Adding thermal overload sense to power supply - Page 1
Forum section: electronics / beginners

author: c4757p
author_rank: Super Contributor
timestamp: November 21, 2012, 03:26:22 pm
content: I have a Mastech HY1803D linear power supply. Not the best, but it works. Anyway, I blew the output transistor on it last night, clearly from thermal overload. Out of curiosity, while I was replacing it, I started looking at the circuit, and noticed it should be really easy to add a sensor to it.(Schematic can be found here -http://opendevices.ru/wp-content/uploads/2011/06/hy1803d.pdf- but it's one of the most poorly drawn I have seen... I have a much better version that I drew last night for practice reverse engineering a PCB, but I don't have it with me presently.)The output voltage appears to be controlled by pulldown diodes D11 and D12, forming a sort of "min" function between the two op amp outputs and the voltage on the pullup resistor and Zener. There are two unused op amp units (they used only half of an LM324, not the HA17324 shown on the schematic). I was thinking I could attach a thermistor to the output transistor, and configure one of the op amp units as a comparator that would drive low when the temperature exceeds a certain value. That would attach through a diode in the same fashion to the junction between the existing ones. This should divert all of the output transistors' base current away, shutting down the output.Are there any unforeseen problems with this - especially when the temperature sense voltage is close to the set voltage, since this is a crappy op amp and not a proper comparator? And what temperature should I set for the overload point - 100°C perhaps? The output transistor is a 2N3055 rated to 200°C, attached to a big heat sink for which I have no data.

author: c4757p
author_rank: Super Contributor
timestamp: November 21, 2012, 03:52:40 pm
content: Actually it seems that drivinghighwould shut down the output, because of the inverse polarity design (the positive terminal is signal ground). Still, same concept.

author: T4P
author_rank: Super Contributor
timestamp: November 21, 2012, 05:32:38 pm
content: A HA17324ISa lm324If i were you i would have tried to attach a fan at the back of the heatsink, helps to improve heatsinking considerably

author: hlavac
author_rank: Frequent Contributor
timestamp: November 21, 2012, 06:05:56 pm
content: I think just lowering the output voltage in the same way as the current limiter/voltage regulator in fact increase the power burned on the transistor due to higher voltage drop on it, making it burn even faster. You would need it to be more like complete shutdown (turn the transistor off completely).With some hysteresis preferably. And a fire alarm siren

author: c4757p
author_rank: Super Contributor
timestamp: November 21, 2012, 10:02:46 pm
content: Update after actually attempting this: You're right, pushing the output to zero doesn't help. It isn't close enough to zero. Even if I get the output down to "0.0006 V" (not really sure if I trust my meter that low...), enough current flows through the ammeter to register around 2 mA, and if I give it areallyhard short by sticking a thick metal bar across the output, the transistor gets quite hot, fast. There doesn't seem to be any easy way to shut it down. I considered adding a relay in the output path, but there was nowhere decent to fit it. I think I'm just going to make it simple: light an LED on the front panel. At least that way I can see that it's hot and shut it down myself.So, do you think 100°C is a safe temperature to do it at?

author: hlavac
author_rank: Frequent Contributor
timestamp: November 22, 2012, 01:56:17 pm
content: Maximum junction temperature of 2N3055 is 200°C, junction to case 1.5°C/W, so if you can keep the CASE of the transistor at 100°C it should be ok to some 66W dissipated. Adjust for case-to-heatsink thermal resistance if you are measuring temperature of the heatsink.Do you have proper contact with the heatsink? Thermal compound?

author: ptricks
author_rank: Frequent Contributor
timestamp: November 22, 2012, 02:01:57 pm
content: Thermistor and transistor to shunt the input of U2B pin 5 and the pot VR2. As temp increases , the transistor will shunt the voltage provided to the U2B opamp current limit setting and it will lower the amps limit , causing current limiting to go into effect until the temp is low. You can use a second transistor off of the thermistor to power an LED that will change in brightness with the thermal conditions.

author: c4757p
author_rank: Super Contributor
timestamp: November 22, 2012, 04:16:29 pm
content: ptricks, I actually tried exactly that, and the output went not to zero, but to something very close to -VBE(yes, negative). It required proper feedback using another op amp to balance the output at zero, and I still couldn't get anything close to zero current flowing through a short. Even when the meter read zero (my proper, accurate multimeter, not the power supply's panel meter) it would still conduct a good 30mA and get hot. I ended up just going with a bright-as-hell LED indicator; I can disconnect the load myself if it gets too hot.hlavac, yep, the transistor is mounted properly with good thermal compound, and the thermistor is on the surface of the transistor (I ended up replacing the 2N3055 with a TIP35C, since I had one available, and it is really easy to get a good connection between them). Maximum power dissipated should be around 60W, since there is approx. 20V across the transistor and the constant current circuit does a decent job at least keeping the current under 3A, even if it can't manage a proper shutdown...Thank you.Edit: I lowered the indicator temperature to 85°C for two reasons: 1) I now have to take the time to notice the LED glowing and disconnect the load. 2) TIP35C is only rated to 150°C, 1°C/W junction-case.

author: hlavac
author_rank: Frequent Contributor
timestamp: November 22, 2012, 04:44:11 pm
content: QuoteThere may be something wrong with the base driving circuitry then? Is the 30mA going thru collector or base?

author: c4757p
author_rank: Super Contributor
timestamp: November 22, 2012, 05:42:21 pm
content: I have no idea. I could probe the voltage across the base resistor R13, but to be honest I'm kind of sick of seeing the inside of this heap of junk. Replaced the dead transistor, added a temperature LED, it works, I'm happy. (I also rebuilt half the circuit board, as whoever assembled it didn't seem able to push a componentall the waythrough the holes on the PCB. Some of the leads were within a millimeter of touching a nearby component and shorting out, with the body of the damn component 2cm off the board...)Out of curiosity, is it actually common for a power supply to have no temperature sense whatsoever, or is that a "cheap piece of crap" thing? It seems like really poor engineering to make a power supply that can be physically damaged by running it within its specified output range.(The thing was loaded with other weird and plain stupid design choices, too. The front panel pots were 6.8k! I replaced the voltage pot because it was terribly scratchy - replaced it with a 50k, changed the other resistors in the voltage divider to match, and all is well. They could have used any old value as long as the others matched, and they chose 6.8k. They used a separate op amp chip for the constant voltage, even though there were two unused units on the main one, and the separate chip is no better in any way. Tons of unnecessary power consumption, especially in the various voltage regulators, which all drop huge voltages from the unnecessarily large input voltage. Some capacitorsreallyclose to their rated voltage, even though they used expensive brands - it's an input filter cap on a 7812 for Christ's sake! Just use a cheaper brand than Nippon Chemi-Con and get the proper voltage rating! Et cetera...)

author: T4P
author_rank: Super Contributor
timestamp: November 23, 2012, 09:01:40 am
content: It's possible those good brand caps are clones -lookcarefully

author: ptricks
author_rank: Frequent Contributor
timestamp: November 23, 2012, 01:06:48 pm
content: Quote from: c4757p on November 22, 2012, 05:42:21 pmMost of the cheap supplies do not have any thermal protection. Even the more expensive supplies may not have protection but rely on what the engineers chose for the heat sink design as the most likely daily usage.One thing I didn't think about is thermal switches , they attach to the heat sink and use a bi-metallic strip that opens when the temp is reached and after cooling close again. They are used quite a bit in appliances and other consumer items that produce heat.Cost is low and they come in various temp limits. They are usually rated to carry more than 10A so you could wire the mains supply through it.About $10 for 6 of them.http://www.ebay.com/itm/Thermal-Switch-Elmwood-3004-12-18-L130-130-degree-QTY-6-ea-/150814012115?pt=LH_DefaultDomain_0&hash=item231d3732d3

author: hlavac
author_rank: Frequent Contributor
timestamp: November 23, 2012, 05:30:06 pm
content: Properly designed supply should not overheat whatever you do.My crappy supply has a mechanical thermostat mounted on the heatsink that bridges a resistor in series with fan power to make the fan blow harder when it gets hot.