Noob builds DIY power supply

rebelrider.mike

rebelrider.mike

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May 25, 2017
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Hi folks, I want to build a power supply as a learning project. Before I started this, my knowledge of a power supply was: 120VAC -> magic box -> 5VDC. I aim to understand the magic box part by building one.

I'll share what I think I know from searches and videos, and maybe you guys can help fix errors and fill in blanks? Here goes:

First, there is a transformer that takes the 120VAC and steps it down to something above 5V.
Then there's a full wave rectifier that converts AC to DC, and also changes the voltage a bit.
Then there is a capacitor that smooths the DC current from super squiggly to kinda squiggly.
Then an IC regulates the kinda squiggly voltage down to a nice flat 5V.

And there's some math involved. Rather than simply recreate a circuit from the internet (there's tons out there) I want to understand the function of the components and learn how to choose them based on the results I want and the math involved. So throwing all that together, here's what I've come up with:


image_xzcrmv.jpg


I guess the equation in the picture should be I x t / dV, but I forgot and put A for Amps. Even though it has to be mA to get uF at the end.
It turns out, that I've got to kind of work backward to figure out what I need. Starting with the final output, 5V 2A. It seems most folks are happy with 1A and use a regulator called a 7805. I want 2A, so I searched around and found an L78S05CV. The data sheet says I need at least 8V in to make it work.

For the rectifier I picked a GBU806, because I already have one. I was able to find a datasheet and it said I can put up to 600V 8A through it, though I'll lose 1V per "element". I'm guessing that's 1V per diode? or 1V per rectifier? I went with 1V per diode. Each path has 2 diodes so that's 2V.

But then there is this constant of 1.414 that everyone seems to use when the rectification happens, so 12V turns into 17V (ish) and loses the 2V to become 15V. So there is a net gain in voltage across the rectifier? The internet seems to think so. I assume there must be some power loss, but I guess it's in Amps.

So with the highest voltage being 15V, and the lowest being 8V (for the regulator) the dV is 7V. The output current I want is 2,000mA, and the time is 8.3ms since I'm using a 60Hz AC. With all the math done, the first capacitor should be a minimum of 15V 2382uF. I have a 16V 2200uF capacitor. I wonder if that's close enough.

I originally thought a voltage off the transformer would be better if it were closer to 5V, so less regulating would be needed. Seems though, the lower the transformer voltage, the more capacitance I need for the smoothing effect.

I also don't understand the purpose of the two other capacitors. The ones just before and after the regulator. People seem to throw those in without any explanation. Why are they there? What capacitance should they have? Can it also be calculated?

Also, If 8V 2A is going into the regulator, and only 5V 2A is coming out, is the difference being converted to heat? Can that be calculated? Does the rectifier also produce enough waste heat to need a heat sink? And is there a way to calculate that?

Where do people go to find all these little components? Especially if you know specs but not part numbers? I found a site called Mouser Electronics. Anyone heard of them?

I have even more questions, but I'll save those for later. :)
 
First up, this is a "linear" type design & it will get quite hot at the 2A rating you're aiming for. Yes, you're right, this type of design dissipates lots of heat.
An approximation would be at about 18W from regulator (approx 14V-5V x 2A) + 2W from diodes (2x 1V x 2A) lost as heat for 10W out (5V x 2A) + transformer losses ?? so overall 10W/30W = approx 30% efficient, not good!
You're going to need a really big heatsink on that regulator chip.

Switchmode designs are usually much more efficient for higher currents like this & that's why they are used extensively.

rebelrider.mike said:
First, there is a transformer that takes the 120VAC and steps it down to something above 5V.
Click to expand...
The volts you need for the "something above 5V" (as you've mentioned later) is the regulation drop out point of the chip or circuit you're using. ie you need "output" (5V) + this drop out voltage to arrive from the earlier part of the circuit.
rebelrider.mike said:
But then there is this constant of 1.414 that everyone seems to use when the rectification happens, so 12V turns into 17V (ish) and loses the 2V to become 15V. So there is a net gain in voltage across the rectifier? The internet seems to think so. I assume there must be some power loss, but I guess it's in Amps.
Click to expand...
So when we measure AC voltage, we usually are measuring the "RMS" voltage, ie a kind of average (which equates to the power a resistor with same value DC would see).
The peak voltage is 1.414 ish times the RMS value.
When you "perfectly" rectify AC the diodes let the peak through, so for "12V" AC RMS, this equals 1.414 x 12 = approx 17V DC.
ie your diagram +12V 0 -12V AC on the left should be 17V 0 -17V.
So losses:
Diodes: A typical silicon rectifier diode has 0.7 to 0.8 "Vf" drop when current id flowing across it. Higher voltage rated diodes like yours can have 1V. Special Schottky barrier type diodes have less drop, maybe 0.3 to 0.6V drop.
Transformer: Since the diodes see the peak, the power to charge up the first capacitor flows in sharp bursts only at the peaks with not much at other times, burst might be 5x the average current. This short burst loads the transformer & pulls down the 17V peak, How much depends mostly on the transformer & the load. A good low impedance transformer might not drop much but a lesser one might drop 1-2V or more. from the 17V peaks.

The power dissipated = approx average voltage x average current across the part, see the efficiency calcs I did above.

So the real peak voltage under load you would likely see with your parts is 17V - 2x Vf - 1.5V transformer sag = approx 13.5V ie your first capacitor is going to need to be bigger.

The two other capacitors close to the regulator are for stability & help with a quickly applied "step" load change.

Have fun experimenting but maybe try doing this at 5V & 1A first, you'll get burnt less !!! :)

Yes Mouser would have parts like these, there are several other parts sources too.
 
Thanks for the reply. Lots to think about!

So, would a transformer rated for 12V output actually read 17VAC? The only thing I know of RMS is that you have to use a "true RMS" meter to read AC voltage that isn't a true sine wave. Like from an inverter.

The datasheet for the GBU806 rectifier says it loses 1V per element. Are they rounding up? Should I use 1V or 0.7V per diodein my calculation?

If the Watts being lost are all converted to heat, is there a way to determine exactly how hot each chip will get? Watts per surface area or something?

If a single 16V 2200uF capacitor is insufficient, could I put 2 in parallel?

How should I determine the two small stabilizing capacitors' values? I assume the one before the regulator should be higher than the peak voltage, and the one after should be more than 5V. What about the capacitance though? I've seen diagrams that show 0.1uF and 0.01uF. After using something larger than 2200uF, how do these tiny capacitors make a difference? I'm not saying they don't, I just want to understand.

I know; tons of questions. Thanks for the patience and answers. I've just bought an Electronics for Dummies book. Hopefully that will help me understand more things. Reading a book though, is harder to retain than videos, or forum conversations. Still, I'll keep at it.
 
Yes, an unloaded transformer will show higher voltage than its "rated" voltage. To get around this, there's usually a load resistor across the terminals to make the voltage stable.

In the datasheets, there's usually a listing for how much wattage is consumed, most of this is expelled as heat.
 
Some useful guides :

Voltage AC (rms) * SQRT(2) = Voltage DC (peak)

Capacitor energy (joules) stored = 0.5 * Farads * Volts * Volts

If you use one large capacitor after the bridge the others are not needed...

Basically think of it as a bucket being filled up with small cups 100 times per second and a small hose pipe drawing the water back out. If the time it takes after for one cup has been added for the water level to drop below your hose outlet then you need a bigger cup. The size of the cup is related to the size of the capacitor and the size of the bucket. The hose is your voltage regulator and output load.

Heat distribution - get a heatsink or bit of aluminium and screw the regulator to it. The amount of heat is approximately (Voltage DC post Bridge - Voltage Ouptut) * Output Current = Watts (Joules per second)

If you have a Joules per second figure then say 15 watts over 10 seconds gives 150 Joules of energy and it is this cumulative total or the Joules figure that you need to work with at times. A heat simk may be able to dissipate 10 Joules per second if it is 20C above room temperature or 3 Joules if it is only 5C above room temperature. Depends on the surface area and if a fan is blowing air on it or it is covered by insulation..
 
rebelrider.mike said:
So, would a transformer rated for 12V output actually read 17VAC? The only thing I know of RMS is that you have to use a "true RMS" meter to read AC voltage that isn't a true sine wave. Like from an inverter.
Click to expand...
A transformer's rated voltage, eg "12V" is the RMS voltage = 12VAC (which has 17V peaks).

rebelrider.mike said:
The datasheet for the GBU806 rectifier says it loses 1V per element. Are they rounding up? Should I use 1V or 0.7V per diodein my calculation?
Click to expand...
Because it's rated 600V, each of the diodes in this guy will have a forward voltage drop of about 1V.
Compare with the specs for a 1N5822 (3A, 40V Schottky diode) or maybe a MBR735 (7A, 35V Schottky diode) which have about 0.5V forward drop.

rebelrider.mike said:
If the Watts being lost are all converted to heat, is there a way to determine exactly how hot each chip will get? Watts per surface area or something?
Click to expand...
If you find the manufacturers specs for each part, they usually give thermal characteristic eg x degrees C rise per watt. Without a big heatsink, the regulator chip is literally going to fry in a few seconds.

rebelrider.mike said:
If a single 16V 2200uF capacitor is insufficient, could I put 2 in parallel?
Click to expand...
Yes you can parallel capacitors like this.

rebelrider.mike said:
How should I determine the two small stabilizing capacitors' values? I assume the one before the regulator should be higher than the peak voltage, and the one after should be more than 5V. What about the capacitance though? I've seen diagrams that show 0.1uF and 0.01uF. After using something larger than 2200uF, how do these tiny capacitors make a difference? I'm not saying they don't, I just want to understand.
Click to expand...
The small capacitors are to suppress high frequency oscillation and help with noise & like I mentioned, step load response. But they're mainly to stop the circuitry inside the regulator chip reacting with the length of tracks or wires which have inductive reactance & it becoming an unwanted oscillator. Ie with out them the chip can go a bit nuts!

Big capacitors are often not quite so good at high frequency response & may have some inductance inherent their construction. Small value devices can be constructed differently to not have much of this unwanted inductance.

So it's not uncommon to have a large value capacitor in parallel with a small one so they combine the best of each one's characteristics, eg big storage with good high frequency performance.

rebelrider.mike said:
I know; tons of questions. Thanks for the patience and answers. I've just bought an Electronics for Dummies book. Hopefully that will help me understand more things. Reading a book though, is harder to retain than videos, or forum conversations. Still, I'll keep at it.
Click to expand...
It's all learning. Learning is good :)
 
the magic box comment made me smile, rest went over my head :p
 
Thanks folks! Still lots to learn, but I feel like I've made huge leaps in the last couple days. I read about RMS Voltage, and while I don't understand the math behind it, the need for conversion between AC and DC makes more sense now. I'm completely happy just using the constant 1.414 though!

Now that I know I've got my math right, I've plugged it into Excel so I can play "what if" with Voltages, and Amps and such. I also looked at 7805 regulators on Mouser Electronics, and found one called UA7805CKCT. Now, I still can't read most of the datasheet, but I did find some interesting info. It has an input of 7-25V, and can handle up to 1.5A. Also, the two small capacitors are recommended right on the datasheet, so I don't have to guess or calculate them.

image_xzlgrq.jpg

I wonder if I should use ceramic capacitors instead of the electrolytic ones?

Also, I tried to read the thermal properties of the datasheet, but don't really understand it. Here's a capture:

image_vrzufa.jpg

Seems to say that there are there is a temperature rise in 3 locations, given in degrees C per Watt. I Also know that the chip can only tolerate temps of up to 150C. So should I add all three values in the column together? Am I reading the table totally wrong?

Here is my latest idea:

image_fhmwll.jpg


I have a couple of 16V 2200uF capacitors in a box, and I'm hoping by paralleling them they'll do an adequate job. I also added the recommended smaller capacitors, but I guessed on the Voltage.

So if I'm only using my 600V rectifier at 12V or so, does that mean I'll have a lower voltage drop per diode? I'd like to have a way to test this stuff as I build it, but even if I had an oscilloscope, I wouldn't know how to use it, LOL.

I'm also looking at total Watts used so I know how much power the transformer will have to handle. Here's what I'm thinking:
The actual device could use up to 5V, 1.5A, so 7.5W.
The regulator will lose somewhere between 12V and 7V. Not sure how to figure that, so I averaged it. 9V, 1.5A, 13.5W.
The rectifier will lose 1.4V, 1.5A, so 2.1W
I suppose the transformer will lose some too, but I don't know how to figure that.

Rounding up, I figure the whole thing will use 25W. I probably should get a transformer that can handle more than 25W just to be safe. 30W?

I have a couple of things that look like transformers, but I can't identify them after a lot of searching around the internet. Maybe you guys might know?
The first one is SPI STA00327 07 HA HI-POT 1421 GP GH-130.
The second is SPI 8TG00364 7 FC2 HI-POT 1423 GP DASH 2 B-5

I've got some transformer questions coming up, but I still have some reading to do before I ask. :)
 
I like magic box idea.. opened up my old workstation and took out 500W PSU.. it provides 24A at 5.5V and 26 at 12V..

Going to add binding posts and call it a day ?
 
Roughly for the KTE :
Junction to ambient (device is not fixed to anything and free air only) 5 Watts would heat it to 5 x 23 = 115C
Junction to case (fixed with top pressed against surface) 5 Watts would heat the fixed surface and only 5 x 3 = 15C heating
Junction to pad (fixed with metal tab - to case / heatsink) 5 watts would then only heat the device 2.7 x 5 = 13.5C

I think this also assumes the fixed surface is an ideal heat conductor and can dissipate unlimited heat...

The KTE / different colums are different packing case types.

KCS is the more typical TO-220 case and why the junction-exposed tab (metal tab) value is a lot lower because the metal fixing tab is effectively the internal heatsink for the device to get heat out of the case.
 
I've had no luck at all with PC PSUs. I've done all the stuff on YouTube, and they either become dead, or they make LED and LCD stuff flicker like mad. I gave up and just bought a power supply dedicated to each voltage I needed.

Reading the datasheet for the 7805 I picked then, the temperature rise would be 19C/Watt. With the regulator losing around 13.5W to heat, that would be 256.5C. Well above the 150C maximum. But with a heat sink, I could expect more like 40.5C. I suppose the ambient temperature should also be taken into account. If the room is around 25C, then that much should be added to these temperatures, yes? So with a heat sink, it's closer to 65.5C. Definitely will need a heat sink!

Also, is it ok to use capacitors that are way above the Voltage of the circuit? The only 0.1uF capacitors I can find start at 25V.


Well, on to transformers. A lot of searching revealed lots of information on the popular TV and movie characters. It took some more digging to find actual electrical transformer info. But I did find a few videos and websites that explain how I might go about building my own transformer. I happen to have a nice core that I have no other use for. Yet another one with writing all over it that I can't identify anyway. And it was weird. The primary and secondary windings were joined at one end, so there were 3 wires total sticking out. How does that even work? I couldn't find any info on any kind of transformer that only uses 3 wires.

Anyway, the copper is gone, and I've got a nice "I E" core with the plastic thing:

image_wkkbpy.jpg


So to use this guy for new windings, I'm going to have to get it apart. I've seen folks saw these open, but the weld them shut afterwards. I'm not a welder. How detrimental is it to cut these things apart anyway? Maybe there is a conductive past out there that could ensure good contact?

Seems there is plenty more math involved if I do use this guy. If I custom wound this guy specifically for my household voltage, I'd do the primary winding for 123V. I've checked many times over many years all over the house, and it's always 123V. So the ratio of turns to get from 123V to 9V would be 13.67:1. So for every 13.67 turns on the primary, I'd want 1 turn on the secondary.

Apparently, I can't just make 13.67 turns on the primary though. I think it has something to do with how big the electromagnetic field gets around the iron core. Like if there's not enough, it will just act like a dead short? Or something. But there's math to tell me how many turns I need.

The equation I've seen is 42 / sqcm = turns per volt. The number 42 is a constant that comes from yet more math that I didn't look into. It also happens to be the answer to life, the universe, and everything. :) Square cm is the area of a cross section of the iron core. Length x Width in cm. The core area also happens to be related to how many Watts you want the transformer to handle.

I would have expected Hz to be in there somewhere, but I couldn't find anything about it. I've heard that if you put 60Hz current through a transformer rated for only 50Hz that it gets really hot.

Anyway, the square root of the total Watts is the minimum area needed in the core. So for me: sqrt 25W = 5sqcm. The core I have is 11.1sqcm, so I'm ok there. As for turns per volt, that would be 42 / 11.1sqcm = 3.78 turns. This is true for both windings. Then I only have to multiply by the Volts on each side. I'd need 454 turns on the 120V side, and 34 Turns on the 9V side.

I also tried to estimate how much wire I'd need. The distance around the core is 13.4cm, so I'll need that much for every turn. I'm also adding 10% because you always add 10% when estimating wire length. It came out to 220ft of primary wire, and 16ft of secondary. Luckily, this is available for really cheap on Amazon.

One thing that I noticed is that there seem to be 2 styles of winding a transformer. One has the primary and secondary windings separated in the middle of the core. The other has the primary winding on the inside next to the core, and the secondary is wrapped around the primary. Anyone know if there are advantages to either style?
 
rebelrider.mike said:
I've had no luck at all with PC PSUs. I've done all the stuff on YouTube, and they either become dead, or they make LED and LCD stuff flicker like mad. I gave up and just bought a power supply dedicated to each voltage I needed.

Reading the datasheet for the 7805 I picked then, the temperature rise would be 19C/Watt. With the regulator losing around 13.5W to heat, that would be 256.5C. Well above the 150C maximum. But with a heat sink, I could expect more like 40.5C. I suppose the ambient temperature should also be taken into account. If the room is around 25C, then that much should be added to these temperatures, yes? So with a heat sink, it's closer to 65.5C. Definitely will need a heat sink!

Also, is it ok to use capacitors that are way above the Voltage of the circuit? The only 0.1uF capacitors I can find start at 25V.


Well, on to transformers. A lot of searching revealed lots of information on the popular TV and movie characters. It took some more digging to find actual electrical transformer info. But I did find a few videos and websites that explain how I might go about building my own transformer. I happen to have a nice core that I have no other use for. Yet another one with writing all over it that I can't identify anyway. And it was weird. The primary and secondary windings were joined at one end, so there were 3 wires total sticking out. How does that even work? I couldn't find any info on any kind of transformer that only uses 3 wires.

Anyway, the copper is gone, and I've got a nice "I E" core with the plastic thing:

image_gtxghb.jpg


So to use this guy for new windings, I'm going to have to get it apart. I've seen folks saw these open, but the weld them shut afterwards. I'm not a welder. How detrimental is it to cut these things apart anyway? Maybe there is a conductive past out there that could ensure good contact?

Seems there is plenty more math involved if I do use this guy. If I custom wound this guy specifically for my household voltage, I'd do the primary winding for 123V. I've checked many times over many years all over the house, and it's always 123V. So the ratio of turns to get from 123V to 9V would be 13.67:1. So for every 13.67 turns on the primary, I'd want 1 turn on the secondary.

Apparently, I can't just make 13.67 turns on the primary though. I think it has something to do with how big the electromagnetic field gets around the iron core. Like if there's not enough, it will just act like a dead short? Or something. But there's math to tell me how many turns I need.

The equation I've seen is 42 / sqcm = turns per volt. The number 42 is a constant that comes from yet more math that I didn't look into. It also happens to be the answer to life, the universe, and everything. :) Square cm is the area of a cross section of the iron core. Length x Width in cm. The core area also happens to be related to how many Watts you want the transformer to handle.

I would have expected Hz to be in there somewhere, but I couldn't find anything about it. I've heard that if you put 60Hz current through a transformer rated for only 50Hz that it gets really hot.

Anyway, the square root of the total Watts is the minimum area needed in the core. So for me: sqrt 25W = 5sqcm. The core I have is 11.1sqcm, so I'm ok there. As for turns per volt, that would be 42 / 11.1sqcm = 3.78 turns. This is true for both windings. Then I only have to multiply by the Volts on each side. I'd need 454 turns on the 120V side, and 34 Turns on the 9V side.

I also tried to estimate how much wire I'd need. The distance around the core is 13.4cm, so I'll need that much for every turn. I'm also adding 10% because you always add 10% when estimating wire length. It came out to 220ft of primary wire, and 16ft of secondary. Luckily, this is available for really cheap on Amazon.

One thing that I noticed is that there seem to be 2 styles of winding a transformer. One has the primary and secondary windings separated in the middle of the core. The other has the primary winding on the inside next to the core, and the secondary is wrapped around the primary. Anyone know if there are advantages to either style?
Click to expand...

7805 is a linear voltage regulator and it dissipates extra energy as heat. For a better solution, you should try a buck convertor.

Video below was helpful.

 
rebelrider.mike said:
Now that I know I've got my math right, I've plugged it into Excel so I can play "what if" with Voltages, and Amps and such.
Click to expand...
Great way to go!

rebelrider.mike said:
I wonder if I should use ceramic capacitors instead of the electrolytic ones?
Click to expand...
Polyester, ceramic or small electrolytic should be fine.

rebelrider.mike said:
Also, I tried to read the thermal properties of the datasheet, but don't really understand it.
Seems to say that there are there is a temperature rise in 3 locations, given in degrees C per Watt. I Also know that the chip can only tolerate temps of up to 150C.
Click to expand...
Like completely charged wrote above, it's three different senarios,
1) no heatsink just air
2) heatsink but attached to poor heat conducting epoxy case of component
3) good heatsink attached to the metal tab of the component

rebelrider.mike said:
So should I add all three values in the column together? Am I reading the table totally wrong?
Click to expand...
No, you'd use the #3 on a heatsink one. You then have to find heatsink & read it's spec for temp rise vs watts (degC/W).
Then it's power dissipated by device x heatsinks C per W = rise from ambient.
So say it's 20 degC in the room now. Say 15W dissipated, say heatsink is rated 2.2degC/W.
like this one: https://www.jaycar.com.au/versatile-heavy-duty-heatsink-72mm-long/p/HH8566
Temp of heatsink would get to 20 + (15*2.2) = 53 deg C (hot but not too bad)
Inside the device, it's junction gets a bit hotter eg 3 degC/W so we add that to the temp like this:
3 degC/W x 15W = 45degC more = junction at 98degC (pretty hot, but still under specs 150C max)

rebelrider.mike said:
Here is my latest idea:

image_ujbdmp.jpg

I have a couple of 16V 2200uF capacitors in a box, and I'm hoping by paralleling them they'll do an adequate job. I also added the recommended smaller capacitors, but I guessed on the Voltage.
Click to expand...
Looks OK but the 16V big caps may be pretty close to or exceed max voltage with no load on the circuit. I'd use 25V ones there. I'd add a 10uF 10V (or more V) cap on the output as well.

rebelrider.mike said:
So if I'm only using my 600V rectifier at 12V or so, does that mean I'll have a lower voltage drop per diode?
Click to expand...
No, the higher voltage rating for each dode in there means they will still have a forward drop of about 1V each even if you're only using them with 12V AC.
This will mean the voltage peaks out of the bridge rectifier will be 17V peak - 2x 1V = approx 15V peak.
Notice this is only 1V less than the 16V cap rating? If the transformer voltage rises a bit with no load, you'll hit 16V & the caps will probably fail soon.

rebelrider.mike said:
I'd like to have a way to test this stuff as I build it, but even if I had an oscilloscope, I wouldn't know how to use it, LOL.
Click to expand...
Your volt meter will tell you most of this, but you can't beat an oscilloscope for visualizing the waveforms :)

rebelrider.mike said:
I'm also looking at total Watts used so I know how much power the transformer will have to handle. Here's what I'm thinking:
The actual device could use up to 5V, 1.5A, so 7.5W.
The regulator will lose somewhere between 12V and 7V. Not sure how to figure that, so I averaged it. 9V, 1.5A, 13.5W.
The rectifier will lose 1.4V, 1.5A, so 2.1W
Click to expand...
Almost! If your caps are bigger like the 2x 2200uF units, the average voltage across them under load will be higher than it would be with just 1x 2200uF.
The "sag" from the peak will be about half that of the one cap. Ie the average input voltage to the 7805 chip will be more like about 13V. (15V peak, sag to ~11V, avg~13V)

You have to use the input volts minus output voltage for the 7805's dissipation, so it's: (13Vin - 5Vout) x 1.5A = 12W dissipated in the 7805

The diodes it will be 1.5A x 2 x 1V = 3W for the bridge rectifier package

rebelrider.mike said:
I suppose the transformer will lose some too, but I don't know how to figure that.

Rounding up, I figure the whole thing will use 25W. I probably should get a transformer that can handle more than 25W just to be safe. 30W?
Click to expand...
Transformers are quite complex & I'd use one that's at least 50W to minimize it's heating, minimize the no load to fully loaded sag & to help it deal with the sharp current peaks those big caps will draw.

My advise on DIY mains transformers: 100% buy/use an off-the-shelf one!
Transformer design is not simple.
Core (steel) heating (from eddy currents), winding heating (from resistance & current flow), getting rid of that heat, magnetic fields being too high & "saturating" the core (causes high current & heating), safety of insulation, insulation vs heat, stopping winding & lamination vibration, the list goes on!

Low power, low voltage, high frequency transformers are more forgiving & I've made a few of those OK.
 
If you can, always use a larger than needed voltage rated capacitor. If you are designing for 12V, 16V cap is usually the next closest. But going with 20V would be even better. It's like having a 2 liter container that will only ever hold 1.75 liters. But, is there a 'chance' that we could get to 2 liters by accident? If so, you use the larger container to make sure you are safe.
Also, never ever use a lower voltage cap than what the design calls for. You'll destroy the cap, and potentially other parts.

The only thing that really matters, much more than voltage, is the farad rating. When the design called for 100uF, you should use 100uF. Don't drop a 200uF thinking it'd be better. When caps are used near ICs (Integrated Circuits), it's usually used to suppress noise at a particular frequency. This can also be true when making a power supply. There will be a certain amount of noise on the output side of things, and you may need to suppress the ripple to make electronics happy. So, if the plans call for a 10uF, 100uF, and a 200uF (values are just examples), don't drop a 310uF thinking you'll be fine. This isn't the case here. Each value suppresses a different frequency on the line.
 
Korishan said:
If you can, always use a larger than needed voltage rated capacitor.
Click to expand...
True, using the next rating up is a good idea especially if you're close to the first rating or there's any chance if it rising, eg unloaded transformer.

Korishan said:
The only thing that really matters, much more than voltage, is the farad rating. When the design called for 100uF, you should use 100uF. Don't drop a 200uF thinking it'd be better. When caps are used near ICs (Integrated Circuits), it's usually used to suppress noise at a particular frequency. This can also be true when making a power supply. There will be a certain amount of noise on the output side of things, and you may need to suppress the ripple to make electronics happy. So, if the plans call for a 10uF, 100uF, and a 200uF (values are just examples), don't drop a 310uF thinking you'll be fine. This isn't the case here. Each value suppresses a different frequency on the line.
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With power supplies and larger "storage" caps, like this circuit, I'd suggest it's actually OK to go up in size. Two caps share the current & bigger caps handle current better.
Only caveat (there's always a caveat!) is inrush current when it's first turned on more current is needed to charge them up and this stresses the diodes & transformer a bit more.

For noise suppression parts or eg audio or radio circuits or even switchmode supplies, yes you're right, some parts are designed to have a specific size & altering up or down can have other undesired effects.
 
Thanks for the replies! I'm almost ready to make a shopping list. Speaking of which, where do you all buy your components?
 
I'm down here in Australia :)
There's a couple of decent retail chains that are handy for many things eg Jaycar & Altronics
+ there's element14, rs components or mouser (and plenty of others) for the serious stuff.
 
I've bought parts from DigiKey. They do pretty good service and quick ship time (when the product is in stock, of course). There's also Mouser that's good, too.

Something like this, I'd only buy the parts from a reputable source. You are needing specific ranges and quality items. Some things like resistors that you can use a +/-5% rating is fine to get from eBay. But when you need closer to +/-1% or less, or good caps, go reputable.
I suppose if you know for certain that the eBay seller has name brand components, you could go that route, too. But that can be tricky to determine, especially if they are in the western part of asia, unfortunately.
 
I'm also looking for a way to estimate the Voltage drop of the transformer once it's under load. I'm reading about K values and X values and math that is a bit over my head. I did see on a datasheet there is a value called Voltage Regulation: 15% TYP @ full load to no load. Maybe that's what I'm looking for? Does that mean it drops 15% below the rated output under load, or it rises 15% above the rated output under no load?

Seems like there would be loss from the wire resistance. Especially the primary, which is small gauge and very long. And I guess there is some kind of loss from field leakage, or whatever that's called. I wonder if the presence of a magnet would affect a transformer's function?

I can figure the Voltage drop on the wire. For example, 120V 0.2A through 220 feet of 28 AWG is roughly 5.7V.
And 9V 2.8A through 16 feet of 18 AWG is roughly 0.57V.

Not sure if I should add windings to compensate for the primary Voltage drop, or subtract them. Or to leave them be.
The secondary winding though, I added windings to compensate for the secondary voltage drop. I ended up with an open Voltage of 9.64V, so the load will drop it to 9V.

I'm skipping a lot of math here, and still don't know about field leakage, but I guess I should be looking for a 10V or 12V transformer if I want 9V under load? Makes sense now, that you guys suggested capacitors higher than 16V. Since they're cheap, I'll go ahead and look for one that is more like 20V, 30000uF.

If I were to build my own transformer, I'd probably use 22 AWG wire on the primary so I wouldn't have to replace it if I ever wanted to use the core to its full potential. About 120W. The Voltage drop would be significantly less too. Only 1.42V.

I think I'm starting to ramble now. But mainly I'd like to find a relatively easy way to estimate Voltage drop across a transformer, if that's possible.
 
A while back, I bought some 5V power supplies to do a test for using TP4056 chargers in series. By mistake, 2 of them were 2A, and the rest were 1A. Here they are side by side:

image_ecbkzw.jpg


And a closeup of both:

image_lxwxdn.jpg


They both have tiny transformers that I can't identify. Would be neat to find some for my project.
Also, I'm guessing they're using a Voltage buckker, as both have a little 3 pin chip, but neither are on a heat sink. These have isolated ground, and I think that is what the black things with the copper windings are for. Interesting to look at a real product that does what I'm designing. It's obviously more complicated. Lots of extra parts, plus a bunch of surface mounted resistors and capacitors on the bottom.

The experiment by the way, worked great. Each cell was brought up to full charge independently, and the TP4056s didn't burn out. The next step will be to build a charger for a 7s battery, but I've got another thread for all that.https://secondlifestorage.com/showthread.php?tid=2077

I also found a heat sink that I hope will work for the rectifier and the linear regulator.

image_oyfntl.jpg


The chips on there now, are the actual rectifier I intend to use, and a stand-in chip that I can't identify. NXP BYC8 T4 PJA1410 B5 7556. Anyone want to take a guess?
The folks on YouTube make part ID look so easy. I can't Identify most of what I find.
 
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