The minimum current is 1. Andy aka k 8 Only then would the regulator start regulating at all. Looking at the datasheet, I would need a 0. Nope, just change R1. Reduce the sense resistor.
|Date Added:||27 July 2013|
|File Size:||58.79 Mb|
|Operating Systems:||Windows NT/2000/XP/2003/2003/7/8/10 MacOS 10/X|
|Price:||Free* [*Free Regsitration Required]|
With LM at 1A you can use low power pot in parallel to high power current sense resistor. Because the LM will regulate the voltage on the adj input allways to 1.
LM constant-current power supply | LEDnique
Eventually I was going to build audio circuits anyway so this is a valuable resource to me. Q1 and Q2 form a darlington pair.
Above all have fun. And a linear pot will give you absolutely no control what so ever.
LM317 / LM338 / LM350 Current Regulator Calculator
Here I’ve used a LM as a constant current source to control the base current Ib through constant current lm317 MJ 15 amp watt transistor. This makes LM useless as a constant current source below 5-volts. So I could use a 1. October 16, Even then, I’m looking at 2mA minimum, which is higher than the 0.
We know that there will be around 1.
You argue for one output stage that hansles both voltage and current regulation, yet you show a design with two separate stages?
October 18, Post as curreent guest Name. With a 5 or cureent load adjust for desired maximum current. A current source is the dual of a voltage source. Vout the load has no effect on the current. Check the ADJ pin current which according to the datasheet is a max of uA. All sorts of constant current lm317 circuits are included in the onsemi datasheet for the LM Page 8 will be of most interest to you.
The full VR1 can easiliy handle this, but the very small part of Constant current lm317 you are using in this 0.
There was an error while thanking. Post as a guest Name.
LM317 constant-current power supply
As 3 rises constant current lm317 Darlington transistor will turn on … … turning on the second transistor. Hey Wouter van Ooijen I managed to create the circuit and using a wirewound type pot.
A solution could be an old-fashioned bulky and costly wire-wound potentiometer.