Voltage regulator as current source? - Page 1
Forum section: electronics / projects

author: Things
author_rank: Regular Contributor
timestamp: March 10, 2013, 12:18:45 pm
content: Hey guys,I'm looking for a buck module to power a laser diode, and one requirement is I need to be able to do current regulation, laser diodes being current driven devices. I'm looking at the PTR08060W from TI, as it has specs suited to what I need.I know you can set up an LM317 as a current source, but does the same idea apply to ALL voltage regulators? If so, how do you implement it? The ratio for finding a resistor is: Rset = 1.182/Vo-0.591 in Kohms.Cheers,Dan

author: Alana
author_rank: Frequent Contributor
timestamp: March 10, 2013, 12:41:11 pm
content: LM317 is good for it because it has low Uref and very low current on adjust pin. 78xx series will work too but they have considerable ground pin currents and no less than 5v of voltage on curret setting resistor. Other chips - look datasheed, i think i have seen few switchmode regulators that can be used both in voltage and current source modes.

author: Things
author_rank: Regular Contributor
timestamp: March 10, 2013, 12:47:38 pm
content: I'm aware the LM317 works good for it, but I'm looking at around 2A or so constant current, and really don't have the room for a large heatsink I'd need on an LM350 or equivalent. This TI module is around the size of a TO-220 package and would be ideal, and I have heard of others using shunt resistors to run them in current regulation, but I'm not exactly sure how you're meant to calculate the resistor values for it?

author: Unknown
author_rank: Unknown
timestamp: March 10, 2013, 01:08:28 pm
content: The way it works is that the regulator will try to adjust the current through its pass transistor so there is a constant voltage (about 1.2 V for the LM317, 5 V for the LM7805) between its output and adjust/ground pin. So if there is a 1 ohm resistor between output and adjust, the LM317 will try to force 1.2 A through it so there is 1.2 V across this resistor. This current will flow through the load, which is connected in series with the adjust pin and resistor. There is also a tiny amount of current flowing from the adjust pin to ground, this current is typically specified in the datasheet, and creates an error, especially at small currents.So as first approximation you can calculate Rset = Vset / I, or Rset = Vset / (I - Iadj) if you want to take the current through the adjust pin into account. Vset would be about 1.2 V for an LM317, Iadj in the uA. For high currents, you might need a power resistor (for 1 A, 1.2 W would be dissipated in the resistor with an LM317, 5 A with an LM7805). You also need to take the dissipation into account. In worst case (output shorted to ground), the regulator will have to dissipate (Vin - Vset)*I, or 29 W for 1 A from a 30 V power supply.

author: SeanB
author_rank: Super Contributor
timestamp: March 10, 2013, 01:28:36 pm
content: Just look for high power LED drivers, they do all that is required for this.

author: Alana
author_rank: Frequent Contributor
timestamp: March 10, 2013, 09:15:21 pm
content: This module works like LM3080 Dave shown in his PSU design. Constant current is passed n Vadjust pin and it causes voltage drop on a resistor that sets up output voltage. From what i see its useless for your appliacation.

author: Things
author_rank: Regular Contributor
timestamp: March 11, 2013, 06:54:29 am
content: Hmm, looks like I may have to find one that has a tracking pin, and use a shunt resistor + an opamp to do CC. Oh well