[SOLVED] Does this power supply look okay?

[tedmalone]

I'm a noob to circuit design, but I am building a custom controller for my pool equipment and need to have 5v DC, 12v DC and 24v DC. The native transformer in the system is a 24v AC unit which I need to keep. To solve my problem, I've created a single circuit which supplies all of the DC voltages that I need. I believe that I've sized the capacitors properly and have chosen a combination of a bridge rectifier and 3 voltage regulators to provide the voltages I need.

I'm assuming that I may generate quite a bit of heat if I put too much load on the 5v circuit, so I'm planning to put a heat sink on that IC and measure temperature of the enclosure to make sure I stay within operating temperature. I also think I'm cutting it a bit close on the 24v DC line. My transformer puts out closer to 28v AC, but there is definitely some signal loss through the circuit. Fortunately, the only thing I need 24v DC for is a few relays and their minimum voltage to activate is 18v DC.

I've uploaded the circuit diagram to Skydrive and made it public. If anyone has time for a quick review, I would really appreciate your comments. I'm getting ready to break out the soldering iron. Also, if anyone would like the Visio of the circuit I'd be glad to share. I've included a parts list in case any other readers would like to try this at home.

Circuit design - https://skydrive.live.com/redir?resid=A7637C7D6CC3DC5!23878

 

[ashugtiwari]

the circuit is fine and will work but with lower efficiency. the regulator u used r linear regulators which r very less efficient of about 60% and dissipate lots of power as heat in case the load is very small. You will end up wasting power.
Instead, u can ideally use one high current (in ur case 6 A) switching regulator with power factor correction to attain maximum efficiency of 98-99% using any small PIC MCU. there r lots of application note on high power buck convertor on microchip website. The op voltage can be configured to 24V. You can use this o/p directly for 24V peripherals. In case of lower voltage like 5V or 3.3V, u can use any good linear regulator with regulated 24V as i/p and regulated o/p, the efficiency will be good enough. If used the single chip switching regulators from linear technology or analog devices, the efficiency will increase and the power dissipation will decrease further.
The circuit will be a bit complex but ur efficiency and regulation will be higher.

hope that helps.

 

[tedmalone]

Thank you for the quick reply. That does help. I was afraid this wasn't the most efficient design. I'll have to research the PIC MCU, I'm not familiar with that. It sounds like it's a better way to get the 24v DC. Once I've done that, it sounds like you're suggesting that I use the same circuit downstream with linear regulators dropping the 24v to 12v and 5v along the way? I guess I could have done that with my existing circuit as well with the 12v and 5v cascading down. But, that would mean that the 24v regulator would have to pass all of the current and the one I have is only rated at 1A.

the circuit is fine and will work but with lower efficiency. the regulator u used r linear regulators which r very less efficient of about 60% and dissipate lots of power as heat in case the load is very small. You will end up wasting power.
Instead, u can ideally use one high current (in ur case 6 A) switching regulator with power factor correction to attain maximum efficiency of 98-99% using any small PIC MCU. there r lots of application note on high power buck convertor on microchip website. The op voltage can be configured to 24V. You can use this o/p directly for 24V peripherals. In case of lower voltage like 5V or 3.3V, u can use any good linear regulator with regulated 24V as i/p and regulated o/p, the efficiency will be good enough. If used the single chip switching regulators from linear technology or analog devices, the efficiency will increase and the power dissipation will decrease further.
The circuit will be a bit complex but ur efficiency and regulation will be higher.

hope that helps.

 

[IanP]

I'm a noob to circuit design, but I am building a custom controller for my pool equipment and need to have 5v DC, 12v DC and 24v DC. The native transformer in the system is a 24v AC unit which I need to keep. To solve my problem, I've created a single circuit which supplies all of the DC voltages that I need. I believe that I've sized the capacitors properly and have chosen a combination of a bridge rectifier and 3 voltage regulators to provide the voltages I need.

I'm assuming that I may generate quite a bit of heat if I put too much load on the 5v circuit, so I'm planning to put a heat sink on that IC and measure temperature of the enclosure to make sure I stay within operating temperature. I also think I'm cutting it a bit close on the 24v DC line. My transformer puts out closer to 28v AC, but there is definitely some signal loss through the circuit. Fortunately, the only thing I need 24v DC for is a few relays and their minimum voltage to activate is 18v DC.

I've uploaded the circuit diagram to Skydrive and made it public. If anyone has time for a quick review, I would really appreciate your comments. I'm getting ready to break out the soldering iron. Also, if anyone would like the Visio of the circuit I'd be glad to share. I've included a parts list in case any other readers would like to try this at home.

Circuit design - https://skydrive.live.com/redir?resid=A7637C7D6CC3DC5!23878

Does this power supply look okay?

No it does not ...

Instead of linear regulators, that will hardly cope with the heat [24*1.4=33.6, (33.6-5V)*2A=57.2W – poor design], use switching regulators for all three voltages.
For example, LM2576-5 will produce 5V, LM2576-12 – 12 and LM2576-adj can be used for any output voltage within its range (5V, 12V and 24V) ..

:wink:
IanP

 

[Fragrance]

I'm a noob to circuit design, but I am building a custom controller for my pool equipment and need to have 5v DC, 12v DC and 24v DC. The native transformer in the system is a 24v AC unit which I need to keep. To solve my problem, I've created a single circuit which supplies all of the DC voltages that I need. I believe that I've sized the capacitors properly and have chosen a combination of a bridge rectifier and 3 voltage regulators to provide the voltages I need.

I'm assuming that I may generate quite a bit of heat if I put too much load on the 5v circuit, so I'm planning to put a heat sink on that IC and measure temperature of the enclosure to make sure I stay within operating temperature. I also think I'm cutting it a bit close on the 24v DC line. My transformer puts out closer to 28v AC, but there is definitely some signal loss through the circuit. Fortunately, the only thing I need 24v DC for is a few relays and their minimum voltage to activate is 18v DC.

I've uploaded the circuit diagram to Skydrive and made it public. If anyone has time for a quick review, I would really appreciate your comments. I'm getting ready to break out the soldering iron. Also, if anyone would like the Visio of the circuit I'd be glad to share. I've included a parts list in case any other readers would like to try this at home.

Circuit design - https://skydrive.live.com/redir?resid=A7637C7D6CC3DC5!23878

The provided circuit is not a good design, produce to much heat dissipation result in thermal shut down of the voltage regulator, you can achieve the same task with little modification in the design.


Fragrance

 

[tedmalone]

Is this the part you suggest:
https://www.mouser.com/ProductDetail/Texas-Instruments/LM2576T-50-NOPB/?qs=X1J7HmVL2ZE00R3IkTMYRw==

 

[betwixt]

Most definitely go with that part or one of it's equivalents. You will also need an inductor, a Shottky diode and some low ESR capacitors but the difference in performance is well worth the price. Even with all three regulators on a small heat sink it should run quite cool at full load. In your existing design it would produce around 90W of heat so you would not only need a large heat sink but a cooling fan as well.

I'm working on pool equipment as well at the moment, just out of curiosity, what is it you are building?

Brian.

 

[IanP]

Is this the part you suggest:
https://www.mouser.com/ProductDetail/Texas-Instruments/LM2576T-50-NOPB/?qs=X1J7HmVL2ZE00R3IkTMYRw==

:idea::idea::idea:

Yes ...
:wink:
IanP

 

[tedmalone]

Most definitely go with that part or one of it's equivalents. You will also need an inductor, a Shottky diode and some low ESR capacitors but the difference in performance is well worth the price. Even with all three regulators on a small heat sink it should run quite cool at full load. In your existing design it would produce around 90W of heat so you would not only need a large heat sink but a cooling fan as well.

I'm working on pool equipment as well at the moment, just out of curiosity, what is it you are building?

Brian.

Okay, it seems like I'm getting close. So, the idea is that the switching regulators turn on/off really quickly and therefore generate less heat in the voltage step-down?

The NTE 578 diodes I have appear to already be Shottky fast-switching diodes, so I think I'm okay there. I'm not sure what a low ESR capacitor is, but I'm hoping that the capacitors I have will do the job. They aren't the critical element in the circuit, right?

For the 24v DC line, do I really need the switching regulator, or will a linear regulator do? I'm only dropping 28v AC to 24v DC and pushing 0.5 amps. That's minimal heat, right?

The pool equipment I'm building is a combination of a Raspberry Pi, Arduino Uno and a variety of relays and sensors. I'm kind of building it as I go. I'd be glad to share more details about it and would love to hear about your project as well. Brian - Drop me a PM with your email address and I'll send you a note with more details.

 

[betwixt]

That's not quite how they work. It's easier to think of them this way in aquatic terms:

You have a water tank and you want to maintain a constant flow from an outlet connection near it's bottom. You don't want the tank to run dry and you don't want it to overflow. There are two methods to achieve it, one is the linear regulator analogy, you pour as much water into the tank as you take from it to maintain the level, the other is the switch mode analogy, you let it almost empty then fill it up in one go. Both achieve the same end but one takes up all your time and effort, the other gives you breaks when you are doing nothing.

Putting it in electrical terms, a linear regulator converts unused energy into heat, a bit like a tank full of water that isn't doing anythng useful because the level is higher than it needs to be. The switch mode regulator tops up the output only when it is needed. That's why the low ESR (low Equivalent Series Resistance = a good capacitor) is needed, it's the tank in the analogy. Just as in a linear regulator, the power loss is the voltage dropped across it multiplied by the current flowing through it. In your 5V circuit, the drop is about 30V - 5V = 25V multiplied by 2Amps = 50 Watts. In a switch mode supply some magic happens! the power is switched on and off in a series pass transistor, when it is turned off, no current flows through it so the power loss is V x zero = zero and when it's turned on it looks like a short circuit so (almost) no voltage is dropped across it and zero x I = zero. In a perfect world no power is lost at all, always either V or I is zero. The voltage is kept steady by controlling the amount of time the transistor is turned on relative to it's off time. The other bit of magic is the diode and inductor, These do two things, firstly the current flowing into the capacitor is restricted by the inductor so the voltage doesn't jump up and down as the transistor turns on and off and secondly, the way it does it is by absorbing some of the current to build up a magnetic field in it's core. When the transistor switches off, no more current is 'pushed' through it and the magnetic field starts to collapse. The magnetism is converted back to electricity and in such polarity that the diode conducts and allows it to flow into the capacitor. so the capacitor is topped up twice per switching cycle. It's a very efficient process and designed properly can reach almost 100% efficiency. Compare that with your linear design, on the 5V rail to produce (5V x 2A) 10W of regulated power you waste 50W in heat.

You might be able to use a linear regulator for the 24V line but the voltage at it's input can reach Sqrt(2) x 24V = 34V so you are actually dropping more than you expected. The actual voltage depends on how well the transformer output holds up under load.

You can contact me through my web site, you can see it in my profile on Edaboard. I warn you that I get several hundred emails every day so it may take a while for yours to reach the top of the pile!

Brian.

 

[tedmalone]

Thanks Brian, I love the physical analogy. Today I scoured the LM2576 guide (28 pages!) and re-designed my circuit on paper. I'm going to update the Visio tonight and see how the group thinks I responded to their suggestions. I've switched over to (3) LM2576 regulators, 2 fixed and one variable for the 24v DC line. Inductor selection was a bit more complicated, but I've landed on (2) 330uH high amperage and 1 220uH inductor. It's really not clear to me how important the sizing is? TI shows a sizing graph with ranges/zones for various voltages and max current. It appears that the lower uH ones are better if you expect to be carrying high current. I've also selected (3) 100uF capacitors for up-stream and (3) 1000uF capacitors for downstream. They're rated for 50v which seems to indicate a lower ESR, although TI claims that any aluminum electrolytic unit will work fine. I also picked up (3) 1N5822 Schottky diodes which seem to fit the bill (50v, 3A diodes). Now, I need to sharpen my pencil and work on the resistor sizing to get 24v on the output of the variable regulator.

That's not quite how they work. It's easier to think of them this way in aquatic terms:

You have a water tank and you want to maintain a constant flow from an outlet connection near it's bottom. You don't want the tank to run dry and you don't want it to overflow. There are two methods to achieve it, one is the linear regulator analogy, you pour as much water into the tank as you take from it to maintain the level, the other is the switch mode analogy, you let it almost empty then fill it up in one go. Both achieve the same end but one takes up all your time and effort, the other gives you breaks when you are doing nothing.

Putting it in electrical terms, a linear regulator converts unused energy into heat, a bit like a tank full of water that isn't doing anythng useful because the level is higher than it needs to be. The switch mode regulator tops up the output only when it is needed. That's why the low ESR (low Equivalent Series Resistance = a good capacitor) is needed, it's the tank in the analogy. Just as in a linear regulator, the power loss is the voltage dropped across it multiplied by the current flowing through it. In your 5V circuit, the drop is about 30V - 5V = 25V multiplied by 2Amps = 50 Watts. In a switch mode supply some magic happens! the power is switched on and off in a series pass transistor, when it is turned off, no current flows through it so the power loss is V x zero = zero and when it's turned on it looks like a short circuit so (almost) no voltage is dropped across it and zero x I = zero. In a perfect world no power is lost at all, always either V or I is zero. The voltage is kept steady by controlling the amount of time the transistor is turned on relative to it's off time. The other bit of magic is the diode and inductor, These do two things, firstly the current flowing into the capacitor is restricted by the inductor so the voltage doesn't jump up and down as the transistor turns on and off and secondly, the way it does it is by absorbing some of the current to build up a magnetic field in it's core. When the transistor switches off, no more current is 'pushed' through it and the magnetic field starts to collapse. The magnetism is converted back to electricity and in such polarity that the diode conducts and allows it to flow into the capacitor. so the capacitor is topped up twice per switching cycle. It's a very efficient process and designed properly can reach almost 100% efficiency. Compare that with your linear design, on the 5V rail to produce (5V x 2A) 10W of regulated power you waste 50W in heat.

You might be able to use a linear regulator for the 24V line but the voltage at it's input can reach Sqrt(2) x 24V = 34V so you are actually dropping more than you expected. The actual voltage depends on how well the transformer output holds up under load.

You can contact me through my web site, you can see it in my profile on Edaboard. I warn you that I get several hundred emails every day so it may take a while for yours to reach the top of the pile!

Brian.

 

[tedmalone]

Okay, I think I've got a much better version of my PS: **broken link removed**

I've got switching regulators, Schottky diodes, inductors and a resistor setup that I think will do the trick for the 24v DC I need.

The resistors gave me a bit of a problem though. How much wattage do they need to carry? I'm not expecting to draw more than .5 amps through the 24v circuit, but I thought I should design it for 1amp max. That's 24W (P = I * V), right? Does that mean I have to use resistors rated at 25 watts? They are huge, and don't seem to come in the larger values. These are supposedly just "programming resistors" to fix the output voltage of the regulator at a specific level. Maybe they don't really need to handle the power of the whole circuit?

 

[betwixt]

The power rating of a resistor is the voltage across it multiplied by the current through it (W=V*I) or the same calculation can be written as (V*V)/R or (I*I*R), they all give the same answer. The current through them is quite small, it's only enough to scale the output voltage to the range needed by the regulator input (FB pin) so it can monitor the output to keep it constant. They can be 0.25W rated.

Brian.