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Power regulator design

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Chris T

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I'm working on a 24VAC -> 5VDC circuit. I'm currently using a transformer to convert 120VAC into 24VAC by using a typical sprinkler transformer (it's rated for ~750mA). I've build the simple circuit to convert AC -> DC by using an electrolytic capacitor rated (100uF/50V) and a 1N4007 diode. I've tested that this works well. I get a non-regulated ~40V out of that. I get problems when I try to rectify the voltage down to 5VDC using an LM2596S-5. I've been following a diagram that I found online to do this:

lm2596_sch.png

There's a few differences between my implementation and what's above.

  • The pins for the LM2596S-5.0 were split & ribbonned and I added a heat sink to the chip (it gets real hot for some weird reason). I've ordered a through-hole version that should help things get less messy here.
  • S1A is changed for a 1N4007
  • I'm using a 5A 150uH inductor
  • SS24 is changed for a 1N5822. Unfortunately the diode has such a large wire that it doesn't fit in the breadboard so I've pried/bent the diode such that it touches the 2 pieces it nedes to touch. Ideally, I'd use an alligator clip but I don't have any at the moment (it touches properly though).
Finally, I'm measuring a DC voltage of roughly 0.4 to 0.5v (It's not a reading mistake, I've checked). About 10x less than I expect. I've also noticed that the chip gets abnormally hot. That doesn't seem right. I can't imagine that the typical implementation use a big heat sink like this so something's fishy. I've attached a picture of my setup on my breadboard to help support what I'm talking about. I measure the voltage between point '4' and GND by touching the inductor and the negative side of the capacitor.

I'd love to hear what I'm doing wrong here. I'd also love to hear if I can get a through-hole equivalent for a SS24 which works on breadboards.

**broken link removed**


Thanks for the help,
Chris
 
Your DC input voltage to the regulator is right on it's maximum voltage rating so you may have damaged it. Also the length of the wires to the regulator chip are much too long. You would be better off using stripboard or a PCB to mount all the components with short connections.

Les.
 
Your DC input voltage to the regulator is right on it's maximum voltage rating so you may have damaged it. Also the length of the wires to the regulator chip are much too long. You would be better off using stripboard or a PCB to mount all the components with short connections.

Les.

Thanks for the reply. That's a good point. The connections are very long and could affect the circuit. Especially when the inductor is that big and affecting the surrounding wires too. Once I get the through-hole LM2596, I should be able to mitigate the first problem. Is there any issue using an inductor rated at 5A? I think that the original circuit used an inductor that's more around 2->3A.

You're right it is on the limit of the voltage when it's unregulated, but once something draws current from the circuit, I would expect the measured voltage to drop between those 2 pins down to a sane levels. For what it's worth, this should work because the application which I am referring is using this chip with a sprinkler with a very similar diagram (opensprinkler.com)
 
SS24 is changed for a 1N5822.
You are using a 40V Schottky. Those diodes don't take over voltage well. I don't like running them above 80%. Is you diode good?
What current are wanting?
LM2596S-5.0 is only a 40V part.
 
You don't say what the design current is, but you may be limited by the size of the input capacitor.
If you are taking 1 A at 5 V, that will mean that the average current is around 200 mA at 24 V. I'll assume 60 Hz mains, and with half-wave rectification, the capacitor has to keep the supply going for around 16 ms each cycle, so it needs to produce around 0.0033 Coulombs of charge. A 100 μF capacitor will drop by 33 V if you take that much charge from it, so basically your circuit will be starved of power each mains cycle.

Switch-mode regulators take more current as the input voltage decreases, so you should have a good reserve of energy in the capacitor. I would aim for around a 1 V drop. If you change to a full-wave rectifier, and the output current is 1 A, then you need around 0.0016 Coulombs of charge. I would suggest that you use something like a 1500 μF capacitor, although you might get away with less.

Layout and good connections are important for power supplies like that. I suggest soldering, not the breadboard.
 
Thanks everyone for the kind help. It turns out that the issue was that my connection to the chip had 2 inverted wires :/ You can see in the picture above that the connection to the chip is not good.

You are using a 40V Schottky. Those diodes don't take over voltage well. I don't like running them above 80%. Is you diode good?
What current are wanting?
LM2596S-5.0 is only a 40V part.

This was just a basic test to make sure that the basics worked without thinking through the details of how big my capacitors would have to be. Now I'm going to do the math properly and size things right.

You don't say what the design current is, but you may be limited by the size of the input capacitor.
If you are taking 1 A at 5 V, that will mean that the average current is around 200 mA at 24 V. I'll assume 60 Hz mains, and with half-wave rectification, the capacitor has to keep the supply going for around 16 ms each cycle, so it needs to produce around 0.0033 Coulombs of charge. A 100 μF capacitor will drop by 33 V if you take that much charge from it, so basically your circuit will be starved of power each mains cycle.

Switch-mode regulators take more current as the input voltage decreases, so you should have a good reserve of energy in the capacitor. I would aim for around a 1 V drop. If you change to a full-wave rectifier, and the output current is 1 A, then you need around 0.0016 Coulombs of charge. I would suggest that you use something like a 1500 μF capacitor, although you might get away with less.

Layout and good connections are important for power supplies like that. I suggest soldering, not the breadboard.

Yah right now I'm not powering anything with it; it's purely used to test the voltage and there's no load on it. But in the future it'll power my raspberry pi & some accessories. So I'm looking at 1-2A. Right now it's a half-wave rect, but I'm looking at using a bridge rectifier to smooth out the current.
 
Looking at the Picture of your Inductor, it seems it is Probably much more than 150uH.
That is a LOT OF WIRE on that Torroid Core for just 150uH.

Possibly it is 150mH?
 
Looking at the Picture of your Inductor, it seems it is Probably much more than 150uH.
That is a LOT OF WIRE on that Torroid Core for just 150uH.

Possibly it is 150mH?

Yah I wouldn't be surprised if it was more. It was bought on eBay and who knows what you're getting there even if marked.
 
On the other hand if it works leave it alone.

He Said: "I've also noticed that the chip gets abnormally hot. That doesn't seem right".

So Possibly the Inductor is causing this.
Excessive Heat is NOT GOOD!
 
The LM2596 data sheet has layout recommendations.
I suggest you follow them as closely as possible.
 
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