How to replace LT3083 with opamp+pass transistor - Page 1
Forum section: electronics / projects

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 20, 2018, 08:40:07 am
content: Hi!I have started a small PSU design that would at first be using an LT3083 (or dual 3080). But I would like to make precision high side current monitoring. The shunt has to be placed at the output of the power stage and not at the input like in Dave's µSupply. Since the LT3083 can't compensate high side shunt directly, I thought about using an external op-amp and pass transistor that could include the shunt in the feedback loop. I originally imagined a schematic that could compensate the LT3080 but I looks quite complicated. As a input supply I have a Vout+3V pre-regulator. The target output is 0-15V 0-2A. The shunt value is 1ohm. Spec of the output should be typically similar to the TL308x option.I'm working on an SMD design which is space constrained and have no active cooling possibilities.What parts could be recommended for this design?Thanks for your help guys

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2018, 09:45:04 am
content: My advice would be used the LT3081, which includes current monitoring. It's limited to 1.5A, but two in parallel could be used to get 2A.http://www.analog.com/media/en/technical-documentation/data-sheets/3081fc.pdfOne problem is these ICs have a minimum load requirement, but that's easilly fixed with a constant current load.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 20, 2018, 01:21:12 pm
content: Thanks, but the Imon output of the LT3081 doesn't meet my requirements. I would liketo be able to monitor current down to the fraction of µA.

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2018, 03:40:10 pm
content: Quote from: AloyseTech on September 20, 2018, 01:21:12 pmThen you'll need different shunts for different ranges. You won't be able to reliably measure the entire range of 2A, with sub-µA precision/accuracy.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 20, 2018, 04:47:54 pm
content: In a linear regulator there will be quite some power loss, no matter what. One can save a little on the power loss to make it lower heat and thus more compact by using 2 or maybe 3 raw voltages.The more common lab supply circuit uses an extra auxiliary supply (e.g. extra transformer) for the regulator.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 20, 2018, 04:53:27 pm
content: Quote from: Kleinstein on September 20, 2018, 04:47:54 pmI'm not sure I understand the power loss issue. Would it be worse using an external circuit (compared to the LT308x) ?

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 20, 2018, 05:22:49 pm
content: Quote from: Hero999 on September 20, 2018, 03:40:10 pmI already plan to use two different range for measurement. High current will be monitored before the regulator, whereas low current will be monitored at the output.My issue is how can I compensate the low current shunt voltage drop?Here is a quick schematic I made using the LT :

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2018, 06:11:34 pm
content: My advice is don't bother with a voltage regulator IC. Make your own using a transistor and an op-amp.How much voltage are you planning to drop across the low current shunt? Unless it's considerable, say >1mV per μA, the arrangement you've posted is unlikely to work well, due to the mismatch in resistors and op-amp offset voltages.How much voltage headroom is there?I think you should use a relatively high value resistor for the low current shunt, say 1k, and bypass it when it's not being used. Three separate shunts for different ranges would probably be ideal: 1k for under 200μA, 2R for up to 100mA and 100mOhm for up to 2A. There would be no point in disconnecting any of the shunts: keep the 1k and 2R in place when measuring 2A, just connect the 100mR shunt in parallel: the other resistors are so high value, it will make little difference, which could even be calibrated out, if necessary.Some kind of switch would need to be used to select the appropriate shunt: relays have a very low resistance, but are slow, a MOSFET would be ideal, but has a much higher resistance and both P and N channel devices may be required, to make a simple analogue switch, unless there's plenty of voltage headroom.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 20, 2018, 08:35:24 pm
content: Quote from: Hero999 on September 20, 2018, 06:11:34 pmHence my initial question : how exactly to replace the LT by a opamp/transistor combo? I'm not really used to this kind of system so I would like some guidance on how to design this subsystem, choose parts and so on.Quote from: Hero999 on September 20, 2018, 06:11:34 pmI was planning to have a 100mR to 1R resistor for the high range, and a 1R to 10R for the low range. I also planned to use the OPA189 opamp for the low range current sensing : ultra low drift 400nV offset typical, 5nV/V PSRR, 168dB CMRR, up to 36V supply (I have +Vtrack/-5V). And for the matched resistors I planned to use Vishay's ACAS06xx series (0.05% relative, 0.1% absolute).What do you call the voltage headroom? The tracking pre-regulator voltage?I also thought about using mosfet to bypass an higher resistor for dual range, but since I would like to sample at 100kS/s, I thought it might be better to avoid dynamic range switching... Why both a P and N should be combined to bypass a shunt? Do you mean that the N channel must be used to drive the P one actually bypassing the shunt?My acquisition system is 16bit +/-5V.

author: David Hess
author_rank: Super Contributor
timestamp: September 20, 2018, 11:13:17 pm
content: Quote from: Hero999 on September 20, 2018, 06:11:34 pmOr keep the voltage regulator IC, not necessarily the LT3083, and insert it inside the feedback loop of an operational amplifier.Quote from: AloyseTech on September 20, 2018, 01:21:12 pmMonitoring 0 to 2 amps with a precision below 1 microamp is better than 2 million counts. Just the shunt's temperature coefficient of resistance is going to clobber that sort of accuracy and that much resolution is going to require precision operational amplifiers and careful attention to noise; a 1 microamp change across a 1 ohm shunt is only 1 microvolt.I would forget using an instrumentation amplifier and instead use a high side current sense amplifier however what are your requirements for measuring output current at low output voltages? High side current sense amplifiers generally only have a compliance equal to the output voltage which is a problem at low output voltages.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 21, 2018, 07:42:52 am
content: Quote from: David Hess on September 20, 2018, 11:13:17 pmCould you elaborate on this solution please?The OPA189 is not an instrumentation amplifier AFAIK. And I don’t need a 1uA precision on the 2A range.I want to be able to measure the current even at low voltage. I haven’t found an high side current sense amp meeting my requirements so I started simulating my own based on the OPA189 + NPN transistor.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 21, 2018, 08:29:41 am
content: Is this what you mean by inserting the regulator in the feedback loop of an opamp?

author: Zero999
author_rank: Super Contributor
timestamp: September 21, 2018, 09:30:53 am
content: To answer your questions:1) Refer to the link below for more information about the op-amp+pass transistor configuration.https://www.petervis.com/electronics%20guides/voltage-regulator-using-op-amp-and-transistor/voltage-regulator-using-op-amp-and-transistor.html2) Voltage headroom is the drop-out voltage, i.e. what's the minimum difference between the input and output voltages.3) Both P and N channel devices may be needed to bypass the shunt. This is because a good 5V or so between he gate and source, is required to turn a MOSFET on. Using an N-channel device alone, would mean its gate voltage would need to be considerably above the output voltage for it to turn on enough, which might not be possible, when the regulator is set to higher output voltages. Using a P-channel devices alone, would mean its gate voltage would have to go negative, when the regulator is set to output voltages, lower than its gate threshold voltage. You need to build your own analogue switch to bypass the shunt.http://www.analog.com/en/analog-dialogue/articles/switch-and-multiplexer-design-for-hostile-environ.htmlIn your circuit, the op-amp is configured as an instrumentation amplifier.Yes, that's the correct way to put an IC regulator in the feedback loop. It has the advantage of safe operating area, thermal and over-current protection, which you don't get with a plain transistor.

author: AloyseTech
author_rank: Regular Contributor
timestamp: September 21, 2018, 09:54:24 am
content: Thanks, I'll read these papers.Since I have a tracking regulator as an input which is configured to be 3V higher than the output, I guess I have 3V minimum of headroom, right?I thought an instrumentation amplifier has some interconnection between two input buffer op amp and an output opamp? I see more or less why this is could be a instumentation configuration you're right. What exactly is wong about using this kind of configuration for my application, and what would be better? In the schematic I provided the "instrumentation" opamp is not used for the precision current sensing, only for compensating the dropout. But I'm quite sure this is not the best solution and will probably not be using it anywayI have some basis in analog electronic but I miss some experience and tips and tricks so thanks for your help!

author: David Hess
author_rank: Super Contributor
timestamp: September 23, 2018, 01:11:11 am
content: Quote from: AloyseTech on September 21, 2018, 07:42:52 amBelow is the example I commonly use. This sort of circuit can achieve a load regulation so good that it takes a 7 or more digit multimeter to even see it. Some care is needed in frequency compensation if the operational amplifier is high bandwidth which might be desirable for low noise.Essentially the integrated regulator becomes an overload protected power pass element to replace a power transistor. The low noise reference and operational amplifier *reduce* the integrated regulator's output noise within the closed loop bandwidth all the way down to DC so that includes drift.

author: AloyseTech
author_rank: Regular Contributor
timestamp: November 16, 2018, 03:24:10 pm
content: Hi,I've been able to work a little bit on this project again.I added the current limiter option. Is this a good solution ?

author: Kleinstein
author_rank: Super Contributor
timestamp: November 16, 2018, 05:10:10 pm
content: The shown circuit is a rather poor solution. It has quite a few bad points:The voltage control loop (U9B) has no added compensation and thus only relies on the speed of the OP. Chances are it will oscillate, especially with the extra buffer and LT3083 used as a "power transistor" and the extra delay via C1. Chances are U9B will work as a kind of comparator, making it a kind of "controlled" triangle, on-off type regulation.The current loop is also poor, with added gain from the transistor. So some kind of oscillation is nearly guarantied.Because of C1 the current limit would also be very slow.The chose OPs are also rather noisy for the small voltage relevant to the current regulation.Precision wise the differential amplifier at the shunt is also bad - it heavily depends on resistor accuracy.

author: AloyseTech
author_rank: Regular Contributor
timestamp: November 19, 2018, 10:40:58 am
content: Thanks @Kleinstein for your comments. What improvements would you suggest ? Would you happen to know where to find a proven DAC controlled PSU design?

author: prasimix
author_rank: Unknown
timestamp: November 19, 2018, 02:39:32 pm
content: Quote from: AloyseTech on November 19, 2018, 10:40:58 amTake into account thatproven DAC controlled PSU designgoes beyond hardware alone for simple reason that DAC require firmware for control. And proven firmware can easily goes beyond simply set output voltage and/or current. We spent much more time on developing solid software (firmware) then on developing hardware for EEZ H24005, despite our core competence is software development.

author: AloyseTech
author_rank: Regular Contributor
timestamp: November 19, 2018, 02:58:21 pm
content: I'm only looking for the hardware part of the system. I don't want the software to be responsible for regulation, only for settings voltage and current. I understand that a lot of work is needed for the software if the regulation is not done in hardware.