Constant current wiring not working on breadboard, puzzled - Page 1
Forum section: electronics / beginners

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 12:51:48 am
content: I puzzled over this until my eyes glazed over, and have no recourse but to ask you guys to help me take a look.I'm trying to do the Constant Current circuit that Afrotechmods has postedhttps://youtu.be/iuMngik0GR8But it doesn't work. I measured voltage across the resistor and there is no voltage?!It should have 1.25 across R1, nominally.Voltages measure OK on all the other parts. Across the power rails, across capacitors, etc.I even removed the LED to just check the circuit.Did I wire anything wrong?I tried with two other LM317 in case it was the chip. Same results.

author: Audioguru
author_rank: Super Contributor
timestamp: September 17, 2016, 01:01:44 am
content: You did not label your wires.The output is supposed to be the ADJ terminal but you do not have a wire connected there, only the resistor. Maybe you wrongly have the output wire connected to the Vout pin by mistake.

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 01:15:23 am
content: Quote from: Audioguru on September 17, 2016, 01:01:44 amThanks but isn't the ADJ and OUT electrically common? I also tried to move the wiring to connect to ADJ but I remember it didn't work either.Still, per my question above, aren't they electrically common, as per the schematic?Thanks.

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 01:18:01 am
content: Sorry, ha! I just realized you're right. I must have shorted it OUT.I will try again. Thanks.(My excuse is I didn't sleep much last night :-)

author: JS
author_rank: Frequent Contributor
timestamp: September 17, 2016, 03:01:36 am
content: No, wait.You shouldn't short them, you should put a resistor between OUT and ADJ and connect the load to ADJ.You also need to connect the load, if not there will be no current. You need a minimum voltage drop across the regulator to start working (depending on which regulator) usually more than 3V between IN and OUT will make it work. So, 3V+1.25V you should expect to have 4.25V less in the load than in the source, to power one LED use at least a 7V source. You could go higher, not lower.ADJ and OUT are NEITHER electrically common NOR should be shorted out. You could short (probably using the ammeter) your ADJ to GND to measure the current and check it's all working.JS

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 08:56:29 am
content: Thanks guys, got it working finally after you both pointed out that oversight on the connection to OUT instead of ADJ.I tested it with different resistances and noticed in fact the R1 voltage is hardly ever the nominal voltage of 1.25, and in fact it went from 1.2 to 0.95 or so. The higher the current you want (the lower the resistance), the lower the voltage across R1. So you'd have to actually measure the voltage across R1 to calculate the actual current (you'd do this anyway but it's very different from assuming the LM317 maintains the same voltage between OUT and ADJ).Here is a pic of my thing working. Interesting experiment.

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 09:02:31 am
content: Actually wouldn't the current actually be LESS than the current through R1?Because the current has to split, to go to the load, and also go back to the chip, so following KCL, part of it would be diverted away from the load current.Unless its super high impedance on the ADJ pin?

author: Dave
author_rank: Super Contributor
timestamp: September 17, 2016, 10:46:40 am
content: No, it is actually more, because you can't have current flowing from a lower potential in a circuit towards a higher potential.The current on the ADJ pin flows out, about 50-100uA of it.

author: Audioguru
author_rank: Super Contributor
timestamp: September 17, 2016, 02:04:38 pm
content: The LM317accurately regulatesthe voltage across the resistor to be about 1.25V. It hardly changes when the resistor value is changed. Then the circuit is an accurate current source according to Ohm's Law.Since you measured less voltage across the resistor when the resistor value is reduced then your input voltage dropped or is not high enough.100mA in 100 ohms is 10V so the input to the LM317 current regulator circuit must be at least 10V plus 3V for the LM317 plus 1.25V for the resistor= 14.25V. The resistor is 1.25V/100mA= 12.5 ohms. If the resistor is reduced to 5 ohms then will the current be 1.25V/5 ohms= 250mA? Only if the input voltage is increased because 250mA in the 100 ohm load resistor needs 250mA x 100 ohms= 25V plus the 4.25V needed for the current regulator circuit.

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 17, 2016, 09:47:38 pm
content: Many thanks for further clarifying the LM317 guys.Looks like I definitely need to use a better power supply to experiment.

author: hurricanehenry
author_rank: Regular Contributor
timestamp: September 18, 2016, 12:45:59 am
content: This is the power plug I have been using for my breadboarding. I had stripped the cable to expose the leads.Since it goes only up to 12V, I will try to experiment with a boost converter to increase the voltage (for the LM317 experiment).For the Constant Current experiment I don't expect to source more than 100 mA, so hopefully this can work.

author: JS
author_rank: Frequent Contributor
timestamp: September 18, 2016, 03:00:59 am
content: A small current comes out of the ADJ pin, about 100µA according to TI datasheet, so, the current in the resistor is lower than in the Being the ADJ terminal is the one at lower potential it has to put some current out, no way it goes in.The LM317 is a voltage regulator, who is doing everything it can to put 1.25V between the ADJ and the out, that way, in conventional use, you have a constant current source in the resistor going between those two pins and the second resistor has that current (plus the Iadj) Then, the voltage across the resistor from ADJ to ground is the final voltage reference at a a constant current. You do need a minimum load for it to regulate, the nice thing is as you have effectively a current source there loading it, you could configure it to chew the minimum current so it does regulate when the complete circuit is unloaded. If you know you will be loaded there is no need for it and you can go for higher value resistors. That's why some applications call for the 220? resistor (when it's going to be loaded) and sometimes go for lower, 120? which already provide the minimum load.Now, for current source you are replacing the resistor between ADJ and ground with your load, minimum load requirements remain the same so you are limited on how low your current could be used. >10mA would be the recommended value, so, don't go over 125? for your setting resistor. Anyway for that ranges makes much more sense to just use a jfet which will do the job quite good, if a single jfet doesn't cut the requirements a second one could be added making the output impedance of the source much higher. There are some devices designed to work just like that, for quite low currents J500 to 511 have fixed currents. Here is something to read if you are into that kinda stuffhttp://www.vishay.com/docs/70596/70596.pdfJS