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Series Power Supply Safety Circuit

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Now that I think about it, I remember a while back somebody in another forum bringing up the diode...I don't remember why but for some reason on our particular power supply the general consensus was that it really wasn't necessary. I am sure it wouldn't hurt to have one though.
 
Maybe I am confused, but if the AC power cord shuts off then why would you need the diode? Once the fan bearings go out and the PSU #2 shuts down because of fan failure, the circuit will detect voltage has dropped below 24v and will then in turn shut down PSU #1 as well = no smoke problem (no power supply either, but no smoke problem).

Edit:
meaning why would you need the diode for this particular problem, not saying there are not possibly other reasons in favor of a diode...
 
The diode is recommended for power supplies used in series, so one supply doesn't backfeed into another. It's just not a good thing. Call it poor design practice, if you will.

===

So, let's say that the power supplies fail because of the fan stopping. Would it not make sense to detect the fans stopping? Now, your right that's a separate problem of the DC voltage too low.

If the DC voltage is too low, the damage may have already occurred to the power supply. If the air flow was detection, it might be able to avert the failure.

==

So, off to solve the original problem:

There is like 4 ways to shut down a power supply:

1) relay
2) Triac
3) Shunt trip breaker
4) Shutdown signal on power supply

There is a few ways to power the system
1) A separate power supply - Maybe $25 for a switching supply at 3-6 Watts
2) Pretend that only one fails and you get a minimum voltage of 1 supply.

There are a few detection schemes
1) look at the series combination
2) Look at each supply separately

There are power-up/shutdown issues.

In some cases if the supplies were less than the minimum, the shutdown circuit would never start.

Then there are logistics
What will this thing look like?
How does one safely get power to the box?
Physical separation of low and voltage and AC circuits. (wiring issues).
Cost.

So, I just took your problem and brainstormed some issues.
 
Here is one solution, although it got a little complicated. But it has the bells and whistles of fault indicator etc.
Here's how it works and a couple of questions?
This would replace your on off switch box for the 2 supplies. It has a momentary type rocker switch that in one direction turns on the ac to the supplies. So you press it and release it and the supplies turn on. If the combined voltage is over about 19 volts the relay turns on and keeps the ac going to the supplies when you release the switch and the green power on light is on. To turn it off in normal operation you push the switch in the opposite direction and it shuts off. If there is a fault and only one supply comes on when you push the switch to start they will power on but not stay on when the switch is released. While the switch is being held the red fault light will be on. If the supply fails in operation the red light will blink briefly and the supplies power down. It would be a pain to build 100's of these on a breadboard so if you go this way a little pcb might be in order though it will probably cost more than the rest of the parts. There is probably $6 in parts.

The 2 relay idea is simple. You could turn it on the same way, but I haven't found a switch type to turn it off. The diode is probably a good idea too, but it needs a pretty big heat sink or it will smoke.
Just to make sure. You still have 12 volts with one supply bad?
 

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Here is one solution, although it got a little complicated. But it has the bells and whistles of fault indicator etc.
Here's how it works and a couple of questions?
This would replace your on off switch box for the 2 supplies. It has a momentary type rocker switch that in one direction turns on the ac to the supplies. So you press it and release it and the supplies turn on. If the combined voltage is over about 19 volts the relay turns on and keeps the ac going to the supplies when you release the switch and the green power on light is on. To turn it off in normal operation you push the switch in the opposite direction and it shuts off. If there is a fault and only one supply comes on when you push the switch to start they will power on but not stay on when the switch is released. While the switch is being held the red fault light will be on. If the supply fails in operation the red light will blink briefly and the supplies power down. It would be a pain to build 100's of these on a breadboard so if you go this way a little pcb might be in order though it will probably cost more than the rest of the parts. There is probably $6 in parts.

The 2 relay idea is simple. You could turn it on the same way, but I haven't found a switch type to turn it off. The diode is probably a good idea too, but it needs a pretty big heat sink or it will smoke.
Just to make sure. You still have 12 volts with one supply bad?

That's a pretty sweet design ronv...I like it. Especially at only $6 in parts. Of course there would be a little more than that for the enclosure box plug, wire, ect...but I am already paying for all those parts. I will have to look into the cost of getting it done up on a PCB. If it's not more than $5 a part (in bulk) it may be worth it in time savings alone.

Yes, if one fails then there is still 12v assuming one is still good, but then if two fail, well, then I guess I don't need to worry about shutting anything off anyway...so there will always be at least 12v.
 
Two comments:

You may want to consider powering the circuit from a pigtail ATC fuse (2-3 Amps or so) directly wired to the output. This keeps the high current from melting wires.

Next, you may likely want a small relay in the actual safety device and from that just outputs a contact closure that would switch 120 VAC to a definate purpose contactor that controls the power. I'd weigh a 24 VAC control circuit as a possible option.

Add the diode.

PS: Glad you figured out that the case, etc costs $.
 
Just dreamed about a possible problem - hate it when I do that. It takes a while for the voltage from the big supplies to decay after you turn them off, so it may turn itself back on after you push the off switch. Maybe you can check to see how long it takes for the voltage to drop with no load.
 
I haven't ever timed it, but I see it fade down every time I power down the dummy load...seems like about 3 to 4 seconds or so.
 
I was afraid of that. Oh, well there are 3 more comparators in the ic, maybe something easy will come to mind.
 
That would be a fun practical joke to play on somebody... :) Not sure it would be so good for business though.
 
Back on the diode topic for a brief detour...I am still not sure I completely understand the necessity for it/risk of not having it. What I do understand (or at least think I understand): The diode protects backward voltage going into another PSU that is turned off, but if you turn them both on at the same time you really don't need the diode...so now, please correct my understanding. In other words, the diode is only there to stop that from happening when one is turned off, but in normal operation when both are on the diode doesn't do anything and is not needed (I think I got might have this part wrong). I am sure I got it wrong, because given what I think I understand, I don't see why it would need to be heat sinked or capable of 50 amps.
 
You understand completely. Here is a a diode: https://www.amazon.com/Amp-Volt-Stu...im_auto_1?ie=UTF8&refRID=1ST2YF163J4HZGP0NGRB Yep, it needs a mounting kit.

You can even use one diode of a bridge: **broken link removed**

===

OK, that over with. As each person is different, each power supply is different and they decay at different rates, So, during the power up/ power down sequence of 3-4 sec as you described it, one or the other supply is bound to be higher or lower than the other.

So, each power up or power down cycle has the POTENTIAL of doing some damage.

50A is easy for a diode. If you can turn off the supply, then really only surge current is more important and the diode could probably be rated at 30 A rather than 50. If the voltage drops to zero on one supply and the load is resistive then the higher supply would only have to deliver 1/2 the current.

So, your right on that count. It does not have to be rated at the supply current with a safety margin.

A huge heat sink probably isn't necessary either especially if your going to turn the supply off.

There is a stud mount package and a large bolt on package (2 screws) that are available for high current and/or high voltage diodes.

Aside:
I caused quite some grief for a manufacturer that we had an excellent relationship with. A lamp power supply failed that needed to supply 40 Amps at 22 V. When I started to look for obvious things, I found the rectifier rated for only 10 AMPS. Naturally, I was furious.

Unfortunately for them, because of space reasons, they had to replace the transformer and the diode, They actually changed it to a full bridge, They did the modification for free.

We helped them a lot and we got lots of engineering changes for free, but they required factory returns.

Our help had to do with traceable spectrum calibration and adjustment of the light source.
 
So are we really only talking about potential damage to the power supply or are we talking about an actual safety issue if I don't have a diode? If it IS a safety issue, then what that is dangerous to the person using the supply could potentially happen with no diode? I just want to fully understand this.
 
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Granted I am as usual late to the party. Would this work? Take a 24 VDC coil relay like this type and power it off your 24 VDC. These relays have a drop out voltage rating of 30% so anytime your 24 VDC drops by 50% you can figure the relay will drop out. The contacts will handle the AC line to both supplies. Would something along those lines work? That is about as simple of a quick and dirty type solution that I can think of.

Ron
 
Ron:

Good point, Made me think of "voltage monitoring relays" https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CEAQFjAA&url=https://www.farnell.com/datasheets/1748204.pdf&ei=oNNFU9dB5s-wBKHygKAG&usg=AFQjCNFCPMbUfkH5IiqLhWk71RdqfWemNw&sig2=bN9D6vRcnz5_bmHwzGx5lw&bvm=bv.64507335,d.cWc&cad=rja

So, you either have to get the thing turned on or supply 24 VDC from an auxiliary supply for a short time during power up. Kinda like a diode OR or the aux supply and the system supply for a second or two during power up. You'd need to combine it with a delay on make time delay relay.

So, three off the shelf parts: A $20 AC-DC converter, REECOM RAC series, a time delay relay and a drop out relay and a diode OR circuit, Once the system powers up, the wire or gets broken.
This system allows auto-restart.

Suggested power supply: https://www.digikey.com/product-search/en?vendor=0&keywords= RAC03-24SCR/277

The TDR could be 24 VDC or 24 VAC. e.g. https://www.ebay.com/itm/24VDC-Powe...405?pt=LH_DefaultDomain_0&hash=item4176e70745 It's purpose is to drop out the fixed 24 VDC supply after say a few seconds. 6 min is a lot. Just meant to be a suggestion.

So, sequence goes like:

120 VAC is applied to RAC 24 VDC supply. A relay generates a 2 sec pulse on power up (AC or DC).

The voltage monitoring relay gets powered from the diode or of the RAC supply and the power supplies in series. After the 2 sec timeout, the or from the RAC supply gets broken.

To turn off, you just have to break the signal going to the voltage monitoring relay. A normally closed push button.
 
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Kiss,

While it would be a PITA the relay could be manually held closed during power up. I have used the relays I linked to and they allow manual operation of holding closed. Been awhile but somewhere in all this junk I should have a few 24 VDC versions, if the OP would like I'll send him one. Trying to keep this cheap and simple.

Ron
 
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