Author Topic: PSU Protection with Output Diodes  (Read 2165 times)

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Offline dannybeckettTopic starter

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PSU Protection with Output Diodes
« on: February 28, 2022, 12:47:24 pm »
Hi all,

I recently blew a bench PSU by overloading the inputs which got me thinking. Could you attach large diodes to each output terminal to ensure you don't accidentally overload the thing? (I know the output voltage would be 0.7V less than expected)
 

Offline Gyro

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Re: PSU Protection with Output Diodes
« Reply #1 on: February 28, 2022, 01:05:34 pm »
It would work, depending on the degree of voltage accuracy that you need... The diode forward voltage will vary with the amount of current draw, from about 0.7V (as you say) up to about 1.1V at full rated current (typical datasheet value).
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Online Ian.M

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Re: PSU Protection with Output Diodes
« Reply #2 on: February 28, 2022, 01:23:55 pm »
I recently blew a bench PSU by overloading the inputs which got me thinking. Could you attach large diodes to each output terminal to ensure you don't accidentally overload the thing? (I know the output voltage would be 0.7V less than expected)
That is almost certainly *NOT* why you blew your bench PSU.   A properly designed bench PSU has effective overload protection, and as long as you don't over-voltage its supply input, wont draw more power than it can handle.   However there are crappy bargain basement PSU designs from the far east pretending to be bench PSUs that don't have effective overload protection . . .

Common causes of loads blowing a bench PSU are reverse current due to connection of the output to a 'stiff' higher voltage source, or power down (loss of supply or switch off) while charging a battery or with very high load capacitance, or voltage polarity reversal due to power down or current limit trip while in a series configuration under load, or due to incorrect connection of a battery. 

A series diode can only protect against reverse current, and as you note degrades regulation by the diode's load current dependent Vf drop.  To protect against polarity reversal you need an anti-parallel shunt diode, cathode positive, directly across the output terminals,  with a current rating suitable for the  max. output current of the highest rated supply in the series 'stack'.  If a battery is involved, you need a fuse and the diode needs to be *much* *much* beefier, so it can clear the fuse before the diode junction fails.  If you also have a series diode, it also needs to have as high a current rating.

Well designed leading brand true laboratory PSUS may have internal reverse current protection and an internal output fuse and polarity reversal protection diode, but it may well not be beefy enough to protect against reverse connection of a high current battery pack.  If in doubt, check the schematic, and if it isn't clear to you, ask here for help reading it.

If charging batteries, I strongly recommend using a purpose designed battery charger circuit, (which may be powered from your bench PSU) rather than simply relying on your PSU's CC mode to limit the charging current and CV mode to terminate the charge.
« Last Edit: February 28, 2022, 01:28:06 pm by Ian.M »
 

Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #3 on: February 28, 2022, 02:54:17 pm »
Thanks guys.

No I wasn't charging batteries, I overloaded the outputs* ;D

I temporarily exposed it to voltager higher than it was happy to see. Anyway, my thoughts were to add hefty series output diodes as a belt as a belt & braces fix initially, then think about creating some form of isolated sense feedback circuit so the supply can regulate post-diode without exposing it to danger.

Good shouts about the voltage drop vs current draw and anti-parallel / fuse situation to protect against reverse polarity. There might be something clever I can do with power MOSFETs in lieu of the diodes.

*edited
« Last Edit: March 01, 2022, 09:11:58 am by dannybeckett »
 

Online Ian.M

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Re: PSU Protection with Output Diodes
« Reply #4 on: February 28, 2022, 06:32:31 pm »
That wont help if you overvoltage the input again.  For protection against overvoltage you need a circuit that disconnects the input very quickly.  One possibility is a crowbar circuit, with a fast fuse 'upstream' of it.  e.g.  the TL431A adjustable shunt regulator has a crowbar circuit in T.I's datasheet, figure 22, (page 27).   If you want the crowbar to be reusable, the TRIAC has to be beefy enough the clear the fuse without damaging itself.  OnSemi had a nice discussion of the considerations for selecting a crowbar SCR or TRIAC starting page 71 of 'Thyristor Theory and Design Considerations Handbook' HBD855/D.
« Last Edit: February 28, 2022, 06:36:49 pm by Ian.M »
 
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Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #5 on: February 28, 2022, 06:37:11 pm »
How come the diode in the attached image won't protect agasint overvolt?
 

Online Ian.M

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Re: PSU Protection with Output Diodes
« Reply #6 on: February 28, 2022, 06:40:11 pm »
How can it?  Apply too much voltage to the regulator *INPUT* and the regulator's pass transistor will fail even if the only load is the regulator's feedback divider. 
 

Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #7 on: February 28, 2022, 06:43:06 pm »
Apologies - I wrongly referred to the PSU output terminals as input (my mind was following the fault current).

Thank you for the triac theory handbook link though - very useful
« Last Edit: February 28, 2022, 06:44:58 pm by dannybeckett »
 

Offline xavier60

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Re: PSU Protection with Output Diodes
« Reply #8 on: March 01, 2022, 06:05:02 am »
Thanks guys.

No I wasn't charging batteries, I overloaded the input ;D

I temporarily exposed it to voltager higher than it was happy to see. Anyway, my thoughts were to add hefty series output diodes as a belt as a belt & braces fix initially, then think about creating some form of isolated sense feedback circuit so the supply can regulate post-diode without exposing it to danger.

Good shouts about the voltage drop vs current draw and anti-parallel / fuse situation to protect against reverse polarity. There might be something clever I can do with power MOSFETs in lieu of the diodes.
What design is it? Most bench PSU's have a diode across the pass transistor for protection from applied higher voltage with the same polarity, all within limits.
Although just about all bench PSUs have anti-parallel on the output, current sourcing deigns like the Harrison topology inherently tolerate brief applied reverse polarity. Assuming that the anti-parallel has been disconnected.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #9 on: March 01, 2022, 09:52:23 am »
OK here's the background -

I was bench testing a small UPS (HP T750 G2) and simulating the internal battery pack with my bench PS (set to ~26V in lieu of 2x series connected 12V lead acids). I connected scope gnd to "battery" -ve, and when I applied power I blew a small mains-connected bridge rect in the middle of the PCB. The fault I created is the same as the one described in this video:



Anyway I fixed the UPS but my bench PS suffered damage. Initially, the PS would turn on and both channels operate, however the chan which was pretending to be a battery developed a weird fault - when left floating it would operate correctly (CV / CC modes worked fine). But if you tied either +ve or -ve sides of that channel to mains ground, the control for that chan would freeze and I'd get no output until I power cycled the PS.

Eventually, the PS died completely, and now when I turn it on all the channels report "0.00" and the front panel is totally unresponsive. No beeps or relay clicks. It is a Korad KA3305D.

I think I did two things wrong in this setup. The first is recreating the BR diode short, the second is trying to use my PS in lieu of a battery (not sure if this is actually a problem). I'm not certain what killed my PS, but I think it now has a fault with the digital electronics rather than any of the analogue stuff. Maybe the PS output temporarily saw a voltage too far away from ground and broke something on the digital board (I did not have the output gnd referenced at the time).

I haven't investigated the fault properly yet however there are schematics for a Korad KA3005P (also attached):

https://www.eevblog.com/forum/repair/korad-ka3005p-power-supply-calibration/msg1109803/#msg1109803

If you assume some similarities, then there are indeed diodes across the pass transistors - I will confirm this when I can.

I'd love to know what caused the fault, fix the PS and put some protection in place to prevent it from happening again.
 

Offline xavier60

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Re: PSU Protection with Output Diodes
« Reply #10 on: March 01, 2022, 10:41:17 am »
Because a mains-connected bridge rectifier is involved, we need to check to see if the battery circuitry is tied to the live side.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #11 on: March 01, 2022, 10:56:18 am »
Indeed, pretty sure the charging circuit is fed from a mains-connected BR with no galvanic isolation. I will test and make certain
 

Offline macboy

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Re: PSU Protection with Output Diodes
« Reply #12 on: March 02, 2022, 03:33:39 pm »
I wouldn't do this.

Some power supplies have an overvoltage protect (OVP) setting which will disable the output if the output goes too high. Better ones have a "crowbar" circuit on the output which will both disable the output and actively short circuit the output in order to protect both the power supply and the load.

Instead of diodes on the output, build a power zener diode, and put it across the output. If the voltage exceeds the safe value (I'd suggest several volts higher than the maximum output of the supply), it will conduct and attempt to limit the voltage to that value, no higher. A power Zener consists of a Zener diode cathode connected to (+) in series with a resistor (e.g. 10 k) connected to (-). The midpoint will be held at zero V until the Zener conducts, when it will rise linearly with the voltage. This is simply connected to gate of a large N-MOSFET or base of a darlington NPN, which is connected across +/-. When the gate/base voltage rises, the transistor conducts, shunting current and preventing further rise. Effectively a simple shunt regulator. The Zener diode itself is chosen based on the desired maximum voltage and the transistor used.

I've attached an example which you can simulate in LTSpice. The power supply sweeps from 20 to 40 V, and is limited to about 32 V by the power Zener.  (Note this is not intended for steady state operation; in this example the Q1 will dissipate about 200 W when the power supply hits 40 V. Also Q1 is an unsuitable device chosen randomly from LTSPice's built in library).
« Last Edit: March 02, 2022, 03:37:19 pm by macboy »
 

Offline dannybeckettTopic starter

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Re: PSU Protection with Output Diodes
« Reply #13 on: March 02, 2022, 07:20:22 pm »
Fantastic, thank you for the thoughts. I found a good article on ESP about the very circuit you've shared:

https://www.sound-au.com/appnotes/an007.htm

I work closely with very powerful (130kV MW) pulsed power supplies and we use huge laser-fired crowbars as trip protection. Another option would be to employ a similar tactic using a relatively cheap OTS thyristor:

https://uk.farnell.com/vishay/vs-50ria20/thyristor-50a-200v-to-65/dp/9104712

I'd enjoy designing and testing a universal power supply protection module to share the schematics etc, it would need to have a programmable max voltage setpoint and respond very quickly to effectively protect. Both power zener and thyristor crowbar options are intriguing!
 

Offline Sal Ammoniac

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Re: PSU Protection with Output Diodes
« Reply #14 on: May 27, 2022, 03:45:54 am »
If you’re working a lot with that kind of application, you might want to consider getting a four quadrant power supply (also known as a bipolar supply).
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