HV linear power supply with safety features - Page 1
Forum section: electronics / general

author: Enceladus
author_rank: Contributor
timestamp: August 23, 2020, 09:05:38 am
content: Hi everyone. I'm currently designing a 160W 380V linear power supply and I wanted to build in some safety features such as foldback current limiting, short circuit protection as well as ground leakage protection. The schematics and LTSpice models attached. The circuit is a modified one (Figure 11) from the AN-2 application notes from Analog Deviceshttps://www.analog.com/media/en/technical-documentation/application-notes/an2f.pdf.So the simulation shows that the devices works pretty much as expected: in the case of short circuit the triac VD2 opens up and the relay disconnects the load and the power supply from +410V. I also added the temperature protection as the transistor will have to dissipate 15-18W at full power and even if I mount it on the same heatsink with the LT3080 it may not be enough (the U1 does have a thermal protection, although it will not dissipate much power during the normal operation).The thing I'm not sure about is the ground leakage. I found a specialised IC which supposed to trigger a triac in this situation. Although the current transformer (TR1) does not really care if AC or DC current is flowing through it as long as the net current is equal to zero I'm not sure if it fine to use this circuit for DC power supply. Has anyone ever tried to use the ground leakage circuits for DC current?PSI'm also thinking to change NTC to PTC instead and put it instead of R18.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: August 23, 2020, 02:31:55 pm
content: A current transformer will not measure DC. It's a transformer. You could get a LEM Hall probe and do the trick.I assume you have a transformer on the input. Otherwise you can't ground the power supply output.Andy

author: Enceladus
author_rank: Contributor
timestamp: August 24, 2020, 12:11:16 am
content: Yes, the diode bridge connects to the transformer 220V->410V. Thank you, I forgot to mention that.The current transformer here is just to detect a leakage to ground, e.g someone touches the +380V and the chassis (it is important that the circuit connects to the chassis where the bridge rectifier is).

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 24, 2020, 02:56:45 am
content: At such a high power level, some form of preregulation makes sense. That could be a switching stage, a SCR in the rectifier, or a variac driven by a motor, depending on your noise requirements.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: August 24, 2020, 03:33:39 am
content: Quote from: Enceladus on August 24, 2020, 12:11:16 amThe current transformer will only detect AC leakage current.

author: f4eru
author_rank: Super Contributor
timestamp: August 24, 2020, 08:47:06 am
content: yes, this circuit only detects AC leaks.For DC, put a current shunt in your connection from GND to chassis (with shunt diodes over it), and put a detection threshold there

author: Enceladus
author_rank: Contributor
timestamp: August 24, 2020, 10:03:52 am
content: Well, the whole point of this current transformer is to disconnect the power supply from the load ASAP in case someone touches the +380V wire and chassis so it should work in AC only. This is just a safety featureNiHaoMike, you are right, I should put an NTC before or after the diode bridge to suppress the inrush current. Thank you!

author: f4eru
author_rank: Super Contributor
timestamp: August 24, 2020, 01:54:56 pm
content: nope, it won't work. A transformer doesn't detect DC.if it is "just" a safety feature, then you have to make sure it is absolutely reliable.

author: f4eru
author_rank: Super Contributor
timestamp: August 24, 2020, 01:59:03 pm
content: next thing: the LT3080 is max. 36V.It will be overvolted and burn short on the first output short circuit.-> concept has to be reworked.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: August 24, 2020, 05:56:16 pm
content: Quote from: Enceladus on August 23, 2020, 09:05:38 amYour current transformer is wired wrong. You have both wires going through the hole in the transformer. The currents will cancel. Also the transformer will not work with DC.

author: Enceladus
author_rank: Contributor
timestamp: August 25, 2020, 01:19:40 am
content: OK I think I didn't properly explain the purpose of this current transformer.So this circuit is similar to the earth leakage circuit barker.During the normal operation the current flowing in and out the current transformer are equal and cancel each other. If someone touches the +380V pin and the ground (not the ground wire that passed the current transformer, it will not work in this case) e.g chassis, then the net current through the transformer is not 0. The RV4145a IC detects that and triggers the SCR. The SCR opens and activates the relay which in turn disconnects the power supply from the load. After activation the SCR stays on till the whole system is turned off (mains switch). According to IEC 60364 the circuit must detect 4-6mA current leakage and break the circuit within 25ms to be considered safe.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 25, 2020, 03:23:29 am
content: Transformers still dont work at DC. The transformer doesn't care that it is "just" a safety device. A DC current imbalance will not induce a voltage in the sense coil because there is no changing magnetic field associated with it.In reality it might or might not trip because a transient is by definition not DC: maybe if the leakage current rises fast enough and the RCD is sensitive enough it would work. It isn't something you should rely on.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: August 25, 2020, 03:09:07 pm
content: Quote from: Enceladus on August 25, 2020, 01:19:40 amBut it's a DC power supply. A current transformer will not work.

author: Enceladus
author_rank: Contributor
timestamp: August 27, 2020, 12:58:19 am
content: Oh, now I understand what you mean. Yes, I think you are right, I can't rely on the transient to be fast enough to be detected by the transformerAre there other ways to detect an earth leakage of DC current? Maybe I can move the current transformer before the diode bridge and use a power transformer with a central tapAs an alternative I can put a small resistor (e.g 0.05Ohm) between the circuit ground and the earth and detect the voltage drop across it, although lifting up the ground doesn't sound very safe to me (although the resistor is very small), not sure if this is allowed

author: bdunham7
author_rank: Super Contributor
timestamp: August 27, 2020, 03:22:07 am
content: Quote from: Enceladus on August 27, 2020, 12:58:19 amA good solution has already been proposed.https://www.digikey.com/htmldatasheets/production/1043763/0/0/1/fhs-40-p-sp600.html

author: Enceladus
author_rank: Contributor
timestamp: August 30, 2020, 08:21:41 am
content: Thank you for the link!I did some research on Hall-based current transducers and found that they are mostly used for high current sensing (e.g >1A). For example LXSR 6-NPS (https://au.rs-online.com/web/p/current-transducers/2003838/) has a nominal current of 6A (smallest I could find) and has a theoretical sensitivity of ~100mV/A. Since I need to trigger the circuit at 4mA leakage current it can be hard to detect.

author: bdunham7
author_rank: Super Contributor
timestamp: August 30, 2020, 05:21:00 pm
content: Quote from: Enceladus on August 30, 2020, 08:21:41 amYou can multiply the current sensitivity by looping the traces underneath multiple times, symmetrically, just like you would with a current transformer. Since your output current is only 500mA, you could use pretty thin traces and on a 4-layer PCB, probably get at least a 10X or 20X multiplication. I don't know of an example of this working, so I suppose you are experimenting. At 20X multiplication you'd have 2mV/mA senstivity so your 6mA trip would be at 12mV. I don't know if that works.

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 30, 2020, 07:51:26 pm
content: Do it the simple way with a resistor to ground as a shunt, albeit one with a much larger voltage drop than would normally be used for shunts. 100 ohms works nicely because that gives 0.6V at 6mA, one diode drop.