Author Topic: Protection for Electronic Load Circuit  (Read 537 times)

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Offline Jwillis

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Protection for Electronic Load Circuit
« on: February 12, 2020, 05:30:24 am »
I got this working well with the new Mosfets I've been waiting for. Originally it had IRFP260NPbF but there was no definition for DC SOA so it was  kind of a crapshoot to determine what maximum current I could pass through them in a parallel configuration. Turns out its ball park 2.5 amps at around 20V but thats the best guess based on failure. The new ones I'm using are the FDL100N50F  https://www.onsemi.com/pub/Collateral/FDL100N50F-D.pdf Can handle much more.

The problem is not really a problem but more a possibility. With the other mosfets ,when one bank goes all the other ones crater to because they have to take more load. I want to add a circuit to protect each driver bank just in case one fails during a high load situation. I thought about fuses but I'm not sure if they would work fast enough. Each bank should be able to handle at least 7 amps at 14 - 15 volts and  at the most 3 amps at 40V . Since the dynamics change ,fuses just don't seem practical. I'm not sure which way I should go on this challenge.
 

Offline GerryR

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Re: Protection for Electronic Load Circuit
« Reply #1 on: February 12, 2020, 12:32:27 pm »
As a first look, I think I would monitor / sum the the three sense resistor voltages for each bank, compare the two banks, and for a defined difference between banks shut the power down.  You could do it on a transistor-to-transistor basis, but with more circuitry.  Just my first thoughts.  :-//
Still learning; good judgment comes from experience, which comes from bad judgment!!
 
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Online RoGeorge

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Re: Protection for Electronic Load Circuit
« Reply #2 on: February 12, 2020, 12:48:15 pm »
Monitor the temperature of MOSFETs, or the output voltage of each of the opamps.
 
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Offline Vovk_Z

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Re: Protection for Electronic Load Circuit
« Reply #3 on: February 12, 2020, 12:49:52 pm »
With the other mosfets ,when one bank goes all the other ones crater to because they have to take more load. I want to add a circuit to protect each driver bank just in case one fails during a high load situation. I thought about fuses but I'm not sure if they would work fast enough.
1) Such situation could be only in Constant Power mode, if you have one. If we speak about only Constant Current mode then others don't get more current.
2) When Mosfet dies it with 90% possibility will die close but not open, so any electronic protection won't help. That means you need a fuse (the simplest way) and it will be OK in 99% situations as for me. Only if you want go another way you can consider some electronic key, but the truth is that first dies power mosfet - then current rises - and after that will you protection work. It means that we can go back to way with a only protective fuse. So you should consider if you really have a need for sophisticated protection.
3) One FDL10050F case can dissipate 100V and 10 A (1000W) simultaneously with adequate cooling.
4) Any of IRFP260, 360, 460 (TO247 case) can withstand about 70-80 W safely. And even up to 100 W depends on cooling quality.
5) A little question to shunt resistance - 0.12 R. 0.1 R is widespread. But if you wan't to put 7 A into one mosfet (and shunt) then I'll use 0.02 R, or 0.025 R or 0.05 R shunt maximum. For example something like OAR5R050FLF or OAR5R025FLF (they are sold on ebay).
« Last Edit: February 12, 2020, 01:29:31 pm by Vovk_Z »
 
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Offline Jwillis

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Re: Protection for Electronic Load Circuit
« Reply #4 on: February 12, 2020, 10:16:44 pm »
Monitor the temperature of MOSFETs, or the output voltage of each of the opamps.

I considered a temperature shut down. If I set a maximum temperature well in an area of SOA and have it shut down the system if that temperature goes above set point. This is a very good idea. Thank you

5) A little question to shunt resistance - 0.12 R. 0.1 R is widespread. But if you wan't to put 7 A into one mosfet (and shunt) then I'll use 0.02 R, or 0.025 R or 0.05 R shunt maximum. For example something like OAR5R050FLF or OAR5R025FLF (they are sold on ebay).

I'll look into those resistors for sure . The ones in the circuit now are only 5W ceramics (don't have anything better than that right now) So adding a resistor in parallel would help. I suppose ceramics would have more drift than a proper shunt resistor .
I did some calculations on the heat sinks I have and with active cooling they should be OK. Providing I did the calculations right . An online   calculator seemed to concur what I found. Fingers crossed and keep monitoring temperatures I guess.
Thank you Sir.
 

Offline Vovk_Z

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Re: Protection for Electronic Load Circuit
« Reply #5 on: February 13, 2020, 05:23:00 pm »
5W ceramic resistors have TCR 10-50 times larger than low-TCR shunt resistors (from my previous post) so they definitely will drift more.
If you need 7 Amps and can't find low-TCR shunt resistors then I'd use two 0.1R ceramics in parallel (0.05R). Because only one will dissipate 4.9 W at that current and will be too hot (about 120-130 C).
If you still need 0.1R then better to use 0.2R or 0.22R in parallel.
 
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Offline David Hess

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Re: Protection for Electronic Load Circuit
« Reply #6 on: February 14, 2020, 02:41:24 am »
A fuse would normally be used in combination with a crowbar circuit to force it to blow.  I would consider using a relay with an active protection circuit.
 
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Offline Jwillis

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Re: Protection for Electronic Load Circuit
« Reply #7 on: February 14, 2020, 04:57:20 am »
A fuse would normally be used in combination with a crowbar circuit to force it to blow.  I would consider using a relay with an active protection circuit.

Now that sounds like something that would work . Could you explain further ?
 

Offline David Hess

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Re: Protection for Electronic Load Circuit
« Reply #8 on: February 16, 2020, 05:10:18 pm »
A fuse would normally be used in combination with a crowbar circuit to force it to blow.  I would consider using a relay with an active protection circuit.

Now that sounds like something that would work . Could you explain further ?

It does not really apply to an electronic load where the source may not be able to supply enough current to blow the fuse.  The idea is that with a known source, like a transformer, rectifier, and capacitor for a DC power supply, an SCR immediately after the fuse between power and ground can be activated to force the fuse to blow.  This might be done for instance if an over-voltage condition was detected at the output.
 

Offline Jwillis

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Re: Protection for Electronic Load Circuit
« Reply #9 on: February 16, 2020, 11:05:19 pm »
A fuse would normally be used in combination with a crowbar circuit to force it to blow.  I would consider using a relay with an active protection circuit.

Now that sounds like something that would work . Could you explain further ?

It does not really apply to an electronic load where the source may not be able to supply enough current to blow the fuse.  The idea is that with a known source, like a transformer, rectifier, and capacitor for a DC power supply, an SCR immediately after the fuse between power and ground can be activated to force the fuse to blow.  This might be done for instance if an over-voltage condition was detected at the output.
   

Hmm sounds complex. What about preset normally closed temperature switches in series. Or Normally open thermal switches in parallel .If any one of them trips it could force a shutdown of the system that has to be manually reset . A warning buzzer could  activate until a reset is done.
 


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