Low-dropout voltage current source - Page 1
Forum section: electronics / beginners

author: mduchalski
author_rank: Newbie
timestamp: March 11, 2017, 08:20:56 pm
content: Hello,I need an in-line adjustable current limiter circuit. My priority is to keep the dropout voltage as low as possible, but overall power loss isn't really an issue. My present idea is to use a circuit that's functionally equivalent to LM317 current source. I replaced the standard regulator with a LDO - I think LM2943. As for the regulation, I used a 0.2 Ohm resistor combined with adjustable-gain differential amplifier. You can see specific values in the schematic. According to my calculations they will account to gain regulation from 6.8 to 56.8x which should correspond to current regulation from 100 to 900 mA.What I'd like to know before I order parts - does this circuit have any chance of working? Are there any better solutions for this application? Or maybe some modifications to the proposed one? Thanks in advance.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 11, 2017, 08:36:04 pm
content: Not sure the shown circuit works. At least the LM358 can have problems working near it's positive supply. This could be a problem. Using the dividers to shift the voltage down adds errors-so the circuit may not be very accurate. There could be a problem if the output voltage goes to low, as the regulator needs a control even lower.If the in voltage is always high enough, one could use the positive version of the standard current sink circuit. The current measuring resistor is at Vin and a P_channel FET (could be PNP if low precision) as a power device, controlled by an OP. The OP needs to work at it's positive supply.

author: danadak
author_rank: Super Contributor
timestamp: March 11, 2017, 08:38:41 pm
content: Take a look at this -http://electronicdesign.com/energy/low-dropout-current-regulator-improves-led-driver-efficiencyhttp://www.edn.com/design/led/4442091/LED-current-regulator-has-low-dropoutOr use LDOhttp://ww1.microchip.com/downloads/en/DeviceDoc/LDOBk.pdfRegards, Dana.

author: Codebird
author_rank: Regular Contributor
timestamp: March 11, 2017, 09:15:13 pm
content: QuoteI use that part quite a lot, and this is certainly correct. The output will never go higher than around supply less one volt, roughly.It works great right down to the negative supply though.

author: mduchalski
author_rank: Newbie
timestamp: March 11, 2017, 09:29:45 pm
content: Thank you all for quick replies. I see there are better options than what I proposed, but since from what I gather it may work I'll probably try it first out of curiosity and maybe share the results.Let me clarify some of the thing you were wondering about. 1. Accuracy is not a major issue, I will use it with a multimeter anyway. 2. The reason I don't want to use some of these LED driver designs is that I want to keep this a 3-terminal device (i.e. no virtual ground). That's because I want to keep a digital signal that's referenced to the same ground point as the limited voltage rail unaffected. It's also why I bothered with a differential amplifier in the first place. 3. I'm sorry for my poor use of words "limiter" and "source". I understand it may have caused confusion, so let me rephrase. What I need is a currentlimiter. I figure that if load can draw desired current at maximum voltage a currentsourcecan provide the two are functionally the same. In this sense currentlimiteris what I need and currentsourceis what I proposed for this task.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 11, 2017, 10:04:25 pm
content: Who needs op-amps?Tim

author: Zero999
author_rank: Super Contributor
timestamp: March 12, 2017, 02:21:38 am
content: You could use the TL072 which works with both inputs at the positive rail.

author: danadak
author_rank: Super Contributor
timestamp: March 12, 2017, 10:37:47 am
content: The TL072 does not have specs for high C loading, they only show a response graphwith 100 pf, and that looks like phase margin starting to exhibit issues. The IR MOSFETshows a typ gate C of 570 pF.You may have to employ some additional comp to stabilize that topology.Regards, Dana.

author: Zero999
author_rank: Super Contributor
timestamp: March 12, 2017, 10:48:52 am
content: Quote from: danadak on March 12, 2017, 10:37:47 amI agree. I didn't perform any AC or transient analysis on the circuit, which may be of limited use, as I question the accuracy of the op-amp model. Having a higher value current sense resistor will help increase the stability, by reducing the capacitance seen by the op-amp, at the expense of greater voltage drop. Another thing to try is a PNP emitter follower to isolate the gate capacitance from the op-amp.I've just run a transient analysis on the circuit. With a 1V step change on the output it did ring but it wasn't full blown oscillation. Adding PNP emitter follower did help.

author: macboy
author_rank: Super Contributor
timestamp: March 12, 2017, 05:30:27 pm
content: I once read an old HP design journal about power supply design that said that an voltage source has zero output impedance, and an ideal current source has infinite output impedance. The output capacitor of a constant-voltage power supply helps it meet the ideal zero output impedance criteria. Think about what the output capacitor does for a current source? Bad things is what. I recently bought an old HP 6186C current source (in need of TLC), and it is designed without output capacitance for a reason. Any output capacitance can provide a current transient much higher than your set point whenever the load impedance drops.