Constant current regulator? - Page 1
Forum section: electronics / beginners

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 24, 2013, 02:09:56 am
content: I'm looking for a true CC regulator that can supply 100mA regardless of the input voltage. (In this case, Vin between 0.5 V ~ 5 V.)I've tried aNSI45035(at max current output, which turned out to be slightly below 60 mA), but as the voltage went down, so did the output current - even though my power source was fully capable of supplying more current.Does anyone have any other suggestions?

author: c4757p
author_rank: Super Contributor
timestamp: May 24, 2013, 02:22:12 am
content: What kind of accuracy do you need? Voltage reference + current shunt + MOSFET + precision op amp (like Dave's electronic load) can be quite good, all depending on the current shunt.LM317 can do a decent job as well, with a bit less accuracy.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 24, 2013, 02:30:58 am
content: Do you have another power rail available? One other than the 0.5V to 5.0V rail? If you don't then it will be a real trick to do what you want.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 24, 2013, 11:21:10 am
content: @c4757p, it doesn't need to be all that accurate, but I would like to minimize the amount of energy loss. (See below for more info.)@TerminalJack505, no, no other rail available.My little project is to see if I can power a tiny weather station data logger with a supercap or two instead of batteries overnight. In order to prevent the supercap from losing too much energy too quickly, I want to clamp the current to a max of 100 mA. I'm using a booster to increase the volts to the active circuit, but what I've found is that even though the supercap has the ability to continue to supply the current required, the NSI45035 (which is in front of the booster) caused the current to sag when the input voltage dropped. (Putting the current limiter after the booster didn't help.) Since every joule of energy is precious, I'm trying to minimize power losses, which means trying not to use resistors that just end up wasting energy in the form of heat loss. I'm also trying to keep the circuit's power usage as small as possible, so I'd rather not start adding op amps and such if there's a more efficient way to do this.

author: lgbeno
author_rank: Frequent Contributor
timestamp: May 24, 2013, 12:01:40 pm
content: These must be some large super caps if the load is in the 100ma range and over night...I think that op amp plus transistor is probably one of the most simple solutions but not the most efficient since you will be dumping the power in the pass element in order to regulate current. You would otherwise need some switch mode circuit to get better efficiency.

author: lgbeno
author_rank: Frequent Contributor
timestamp: May 24, 2013, 12:05:05 pm
content: Compute '200farad*3.5v/200ma to hours' with the Wolfram|AlphaIt's theoretically 1 hour.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 24, 2013, 03:40:27 pm
content: Since putting the limiter after the boost converter didn't work my guess is that the boost converter is the problem.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 24, 2013, 04:20:33 pm
content: @lgbeno - Actually, two 350F supercaps, 2.7 V @ 100 mA. I calculated that theyshouldlast ~8 hours. (Perhaps longer if I can squeeze the system down to drawing only 60~70 mA.)@TerminalJack505 - I don't think the booster is the problem. I tested the NSI45035 with my bench power unit. As I dropped the voltage down, it's output current dropped - even though my bench power unit was set to supply up to 200mA. I confirmed this same behavior with the supercaps themselves, letting them run the circuit over several hours. My meter showed the same drop in output current. When they were at ~1 V, I bypassed the NSI45035 and fed the booster directly. The supplied current went well above 100 mA instantly, so I know the supercaps can still deliver that amount of current, and the booster can still work just fine at that low voltage.I think what I need is some chip or circuit that can regulate the voltage to maintain the current output. I've been searching for such a beast, but so far I haven't found any candidates yet.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 24, 2013, 04:28:20 pm
content: You might try using a constant currentsinktype of circuit then. Right now you are using a constant current source.With a constant current sink you can use the booster's higher voltage to run the components rather than the potentially low input voltage. It's hard to make a lot of ICs work at just 0.5V. You'll have better luck finding components that work at 1.8V or whatever you are boosting the input voltage to.

author: IanB
author_rank: Super Contributor
timestamp: May 24, 2013, 05:51:41 pm
content: Quote from: SoftwareSamurai on May 24, 2013, 11:21:10 amI think you have a bit of confusion about what you are trying to achieve.Let's try an analogy.You have a car that you want to cruise at 60 mph.However, to prevent the fuel tank emptying too quickly, you can put a limit on the rate at which the engine can draw fuel from the tank. This way, you can improve the fuel economy, and increase the range of the car. Obviously that greedy engine is sucking too much fuel so you have to limit it. Then you can achieve 200 mpg at 60 mph in freeway driving. This is so obvious it is a wonder the car manufacturers have not already done it?

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 25, 2013, 01:55:24 am
content: Quote from: TerminalJack505 on May 24, 2013, 04:28:20 pmYep, tried that. The NSI45035's current still sagged when the input voltage went down.

author: IanB
author_rank: Super Contributor
timestamp: May 25, 2013, 02:08:17 am
content: I think your brain is fried. You should stop smoking whatever you are smoking.Draw a block diagram of your system. You have something like:(Power source) -> (Thing) -> (Thing) -> (Data logger)Write an efficiency against each Thing you place between the power source and the data logger. All Things have a power conversion efficiency < 100%. The overall power conversion efficiency is the product of each efficiency, so that:(overall efficiency) = (Thing 1 efficiency) x (Thing 2 efficiency) x (...)You will see the best result is obtained with a minimum number of Things between your power source and your data logger.In addition to efficiency, every Thing has a requirement on the minimum input voltage it can work with (including the data logger). So what you need is a single Thing that can raise (or lower) the power source voltage to the data logger's most efficient supply voltage at maximum efficiency.I have no idea where you get this constant current idea from. Mostly I think you are crazy, but maybe you are just misguided...

author: David_AVD
author_rank: Super Contributor
timestamp: May 25, 2013, 02:53:42 am
content: Yeah, there's some serious issues with the OP's theory for sure. How about you start over?What voltage is available to power the end device?What voltage does the end device need to run?What's the typical operating current of the end device?

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 26, 2013, 01:54:35 am
content: Okay, looks like I need to greatly simplify things so everyone can understand what I'm trying to do. Forget about the tiny weather station data logger for now and pretend I've only got three basic components: A supercap, an LED, and a voltage booster.For those that need specific components in order to understand this example circuit:1) A Maxwell 350F 2.7V supercap.2) An NCP1402 @3.3V voltage booster (with necessary external components).3) A Luxeon 3535L LED (specifically, MXA8-PW65-0000).The goal is simple: Power the LED from the supercap for as long as possible, keeping the amount of lumens constant until the supercap's voltage drops below the NCP1402's low-cutoff limit.LEDs are driven by current (as long as the voltage is above the LED's threshold). More current equals more lumens and vice versa. The supercap is capable of supplying tons of amps. The NCP1402 is capable of supplying 200mA. (Actually, it can peak up to 400mA.) Now, let's say I've decided that the LED is bright enough for me when I feed it just 100mA.So given this example circuit, I'll ask again:Does anyone know of a constant current regulator (that can be inserted into this circuit) that will:1) limit the current to 100mA, and2) provide that 100mA constant currentregardlessof the voltage level?

author: c4757p
author_rank: Super Contributor
timestamp: May 26, 2013, 02:13:05 am
content: Piece of cake to turn any boost converter into a constant current driver. See the attached screenshot. (I used different voltages and whatnot, just because I didn't have a part in my library that works at 2.7V. You can adjust to your liking. It still gets the message across, I think.)D2 and D5 are the LEDs (plural in this case, just to give enough forward voltage for this particular simulation)R1 measures the current flowing as a voltage drop.R3 and D4 add a diode drop to this, so that the entire feedback voltage (1.25V) doesn't have to be dropped in the resistor. This would be a lot of wasted power.D3 sets a maximum voltage so it doesn't reach for the stars if the LED is disconnected.You should have a resistor at Ipk for this particular chip, but I didn't bother because it's just a sim.You can trim the set current by trimming R3 (diode drop) instead of R1 (sense resistor).Edit: Seems the same exact setup should work just fine with NCP1402. Just adjust the voltages.Second edit: Actually, 100mA at 1.25V or 0.9V probably wouldn't be too much wasted power, so you can omit the diode. You still need D3 and R4, though.

author: IanB
author_rank: Super Contributor
timestamp: May 26, 2013, 06:00:47 am
content: Quote from: SoftwareSamurai on May 26, 2013, 01:54:35 amThe answer to this question as indicated by c4757p is to use a single component to provide both the voltage boost and the current regulation in the same circuit. You need a current-regulated boost converter that operates as efficiently as possible. This is the kind of circuit used in every LED flashlight that runs off one or two AA cells.There is of course a minimum input voltage below which the boost converter will no longer function. Even when it does function, the conversion efficiency will typically drop as the input voltage falls. You may want to design so your supercap never falls below about 1 V or something.

author: IanB
author_rank: Super Contributor
timestamp: May 26, 2013, 06:07:20 am
content: Quote from: c4757p on May 26, 2013, 02:13:05 amI'm not sure why you think voltage dropped in a diode wastes less power than voltage dropped in a resistor? The power formula is voltage difference times current flowing. If a diode and a resistor are each dropping 0.7 V at 100 mA they are both dissipating exactly the same 70 mW.

author: c4757p
author_rank: Super Contributor
timestamp: May 26, 2013, 06:15:34 am
content: Because the diode isn't carrying the sense current. It's just carrying an additional mA or so into the sense resistor, sitting 0.6-0.7V above it and reducing the actual sense voltage that is required. It was a sort of a "poor man's current sense amplifier", additive rather than multiplicative.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 26, 2013, 03:01:33 pm
content: Quote from: c4757p on May 26, 2013, 02:13:05 amAh! I was thinking along the lines of anexternalcurrent regulator. Didn't think of using the boosterasa constant current driver.Thanks c4757p. I'll have to breadboard this and test it out.

author: c4757p
author_rank: Super Contributor
timestamp: May 26, 2013, 03:17:21 pm
content: Again, don't forget the Zener diode to limit the voltage output, unless you've calculated the maximum open-loop output of the regulator and your circuit can handle it. It'll climb as high as it can without a loop. I wouldn't be surprised by 100V. And put it in the feedback loop, not as an actual load, unless you really want it to take 100mA.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: May 26, 2013, 03:57:18 pm
content: @ c4757p: Point of clarification: R2 doesn't make sense to me as diagrammed. It appears to be shorted via the direct connection, yes?

author: c4757p
author_rank: Super Contributor
timestamp: May 26, 2013, 04:11:22 pm
content: Oops! That was me being clumsy drawing the connections. It should go straight to the DRC pin and nothing else; the line northbound from Ipk just touched it accidentally.R2 is particular to MC34063, though, and that chip doesn't work under 3V. I just used it because it was all I had in my library. (I'm sure there's some LT part in the built-in library that will work, but I'm not familiar with their part numbers.)