LT3080 Hell - Page 1
Forum section: electronics / beginners

author: DanInvents
author_rank: Contributor
timestamp: September 19, 2024, 11:15:05 am
content: Hi!I have been trying to implement a current sink based on the LT3080 without success (see image). After many hours I was desperate and I cut the trace that goes from the op.amp to the set pin. When I measure the current with my multimeter from SET to GND I get 660mA!. That is 330mA per regulator!Clearly there is something wrong but I don't know what it is. The set current from the regulators should be 10 microamperes.Any help is appreciated!

author: AndersJ
author_rank: Frequent Contributor
timestamp: September 19, 2024, 11:25:48 am
content: The set pin is a constant current output that wants a resistive load.It might not appreciate an external voltage pushing current into it.

author: DanInvents
author_rank: Contributor
timestamp: September 19, 2024, 11:47:56 am
content: LTSpice simulations indicate that my circuit should work. Mr EEVBlog has a series of videos showing how to drive an LT3080 with operational amplifiers.https://www.eevblog.com/projects/usupply/

author: ArdWar
author_rank: Frequent Contributor
timestamp: September 19, 2024, 11:58:12 am
content: You can "overpower" the set pin just fine. It then act as voltage follower.Tho I'm not exactly sure if your circuit is constant current. Hmm

author: magic
author_rank: Super Contributor
timestamp: September 19, 2024, 12:08:10 pm
content: Maybe fake chips or a pinout oopsie?If not then it looks like they are blown; shorting SET to GND should simply give you 0V output. I see that there is a fairly restrictive absolute maximum rating on SET pin voltage, maybe your opamp circuit violated it. But I don't know if that would be enough to completely ruin the chips like here.

author: DanInvents
author_rank: Contributor
timestamp: September 19, 2024, 12:26:58 pm
content: The chips are original and the pinout is correct. I find it strange that they could be blown, these are pretty robust components.

author: DanInvents
author_rank: Contributor
timestamp: September 19, 2024, 12:42:51 pm
content: I have now removed the regulators, and disconnected the SET pin so that it is floating. When I place a new regulator it behaves the same way as the previous ones.

author: MrAl
author_rank: Super Contributor
timestamp: September 19, 2024, 01:18:33 pm
content: Quote from: DanInvents on September 19, 2024, 11:15:05 amHello there,Apparently, the SET pin is clamped to the output with diodes. That means if the SET pin goes too much higher than the output, the current into the SET pin will increase. I think that's about the only way it can draw excessive current with a working chip and everything else connected right.To test for this condition, measure the SET pin voltage relative to OUT. See what voltage it gets clamped at, that will tell you the approximate clamping voltage which you can avoid in the future. Maybe a lower value series resistor would help with this if nothing else works, maybe 1k but you'll have to decide that.That's an interesting chip. They provide a separate pin so you can power up the internals to a voltage level that will better drive the output transistor and thus obtaining a lower dropout voltage from IN to OUT.

author: AnalogTodd
author_rank: Regular Contributor
timestamp: September 19, 2024, 01:26:57 pm
content: I see two heat sinks on your schematic and you are using 5-pin LT3080 parts. Are you using the TO-220 versions of the LT3080 and bolting them to the heat sinks? If you are, then having the heat sinks grounded will short the output of the parts to ground as the tab of the LT3080 is OUT, not ground.

author: DanInvents
author_rank: Contributor
timestamp: September 19, 2024, 01:31:21 pm
content: Yes, I am using the TO220 version and the components are attached to the heat sinks. I used proper insulation to make sure that the output does not short to GND.This device is for a Physics experiment for which I need adjustable current. Considering that I have wasted the whole day with this, I give up. I will just use an external switching power supply.Thank you for all the help. Sometimes things just don't work even when everything else points in the right direction.

author: SuzyC
author_rank: Frequent Contributor
timestamp: September 19, 2024, 02:31:02 pm
content: When everything seems to be setup right and yet doesn't work, you first troubleshooting step should be to carefully check your wiring, make sure you are counting ic pins in the right direction, so look carefully for wire shorts, even a wrong concept of circuit operation.Read the specsheets to make sure you are identifying and connecting pins correctly.Step2: Get a scope or DVM and trace the power input to output and see which point shows unexpected readings.With just a quick glance at your design, your circuit looks to be a current/voltage source rather than a current sink.Why so complicated?Why use the LT3080 for this purpose??With just one transistor, or one power darlington, one LED and two resistors, one variable, you can make a fairly good current sink.Use an op-amp to measure the current through a shunt resistor to control the current through a power pass transistor to regulate current. Google current regulators.The circuit must follow the laws of physics!If it doesn't, you've made some mistake in expected circuit operation, wiring or component values, etc.Fascinating, you don't describe the current range you want to control..nor the voltage range of compliance.Something about your goals in creating a current sink is important to reveal to help anyone possibly understand your circuit.Could you possibly explain how your circuit works to another person?

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 19, 2024, 03:09:24 pm
content: Quote from: SuzyC on September 19, 2024, 02:31:02 pmThat was exactly my thinking. The LT3080’s block diagram clearly shows NPN’s emitter output stage, which can only source current, connected as shown.

author: SuzyC
author_rank: Frequent Contributor
timestamp: September 19, 2024, 03:21:23 pm
content: U404B op-amp section circuit complete SNAFU.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 19, 2024, 06:38:23 pm
content: I think it's just a horrible design. You already have a compensation loop around the LT3080, and as a result it is nothing more then a very expensive BJT. Also consider you can get BJT's that can easily deliver 10A or more, but of course you have to watch out for dissipation. 2SC5200 is a popular BJT these days (Costs EUR 3 for 230V 15A and 150W), although it's Hfe is a bit low to drive it directly from an opamp.LT3080 is a nice IC in itself, but it's horribly overpriced. I just checked a local european store and it's EUR 5.5, while an LM317 costs 40 cents (both 1+).

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 19, 2024, 06:42:16 pm
content: Quote from: Doctorandus_P on September 19, 2024, 06:38:23 pmI am guessing here, but it's probably also the main reason that Dave never finished his nice little power supply. He put it chock full of easy to design around but expensive parts, and then after normal profit margins the thing turned out to have to be very expensive.If you want to learn proper electronics design, then also always keep an eye on your BOM cost. There is also a huge difference between a one-off for internal use and designing a kit that does have some commercial value.

author: magic
author_rank: Super Contributor
timestamp: September 19, 2024, 07:02:08 pm
content: Hmm, does it behave more reasonably if you put a terminator (50Ω or whatever) on the output BNC? Or even an ordinary THT resistor? Basically, any sort of load to meet the "minimum load current" spec (mind that you have two chips there).One possibility I can imagine is that perhaps the voltage follower in LT3080 gets phase reversal when SET goes 0.5V or so below OUT and then the output NPN is turned hard on. IIRC we have seen some bizarre behavior of voltage regulators due to error amplifier phase reversal before.

author: MrAl
author_rank: Super Contributor
timestamp: September 20, 2024, 05:37:22 am
content: Quote from: DanInvents on September 19, 2024, 01:31:21 pmYou do not seem to be reading all of the replies.The SET input is connected to the OUT terminal through DIODES, which means if the SET input voltage is greater than the OUT terminal voltage the diodes will conduct, and possible with a lot of current. Therefore, you need to measure the voltages to see if this is really happening or not.That seems to be the only way that the SET input can draw a lot of current, so check for that first.

author: magic
author_rank: Super Contributor
timestamp: September 20, 2024, 05:43:52 am
content: It certainly is the most likely origin of the excessive SET pin current.As for how such current may be available on the OUT pin when the output is already too high, see my speculation above.These parts are not supposed to work without some minimum output load, so start with fixing that.(The opamp circuit also makes no sense to me, but that's another story and it doesn't explain the bizarre behavior of the regulators.)

author: exe
author_rank: Unknown
timestamp: September 22, 2024, 08:32:53 pm
content: What's the objective? If you need to pull just 10uA, then it can be done with an opamp, resistor and a small fet. With a slow opamp will get away without compensation.If you aiming at pulling a few amps, I suggest this design:https://www.eevblog.com/forum/projects/dynamic-electronic-load-project/msg288313/#msg288313. It's not just proved to be working, but the topic also includes thorough design procedure.

author: MrAl
author_rank: Super Contributor
timestamp: September 22, 2024, 09:56:38 pm
content: Quote from: magic on September 20, 2024, 05:43:52 amIf the output is greater than the SET input then it's not the diodes conducting.However, the data sheet gives no indication that there would be any other mechanism that could cause excessive SET pin current if the chip is still working properly.Another possibility is that the company makes chips with different pinouts.What I have to wonder though, is how can he measure 600ma from the output of a sort of regular op amp. That doesn't make too much sense either.

author: magic
author_rank: Super Contributor
timestamp: September 23, 2024, 05:30:40 am
content: I assumed (maybe incorrectly, but probably not) that the SET pinsources600mA of current so OUT voltage is higher than SET.And this supposedly happens when the pin is shorted to ground by an ammeter, not connected to any opamp.

author: ArdWar
author_rank: Frequent Contributor
timestamp: September 23, 2024, 06:50:55 am
content: Quote from: magic on September 23, 2024, 05:30:40 amThat situation should never happen in normal operation. OUT follows SET, so shorting SET to GND should result in OUT going to zero V. |VOUT- VSET| < 0.3 V shouldn't get violated. Unless OP backfeed OUT, or the whole silicon got locked up due to something else. I'm still not sure what OP meant by "trying to implement a current sink"

author: magic
author_rank: Super Contributor
timestamp: September 23, 2024, 07:04:38 am
content: Passing more than 10mA through SET pin diodes isn't normal operation, and once it happens for any reason, absolute ratings are violated and the chip is "allowed" to do whatever stupid thing it wants, including continuing to pass more than 10mA through SET pin diodesI don't think the regulator was backfed, because there is nothing besides some 10kΩ resistors on the output, if the posted schematic is to be believed.

author: ArdWar
author_rank: Frequent Contributor
timestamp: September 23, 2024, 07:18:05 am
content: Shorting SET is normal, that's how you get true 0V out. I've tried that multiple times (well, using LT3097, and thru ~10R min resistance, but ISETnever went out of spec).The only thing I yet to figure out is the following conditions:Quickly hard-shorting SET when regulator is already powered up instead of letting it ramp upDrive SET with fast slewing signalRelated observation to #2 point is that using LT30xx as voltage follower is just like using composite opamp, including its caveats. Its unity gain bandwith is about 1MHz, so theoretically you should only drive it with outer loop significantly slower than that.

author: MrAl
author_rank: Super Contributor
timestamp: September 23, 2024, 11:01:45 am
content: Quote from: magic on September 23, 2024, 05:30:40 amHi,Yes, that's right. So my only guesses left are that the company makes chips will different pinouts, or the user was following the pinout for a different package type. Different package types often have different pin numberings. This would mean any new chip tried would show the same 'problem'.

author: mtwieg
author_rank: Frequent Contributor
timestamp: September 24, 2024, 11:38:57 am
content: It's unclear how the circuit in the first post is intended to operate. Is it meant to sink a regulated current from the input supply, or source a regulated current into the BNC output?Either way, I can't see it doing either. U404B seems to be creating positive feedback for U401/U403... no surprise if it latches into a state where U404B drives both SET pins very high.As a first step I would put a diode between the SET pins and U404B's output, thus ensuring they will only sink current from the SET pins. But that still won't make this circuit "work".