How to improve LT3080 regulation in high voltage application? - Page 1
Forum section: electronics / projects

author: JoergR
author_rank: Contributor
timestamp: May 31, 2025, 01:40:30 pm
content: Hi all,The LT3080 is a phenomenal regulator that can keep output voltages to within mV accuracy - regulating 10V from a 16V current limited source.Being a floating device that does not sense ground, it can also be used as a high voltage regulator. The data sheet (https://www.analog.com/media/en/technical-documentation/data-sheets/lt3080.pdf) offers this approach:The MOSFET M1 is supposed to keep the voltage at the input of the LT3080 about 6V above its output.But implementing this leads to excursions of up to 150mV at 10V target regulation under very mild load steps (200mA). About 300 times as high as with the low voltage input!Also, looking the startup of the circuit, the MOSFET does not provide protection from excessive input voltage to the regulator.One can correct these problems to some extent by increasing the resistor R5 on the gate of the MOSFET to something like 50k.Hence, the source of the problem is the direct current path from the V1 through R5 and D4 to the output.But I'm a bit lost as to why this is and how this in be improved if I'd like to keep the excellent regulation of the LT3080 also in a high voltage application like this.Thanks for any suggestions! The LTspice file is attached if anyone is interested in reproducing this.LT3080_HighVoltage.asc(5.77 kB - downloaded 27 times.)Cheers,Joerg

author: Kleinstein
author_rank: Super Contributor
timestamp: May 31, 2025, 03:58:35 pm
content: One could replace the enhancement type MOSFET and zener diode to create the shift with a depletion mode FET. The problem here many be that the threshold voltage is often quite small and thus only little voltage (like 1 V) for the LT3080.One could imporve on R5, by using a constant current source - so a more constant current and thus easier to add a minimal load to the output.

author: iMo
author_rank: Super Contributor
timestamp: May 31, 2025, 04:18:18 pm
content: .. and mind the TC of the RSET as well (when you are chasing sub mV stability)..

author: JoergR
author_rank: Contributor
timestamp: May 31, 2025, 05:35:11 pm
content: ... hm, that I don't understand, can you pls. explain what you mean?To my understanding, the LT3080 pushes 10micro Amps out of through Rset and regulates the output to 10uA * Rset, so I don't have a choice in the R-Value. As for the parallel capacitor, that doesn't seem to make any difference. 1uF is just as bad as the 47pF in the schematic or no parallel capacitor across Rset. The dominant factor really seems R5.Cheers,Joerg

author: mtwieg
author_rank: Frequent Contributor
timestamp: May 31, 2025, 05:39:39 pm
content: Quote from: JoergR on May 31, 2025, 01:40:30 pmAs you said, the issue is the direct path through R5 and D4. What this effectively does is raise the minimum load current for proper regulation. This will be equal to Ioutmin = (V1-Vout-Vz)/R5 + 1mA, where V1 Vz is the zener voltage of D4 and the 1mA comes from the LT3080 itself. For your simulation, Ioutmin = (60-10-6.2)/10k +1mA = 5.38mA. So if you change the value of R4 to 1.8K, then the regulation will be good again.This is the downside of regulators which don't have a GND connection. All the quiescent current has to go somewhere, and usually if there's no GND connection it has to go to the output. That's how you end up with a significant minimum load current.QuoteI believe what you see here is an artifact of how LTspice determines initial conditions. If you start the sim with all sources at zero in the analysis settings, then the output voltage will not overshoot like that.Quote from: JoergR on May 31, 2025, 05:35:11 pmI think they're just pointing out that the temperature coefficient of resistance for Rset will determine the overall temperature coefficient of Vout.As for the capacitor Cset, this won't really affect line/load regulation, but it might reduce output noise. It will also cause Vout to ramp up slowly on startup (basically a startup).

author: JoergR
author_rank: Contributor
timestamp: May 31, 2025, 09:27:21 pm
content: thanks so much, that is really insightful!First, yes, adding startup to the .tran command fixes the overshoot at the beginning.Second, and more importantly, the minimum load current for the LT3080 from the data sheet is indeed .5mA - so in my initial design, all of the load current was already coming from the path via R5 and there was nothing for the regulator to regulate.In an actual high voltage design, where the output voltage is going to be higher, it's going to be easier to maintain this minimum output current also at zero load by using the trick with the lower Rset, where some of the output current is diverted to Rset and so one effectively has a minimum load. For a wide range of regulation, I'll probably want a constant current minimum load, but I'll leave that for later.I've changed the value of R5 to 250k now, which seems to be the maximum one can do without sacrificing regulation performance - that gives (60-10-6.1)/250k = 0.18mA contribution to the output current. The Rset configuration now draws 1mA at zero load, so the LT3080 needs to provide .82mA which proves to be enough. Peak-to-Peak regulation error is now 1.2mV and that is about the same as I had before from a low voltage source.Thanks again for this insight!Cheers,Joerg

author: JoergR
author_rank: Contributor
timestamp: May 31, 2025, 09:44:17 pm
content: Thanks for the suggestion with the depletion device. That's sounds like a good idea, too.I'll need to find a depletion NMOS and corresponding model for LTspice first to try this out.For the moment, I'll stick with making sure I have sufficient load on the regulator.Cheers,Joerg

author: Kleinstein
author_rank: Super Contributor
timestamp: June 01, 2025, 07:38:17 am
content: For relatively low voltage, low power one could use a JFET as a depletion FET. However there are few (and rather expensive) power JFETs.In principle a JFET (e.g. mmbf4392) and BJT as a kind of darlington like configuration should work for more current.Depl. MOSFETs are something like DN2450 or BSP135. However these 2 may be to high in resistance. At least in the older LTspice version I have, there are no models for depl. MOSFETs included.

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 01, 2025, 12:10:53 pm
content: Could also divert most of the current through D4 to GND using a transistor like this:

author: JoergR
author_rank: Contributor
timestamp: June 02, 2025, 06:58:34 am
content: Quote from: mtwieg on June 01, 2025, 12:10:53 pmOh, that is also very nice! Thanks so much!

author: MisterHeadache
author_rank: Regular Contributor
timestamp: June 20, 2025, 03:28:49 pm
content: What is the purpose of diode D3 (1N4148?) Is there some transient condition that it is protecting the MOSFET against?I'm also struggling to understand how this circuit works. Initially I thought the Zener D4 would set the differential voltage across the LT3080 (Vin-Vout) to the voltage rating of the Zener when a higher voltage was fed into the MOSFET. But judging by JoergR's simulation results, that's clearly not the case because Vin-Vout is about 3.3 volts.

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 20, 2025, 06:10:03 pm
content: Quote from: MisterHeadache on June 20, 2025, 03:28:49 pmIf the output is suddenly shorted, then the change in Vout would couple directly to the Vgs of M1 (especially if C1 is large). If Vout is normally high (like >20V) then this could easily kill M1. D3 is likely to protect against that.QuoteThe voltage across the LT3080 will be Vz - Vgs.

author: MisterHeadache
author_rank: Regular Contributor
timestamp: June 20, 2025, 07:22:37 pm
content: Quote from: mtwieg on June 20, 2025, 06:10:03 pmAh, OK I see it now. So for example, if I wanted the LT3080 to share more of the heat dissipated (and reduce the heat dissipated by the MOSFET by the same amount), I should increase the Zener voltage of D4.New question - what is the minimum V1 at which this circuit will work? At present it is 60V, but if it were reduced to below Vout + Vz would it fail to start?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 21, 2025, 08:25:51 am
content: The circuit should still start with a low input voltage. The output may just be a bit lower. In addtion to the drop out of the LT3080 there is the gate voltage (a bit above the threshold, but a bit less than the zener voltage) of the MOSFETs. Depending on the FET used this would be some 3 volts as total drop out voltage. It could be a bit less (like 1 V resistive drop) with a depletion type FET.