Using a Zener + resistor to guarantee a minium load current for an isolated DCDC - Page 1
Forum section: electronics / projects

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 29, 2015, 11:53:49 pm
content: As many isolated DC/DCs need like 10% load current to deliver the specified output voltage, I'm trying to figure out what would be the most elegant way to guarantee this load current without just wasting the current and thus reducing the possible output current. I wonder if it would be a good idea to use a Zener with the specified output voltage (or a bit below that) and connect it via a resistor to the DC/DC's output. Like: output - resistor - Zener - ground.So when the voltage gets too high because there is no sufficient load current, current will flow into the Zener and thus the output will stabilize since the current flow will improve output voltage regulation.I would assume that (given that the Zener's breakdown voltage is exactly the desired output voltage or slightly less), the maximum current flowing into the Zener will be <= ~10% of the (maximum) load current. So e.g. for 400mA maximum output current, the current through the Zener should not exceed 40mA. Well, at least not permanently. So I guess the resistor could be small. Like 20Ohm for a 5V output.I simulated this in LTSpice and it seems to work as expected, so is this a valid approach or did I overlook some potential issue?Are there maybe better approaches to guarantee a somewhat stable output without permanently sacrificing 10% of the maximum output current?

author: DanielS
author_rank: Frequent Contributor
timestamp: June 30, 2015, 03:06:39 am
content: The fancy way to do it would be a TL431 or equivalent (programmable reference/zener) with a PNP or other transistor to boost drain current if necessary. This is what they call a shunt regulator.You set the reference voltage to your maximum allowed output voltage so the shunt only draws significant current when the output is above normal, which usually only happens when the output is under-loaded. Once you put external load on it, the voltage drops and most of the shunt regulator current stops.I used that circuit to get around PSU overshoots earlier this year until I could investigate the circuit and come up with a proper fix.

author: daqq
author_rank: Super Contributor
timestamp: June 30, 2015, 06:14:30 am
content: I had to deal with small DC DC converters like this mostly - 150mA down to 66mA on the output and such. I just use an LED with a series resistor - it works, and uses the required power as well as doing something useful-ish - indicating that the part of the circuit is alive.

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 08:08:50 am
content: Quote from: DanielS on June 30, 2015, 03:06:39 amI used a TL431 as replacement for a Zener before , but only to get a more precise voltage limit. Now for this case, I don't get care so much if I get 5.0 or 5.5 Volt on the output. It's just that I don't want 7V or 10V on the (more or less) unloaded DC/DC. So I wonder if this additional effort is worth it.Quote from: daqq on June 30, 2015, 06:14:30 amYeah, but for an USB Isolator, I'm more in the 400mA output current range. So I'd need to burn 40mA which feels wrong and reduces the current capability quite a bit.

author: DanielS
author_rank: Frequent Contributor
timestamp: June 30, 2015, 02:24:09 pm
content: If you use power adapters with class V-VI efficiency rating, these often have no minimum load rating since they cannot afford to waste output power in their standby power budget.

author: Ian.M
author_rank: Super Contributor
timestamp: June 30, 2015, 04:11:23 pm
content: A TL431 can easily switch 40mA. The switching point is very well defined and if you get a TL431A, is 1% accurate, much better than a Zener. If you set it up in the shuntt regulator circuit to switch at lets say 5.3V (just above the upper limit of the USB spec), with an extra 75R resistor to limit the current it can sink in series with its cathode it will only sink the 40mA if the rail is higher than normal, with approx 60% of the dissipation in the resistor.OTOH it would probably be better and certainly more energy efficient to add a 5V LDO regulator on the DC-DC converter output. If the converter output dips below nominal the final output will droop with it, but if it rises, the final output will be held steady at 5V.

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 04:54:14 pm
content: In the meantime, I decided anyway to use a TLE431 or to be more exact, a TS431BCX or TS431ACX. I already updated the schematics, did the layout and all. For the moment I set the TLE431 to 5V (R1=R2=22k) and used a 22Ohm series resistor. I still have to think about both, the value for the series resistor and the voltage level. For the moment, my idea was like: at 5.5V, and with the TLE431 set to 5V, 0.5V are dropped on the 20Ohm resistor. 0.5Volt/22Ohm = 0.023A = 23mA. At 6V it would be 45mA, so my guess it that I can never really reach or exceed 6V. Well, and there's also a LED which always draws 5mA at 5V, so there's always a bit of load.Problem with a 5V LDO is that the DC/DC is an isolated 5V to 5V converter, so I can't really guarantee that the dropout voltage is high enough for the LDO to work correctly. Might still work though, but feels like a hack. Besides, very low dropout LDOs are usually a little picky about their input voltage. E.g. a 3.3V TLV1117LV has an maximum input rating of 6V while a DC/DC without load can reach twice its output voltage (e.g. 10V instead of 5V). Besides, I don't really get why adding an LDO after the DCDC would be more efficient. It's the same principle after all: it drains current if the voltage gets too high. More or less the only difference is that the main current is not going through the TLE431 while it would go through an LDO after the DC/DC. Which means some additional losses due to the internal resistance.

author: Ian.M
author_rank: Super Contributor
timestamp: June 30, 2015, 05:19:42 pm
content: If 5.5V is your upper limit, set the TL431 REF resistor ratio to get exactly that voltage. The top end of the divider should go to the rail, not the TL431 cathode. The extra resistor goes between the cathode and the rail and should be chosen to drop Vlimit-2V at the desired current. The result is a sharp switching action at the upper limit and negligible load at lower voltages

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 06:11:56 pm
content: Quote from: Ian.M on June 30, 2015, 05:19:42 pmOk, what about this?2.5V*(1+22k/20k) = 5.25V3.25V/40mA = 81.25OhmWell this leaves me with the 75Ohm you suggested earlier.In my LTspice simulation with a LTM8046 as DC/DC dummy, this results in a ~50mA peak for one millisecond where the voltage shortly reaches ~5.5V at the begin of the current peak. After this 1ms, current drops to like 2mA and voltage is stable at ~5.3V. I might need to do some experiment with the real DC/DC, but this looks good enough for me.Thanks for the hints!

author: Ian.M
author_rank: Super Contributor
timestamp: June 30, 2015, 06:31:21 pm
content: Be aware that this may not work when powered by a laptop as it certainly violates the USB suspend current specs and the controller may detect a power error if it ever tries to suspend the device.. Most desktops don't monitor the load on each port vary closely, so it will probably be OK there, but it is wasting approx 1/4W when unloaded.

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 06:52:15 pm
content: This is the secondary/slave side of an USB isolator. I somewhat doubt that suspending the device will work correctly with this configuration anyway.Besides, honestly I don't expect the full 40mA to be needed. The 10% (40mA) number is usually specified to ensure the line regulation specification to be fulfilled. E.g. you have +/-1.2% line regulation only guaranteed if the load current is at least 40mA. I would suspect that I need much less than 40mA to stay below the 5.3V where the shunt regulator kicks in.Anyway, there aren't too many alternatives, are there? The shunt regulator approach at least doesn't waste current when the voltage is below 5.25V.And even if there should be issues with suspend mode, I guess this is still much better than having a somewhat unpredictable Vbus voltage for low or no loads like e.g. in this approach:https://github.com/watterott/USB-Isolator/tree/master/pcb

author: DanielS
author_rank: Frequent Contributor
timestamp: June 30, 2015, 06:57:34 pm
content: Quote from: AcHmed99 on June 30, 2015, 05:04:27 pmThe shunt is only there to provide the DC-DC converter with a minimum load to prevent the output from floating to an excessively high voltage if the converter requires a minimum load. Once you connect a load, the output voltage gets dragged down and the shunt regulator current drops. When the load is large enough, the output voltage drops below the shunt's threshold and the shunt becomes effectively disconnected.The only time where the shunt is drawing power is when the load is below the DC-DC converter's minimum requirement. Beyond that, the shunt regulator is almost 100% efficient since it is effectively turned off.

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 07:03:53 pm
content: Quote from: DanielS on June 30, 2015, 06:57:34 pmOr to be more exact: the shunt current is only flowing when the output voltage is over the shunt regulator's voltage threshold. Which again is usually caused by too little load current.That was the idea from the beginning. Actually the initial approach with the Zener was exactly the same idea. It's just that the shunt regulator has better accuracy and you can set any threshold voltage you like.

author: tszaboo
author_rank: Super Contributor
timestamp: June 30, 2015, 07:26:11 pm
content: Quote from: 0xdeadbeef on June 30, 2015, 07:03:53 pmYes, it works. I did this with those 1W Tracos, they dont like light load. I had a Ti DC-DC where the output voltage was something like 43V when there was no load (only capacitive) instead 30.If it really bothers you, they make it also with regulated output, or better yet, just put a 78Lxx on the output.

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: June 30, 2015, 08:53:41 pm
content: Quote from: AcHmed99 on June 30, 2015, 07:22:13 pmLook at the circuit I posted somewhere above. The load (which is not visible there) would be connected directly to Vbus2, so there is only current through the 75Ohm resistor (and the shunt regulator) if the voltage is above ~5.3V. However this current will improve the load regulation of the DC/DC, it will come closer to 5V which again reduces the current through the resistor + shunt regulator.If the load current is > 40mA, the DC/DC's regulator will meet its line regulation spec, the voltage will not raise above 5.3V and there will be only a neglible current going through the shunt regulator and voltage divider.

author: DanielS
author_rank: Frequent Contributor
timestamp: July 01, 2015, 12:47:07 am
content: Quote from: 0xdeadbeef on June 30, 2015, 07:03:53 pmAnd those are exactly the two main reasons why I suggested the 431 or equivalent circuit - OP did not initially mention how tight his output tolerance was nor how much current was in question here. A 431 can be made to fit just about any possible idle-loading requirement.