Question about how voltage supply affects constant current driver

[mkoll4]

Hello, first post here. I'm not very knowledgeable in electrical engineering and haven't been able to clarify my understanding of a constant current driver.

I have set up a typical constant current source with an LM317 and can set it to output the current I want. The circuit I followed uses a 9V battery, and I am wondering how changing the source voltage affects the power on the diode. If the voltage decreases so does power if the current stays the same, but what voltage is used in P = IV for the diode power?

An extension of my question is what kind of power could you get a diode from an arduino. It can supply 5V, but only about 40mA from what I understand. I'd need a constant current source and I'm wondering if I can get enough power if I used a 5V power supply. Basically I don't understand a constant current source well enough to reliably predict the power/current. I guess I can just experiment with it before testing it with diodes.

Thanks for any help!

 

[diachi]

The LM317 is a linear constant current source. With linear power supplies the input current is around the same as the output current.

How this works in regards to power is as follows (we'll just make up some numbers here)

Current output: 1A
Input voltage: 10V
Output voltage: 5V

So power input is 10W (P=IV, P=1*10, P=10W). Power output is 5W (P=IV, P=5*1, P=5W). So where does the missing 5W go? In a linear supply it's simply turned into waste heat. For this reason, it's best to have your input voltage as close as possible to your output voltage, as this reduces the amount of power wasted and the amount of heat produced.

As for lowering input voltage, you shouldn't see a decrease in output power until you reach the minimum input-output voltage differential, which is up to around 3V for the LM317. Meaning your input voltage must always be 3V or more than the output voltage.

In regards to your Arduino question, you won't be able to get much power out of one of those, about 0.2W. By the time you account for driver losses and such your available output power after the driver is very small. If you're looking to modulate a laser you'd power it from a separate power with a driver capable of modulation (it'd have a modulation input). You'd then use the Arduino to modulate the diode using the modulation input on the driver.

Hope that clears things up.

 

[mkoll4]

Thanks for the response. To clarify, what determines the output voltage? How would I predict it or should I just use a multimeter?
Thanks!

 

[diachi]

Thanks for the response. To clarify, what determines the output voltage? How would I predict it or should I just use a multimeter?
Thanks!

Output voltage depends on the set current and load.

For laser diodes you'll want to look at a datasheet or current/voltage graph for whatever diode you're using, that'll tell you roughly the voltage drop across the diode at a specific current. You'll want to add 1.25V (voltage across the current set resistor) to the diode's voltage drop to get the final output voltage from the LM317.

That or just use a multimeter!

Keep in mind, when setting the current on an LM317 or any other laser diode driver it's best to use a test load, this lets you verify the current is set correctly and that everything is operating correctly.

 

[paul1598419]

Really, as long as the current is constant, it doesn't matter what the voltage drop across the diode is. Knowing the Vf is only necessary when you are trying to set up the driver's supply as you don't want to under supply it. However, on a linear driver any excess voltage delivered to the driver will be wasted as heat.