated?

using a simple little meter how would I determine what the voltage goes to under load? would I use the meter to complete the circuit between the power supply and the load?

The power supply comes bundled with an HP-Jet-direct like device that is rated for 13V-300ma, what is the likelyhood that my IR array will not blow up since it is rated for the same voltage and amperage?

[Slik Geek]

what is the difference between regulated and unregulated?

Unregulated: The AC line is rectified (AC turned into DC) and probably filtered (the peak of the AC waveform, or ripple, is smoothed so the output is nearly pure DC). Nothing more is done to the output. Any variation in the input line voltage will be transferred to the output. Another source of variation may be due to a load current change (which will cause a voltage drop due to resistances in the circuit).

Regulated: The same as unregulated, plus an active circuit attempts to maintain a constant output voltage. Think of it as a very fast, electronically controlled resistor that rapidly varys its resistance to control the voltage drop it causes. The voltage drop is varied such that the output voltage is constant. (In reality, the control isn't perfect, so there still is some variation, but the variation is significantly reduced). With a regulated output, the voltage won't change significantly between "no load" and "full load" conditions.

using a simple little meter how would I determine what the voltage goes to under load? would I use the meter to complete the circuit between the power supply and the load?

Measure the voltage across the terminals with a load connected. The measurement is the same as you already have done for the "unloaded" measurement. The meter isn't part of the circuit between the power supply and the load, it is just connected in parallel with the load (or could simply be connected across the load).

The power supply comes bundled with an HP-Jet-direct like device that is rated for 13V-300ma, what is the likelyhood that my IR array will not blow up since it is rated for the same voltage and amperage?

The probability is low that it will cause damage, but why take the risk? Find a way to test it without risking damage. Can your meter measure current? For that measurement, the meter has to be in series with load... it will be part of the circuit for the flow of current to the load.

[LCHIEN]

My opinion:
You have an unregulated supply, which, while the label says is 13V, 300mA, the "unloaded" output is 18.16V at zero mA.

An unregulated supply is for loads which do not require a constant voltage at various currents or clean power (or perhaps for a device which has a built-in regulator, like a printer). It will have variations both every 8 milliseconds as the input chages at twice the line frequency and whenever the load changes.

A regulated supply has a variable resistance semiconductor element and active feedback to make the output voltage exactly 13.0 volts varying only a few millivolts and correcting constantly, can track changes in load as fast a say a million times per second.

I think in your case, the LED array only requires an average output of 13V, regulation is not critical. Its only output is light, and you eye cannot perceive the 8 millisecond changes. Once you plug it into the power supply, the voltage from the power supply will drop to about 13V, one or two volts either way is close enough.

I think you can safely use the power supply with no resistors. Check the voltage when you have it on. I believe with its 12-13.8V specification your LED aray was made to be used with a car electrical system. Most auto electronics are designed to operate safely from around 10.5V to 16V which covers low batteries and high alternators, but most properly tuned auto systems are within 12-14 volts.

And you are right the little diagram is telling you the center contact is the positive one.

[scorrpio]

Some notes: each device has a certain working resistance R. But device is usually rated for certain voltage V. What this means is this device will consume V/R current, and therefore dissipate V*V/R amount of power. Notice that V in that formula is squared. Means that as voltage is increased lineraly, power dissipation increases parabolically. Doubling voltage quadruples power dissipation. In case of your array, going from 13 to 18V almost doubles the wattage the array has to dissipate. Whatever device the supply was used for likely had an internal voltage regulator that took care of the excess. So, I would not recommend running it without a resistor. Btw, it should not matter on what side of the array the resistor is. You just need to make sure this resistor can safely dissipate 1.5 watts.