How To Choose The Right Rectifier & Capacitor For Rectifing AC to DC
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- čas přidán 20. 06. 2024
- (Please read calculation examples below) In this video I discuss how to choose the right size & spec'd full bridge rectifier & electrolytic capacitor.
In the interests of avoiding confusing calculations please carefully study the below examples. Note in "Example 2" how the results = 200 this could easily be mistaken for 200uf. However, it is infact 200,000uf
Example 1:
10 amps x 8.3ms / 5.5volts = 15.0909 (15,0909uf)
Example 2:
10amps x 10ms /0.5volts = 200 (200,000uf)
Information provided in this video is for educational purposes only.
If you attempt to recreate/replicate anything you’ve seen in this Or any other video, you’re doing so at your own risk.
- Schematix - - Jak na to + styl
Please study the calculation examples in the video description to better understand how to read the results of any calculations you make for capacitance :)
Schematix Hi there I was wondering if how you find out the minimal voltage drop and the peak voltage after rectification.
thanks
I love your tutorials but I don't understand your capacitance formula. Result say 15.090mF but you call it 15 thousand mF. Does higher capacitance mean more voltage drop?
Raks Electric; OK, first you have to find the peak voltage after rectification. You calculate that simply by multiplying the AC output voltage of the secondary of your power transformer by the square root of 2. That gives you the peak DC voltage output of the bridge rectifier, before adding any smoothing/filter capacitors to it. (A very close example to this video would be a secondary transformer voltage of 50V X 1.4142 = approximately 70.71VDC out of your rectifier bridge.)
Then, *you* as a designer of your circuit, have to determine the "minimum acceptable voltage" that your circuit can safely or effectively operate from, which, depending on circuit design and type, can vary *significantly* , however, for *most* electronic items, if you stay within a 5-10% drop from the Peak DC output voltage of your rectifier then you should be fine. Your specific design *may* require tighter tolerances than that, but if so, that increases your required capacitance ( *and cost* ) quite significantly!
So, just for an example that's again relatively close to this video for ease of comparison, let's say that in order for your circuit to work effectively and to perform at an optimum level it needs an absolute "minimum acceptable voltage" of 65 volts DC. If you do the math, that voltage would be within about -8% of the peak DC volts out of your rectifier, so that should be just fine in most cases.
The *difference* between the peak voltage out of the rectifier and the "minimum acceptable operating voltage" for your circuit would then be the "acceptable voltage drop" that is used in the final equation for determining the required capacitance value. In this example it would be:
70.71 - 65 = 5.71V "acceptable voltage drop"
Now comes the (slightly) confusing part; This video didn't explain the discrepancies in the *units* , and *that* is why the answer is not exactly what is shown, but rather off by a factor of 1000 in each case... The (correct) formula for calculating the required capacitance is, (should be);
Circuit Current (IN AMPS) x Half Cycle Time (IN SECONDS) / Acceptable Voltage Drop (IN VOLTS) = Required Capacitance (IN FARADS)
If you use the above formula with the above shown units then your answer will always be correct... Using the above example's first Voltage calculations, and assuming (again *for example only* ) that your circuit current is 10 Amps and you are in a location that uses 60Hz frequency of AC power, like in the USA, then the correct way to use the formula is as follows;
(First calculate the last needed parameter to use in the final formula, which is the *Half Cycle* Time of a 60Hz line frequency, but calculate it in SECONDS, instead of mS, as is the error in the above video... This then equals 0.008333 Seconds.) (1/120th of a Second)
Now finally put all of your (correct unit) numbers into the correct unit formula which I showed above and you get;
10 (AMPS) x 0.008333 (SECONDS) / 5.71 (VOLTS) = 0.0146 (FARADS) of required capacitance. (Approximately anyway, due to rounding of answer)
Now, since 1 FARAD = 1,000,000 MicroFarads (or uF), then just multiply the above answer by 1,000,000 and you get the *correct* answer of 14,600 uF.
In this example's case, I would just use a 15,000 uF capacitor for a filter capacitor, which is the next largest commonly available value above the "minimum required" 14,600 uF.
The above example may seem complicated at first, but once you do it a few times with different values and for different electronic projects, you'll get the hang of it quickly. And if you always use the consistent (whole) unit formula above, and then just quickly convert your answer from Farads to MicroFarads (1/1,000,000th of a Farad), then you will *always* calculate the correct answer when doing this!
FYI, the formula in this video gives answers in MILLIFARADS (or mF) (1/1000th of a Farad), NOT MICROFARADS, (or uF) (1/1,000,000th of a Farad)!... Simply because they are using MILLISECONDS (mS) (1/1000th of a Second), rather than full Seconds... Somewhere along the line, the units got switched or dropped, (i.e. "lost in translation"), in the formula shown in the video, from the original, correct formula which I gave above... Possibly because oftentimes mF (millifarads) is commonly mistaken for uF (microfarads), simply because of the letter "m"!
Anyway, now you know the "rest of the story", LOL!... I hope this detailed explanation helps *someone* out there to better understand this at least, because knowing what is the correct size of capacitor to use to filter your AC-DC power supply is a VERY important part of proper electronic equipment design!
Accumulator1; I totally understand your confusion, however, you made the same exact (very common) mistake that this video makes, in confusing the units of mF (MILLIFARADS) and uF (MICROFARADS)... The answer actually IS 15mF, however, it is ALSO 15,000uF, because they are a ratio of 1:1000 ... The full, detailed explanation is below;
(BTW, A Higher capacitance used for the filtering of a bridge rectifier actually results in LESS of a voltage drop, not more, and therefore a higher capacitance also results in a more stable final DC output!)
Now comes the (slightly) confusing part; This video didn't explain the discrepancies in the *units* , and *that* is why the answer is not exactly what is shown, but rather off by a factor of 1000 in each case... The (correct) formula for calculating the required capacitance is, (should be);
Circuit Current (IN AMPS) x Half Cycle Time (IN SECONDS) / Acceptable Voltage Drop (IN VOLTS) = Required Capacitance (IN FARADS)
If you always use the above formula with the above shown FULL units then your answer will *always* be correct... Using this video's "Acceptable Voltage Drop" calculations for an example, and assuming (again *for example only* ) that your circuit current is also 10 Amps and you are in a location that uses 60Hz frequency of AC power, like in the USA, then the correct way to use the (correct unit) formula is as follows;
(First calculate the last needed parameter to use in the final formula, which is the *Half Cycle* Time of a 60Hz line frequency, but calculate it in SECONDS, instead of mS, as is the error in the above video... This then equals approximately 0.0083 Seconds.) (1/120th of a Second)
Now finally put all of your (correct unit) numbers into the correct unit formula which I showed above and you get;
10 (AMPS) x 0.0083 (SECONDS) / 5.5 (VOLTS) = 0.015091 (FARADS) of required capacitance. (Approximately anyway, due to rounding of numbers)
Now, since 1 FARAD = 1,000,000 MICROFarads (or uF), then just multiply the above answer by 1,000,000 and you get the *correct* answer of 15,091 uF.
In this example's case, I would just use a 15,000 uF capacitor for a filter capacitor, since it is so very close to the calculated value, and that value is the closest commonly available value. Plus, since most electrolytic type capacitors generally tend to measure slightly higher than stated anyway, especially when new, then you should be just fine!
The above example may seem complicated at first, but once you do it a few times with different values and for different electronic projects, you'll get the hang of it quickly. And if you always use the consistent (whole) unit formula above, and then just quickly convert your answer from Farads to MicroFarads (1/1,000,000th of a Farad), then you will *always* calculate the correct answer when doing this!
FYI, the formula in this video gives answers in MILLIFARADS (or mF) (1/1000th of a Farad), NOT MICROFARADS, (or uF) (1/1,000,000th of a Farad)!... Simply because they are using MILLISECONDS (mS) (1/1000th of a Second), rather than full Seconds... Somewhere along the line, the units got switched or dropped, (i.e. "lost in translation"), in the formula shown in the video, from the original, correct formula which I gave above... Possibly because oftentimes mF (millifarads) is/are quite commonly mistaken for uF (microfarads), simply because of the letter "m" being common between the two unit names! In addition, the "u" in uF isn't actually a "u" anyway, but instead it is only the closest resemblance on a common keyboard to the real unit measure, which is actually the Greek letter "mu", shown in lowercase, but to make matters even worse, that same Greek letter in Uppercase is also "M", and sometimes people state MILLIFarads as MF instead of the more proper mF!... This all results in *very much* confusion between the MICRO- and MILLI- prefixes of the FARAD capacitance unit!! (I see that mistake being made almost all the time, so it helps to understand the difference!)
Anyway, now you know the "rest of the story", LOL!... I hope this detailed explanation helps you to better understand this, because knowing what is the correct size of capacitor to use to filter your AC-DC power supply is a VERY important part of proper electronic equipment design!
JoeJ8282 thankyou very much now I understand
You are very good at explaining what can seem a complex subject. 50 years ago I was into electronics (valve amplifiers) it's been quite a struggle to bring myself up to date. I did try and tighten a connection with 240V on it whilst well earthed. Seems at 70 odd years old my heart is pretty sound. Good videos young fella me lad, greetings from the old country.
thank you. I've been trying to sort this aspect of AC-DC rectification for a number of days. YOURS was the first well explained, thorough description I've found.
Thanks, again..
GeoD
Id like to agree with this persons statement, thank you for explaning this and walking us threw how to do it! Btw popping caps can be fun I got a good chuckle outta that
The instructor has calculated along the same lines as I always taught and here is the simple
way to remember:
The basics of a capacitor connected to a bridge rectifier is
this: Normally you use a 1,000u electrolytic for each 1 amp of current
that you will be taking from the circuit. When NO CURRENT is taken, the
output will be completely smooth and appear as the peak value of the AC
that is delivered to the circuit. As you take current from the circuit,
the output voltage will dip and then rise again at the rate of 100
times per second. This is because the electrolytic cannot keep the
output voltage totally smooth as the electro is discharging and
supplying energy to the load at the rate of 100 times per second. As
the load increases, these dips become larger and when the current
reaches 1 amp, the dip will be 5v.
For instance, if you supply the circuit with 15v AC, the output of the bridge and filter electrolytic
will be: 15v (minus 2v through the bridge) and then 40% more than 13v to
create the peak value of 18v. This will dip to 13v with full load.
These are the FACTS. No mathematics is needed. If you put the output voltage
(which has a ripple of 5v) through a 3-TERMINAL REGULATOR that outputs a
voltage of 13v, you will never see any ripple.
That's all you need to know.
Colin Mitchell www.talkingelectronics.com
Can you damage capacitor if amps reach 5 or 10A but you calculated for 1A draw or it just won't do anything?
I was always taught to use 2000uf per amp just to be safe.
Good Job Colin! Well explained too. I have a question. I have a transformer rated at 120VAC stepped down to 12VAC (180mA written on transformer label) with a full bridge rectifier (1n4007) and capacitor. My circuit requires 12VDC. How much capacitor (capacitance and voltage) do I need to ensure no problems occur? Thank you in advance :)
I have always wondered how the capacitor value was calculated, a clear and concise explanation.
I learned a lot here. Great video! Thank you for including an example calculation for 60Hz too that was very helpful. Now to go find a huge capacitor and pray I don't wire it up backwards. :)
Excellent explanation and demonstration, I managed to follow every part of it, very clear and at a speed that someone like myself can follow. I will definitely watch more of your videos.
Thanks for posting this. It takes me back to basic electronics I used during my military service. Well thought out and explained. You have a new follower!
Thanks kindly for your support ;)
GREAT video! I learned a lot from it. You're video is the first I've come across that not only talks about choosing the right capacitor, but HOW to choose the correct capacitor. You do a great job in your explanation of things! Keep up the good work.
You have a wonderful way of explaining
Hi Schematix, hope you're doing well! I've enjoyed the couple of videos you've posted on transformers and other power supply components.
Very informative. Thanks learned a lot on transformers and rectifiers.
Great videos and very well explained, cheers mate
perfect,, this is exactly I have been looking for .. you do really nice job. thanks !
Thank you this really helped me understand the calculations.
very interesting and well descriped I am from Malta, and I really enjoy your lessons Thanks for your time
This was great. I am ok with transforming and rectification etc but was never too sure about a good capacitance selection. Now I have it. Subscribed for good clear presentation of material.
PS. Kiwis rock!
Great video, thank you for the calculations
Thank you sir. I was planning to build myself an amplifier PSU this video helped me a lot :)
Great job all around. Quality and content.
Great video. I also appreciate your humor.
You answered so many of my questions but show us what ur wiring up for ur example so I can see a finished setup thanks keep making videos!!!
Short, simple and well explained
The best explanation out there on this topic! Thank you!
Thank you that was exactly the info I needed. Very informative, I am subscribed. Keep up the good work
These are really interesting and well explained videos. Your arithmetic at the end fired my awkwardness a bit, multiply the result by 1000 to get the microfarad result. (Because we don't say the mili word with capacitors!)
yes the cycle time input in the formula is in ms and the capacitance im mF. I specially enjoyed the definitions of a poor power supply.
It is such a shame you decided to stop your videos. Giving that microwave transformer 'what for' brought a smile to my face. For gawds sake keep this bloke away from a power station. However I am about to do exactly the same to a microwave transformer. I need 40V so my wire choice in amps and insulation will be a bit tight. Perhaps you could add a comment as to whether you will do any new videos. I hope you do. Happy New Year from Pomland.
I am so pleased that I have found your channel.
I am going to make a battery charger as I have been told that the new type solid-state chargers are not much good.
I want a charger that will charge a tractor battery, so I am looking at 4-5 Amps output current.
I can get a transformer from Jaycar, now I know what size rectifier and capacitor I need to get.
I really like your content, very informative and to the point, really cool to see a fellow kiwi sharing his knowledge! :)
Thanks mate :)
Old info, bin there, done that 25 years ago.
Very interesting ! The method of calculating the required capacitor is very clear and easy .. thank you so much I really enjoyed your video it was a wonderful morning :)
Thank you. Best explanation I have yet come across on YT!
Thanks for your videos Schematix, they are Excellent.
Cool I needed a refresher course on rectification and this did the Job. especially about how to determine capacitance!
Your information was a BIG help to me. Thanks
Great job I understand more now than I ever have
Good video. So the result of cap with not enough uf will not fully smooth out the current as well as a cap with the proper uf required by your calculation? Does a toroid core wrapped with copper wire, and connected to the cap, help smooth out the current even more than without it?
I really appreciate that you are sharing your experience worth us
Excellent explanation and demonstration
Awesome explanation!!
Very easy to understand,thank you 👍
Thank you so much i learn allot with this video.
The forward voltage drop shown at 4:24 in your video (Vf = 1.1.V) is the forward drop per diode in the bridge. The positive and negative paths through the bridge each have 2 diodes in series. You need to make allowance for double the voltage drop i.e. 2.2 volts and consequently double the power dissipation. Selection of the "right" capacitor also requires consideration of the load current. If the ripple current rating of the capacitor is inadequate it will have a short life.
Excellent video!! Thanks very much
Hi a very explanatory video, its appreciated greatly.
The question it brings to mind is, how did you decide on the minimum acceptable voltage and for a long service life how does this acceptable voltage effect the life of the capacitor. From the data sheet I only see the recommended ripple current and not ripple voltage.
This was so clear! Thank you! Thumbs up and suscribed!
thank you very much for making the film
Thanks for the explanation. I don't understand frequency, farads etc.
Still very engrossing tutorial.👍🏻
These vids are the king keep them coming teacher many thanks.
Sir,
Excellent video! I just found what I was looking for.
But, I have few questions.
On 12-6-0-6-12 CT transformer, I'm getting 32V from first and last terminal by reading from multimeter without any load. It is the RMS right?
If it is RMS, then peak should be 45.25V. If I use a 35V capacitor, I'm getting 45V after adding capacitor. is it okay/safe?
Whats does 35V mean on that capacitor - maximum of RMS of maximum of peak?
Thanks in advance :D
Great video great teacher you are
This is the stuff why i subbed to this channel, good for people like me who have to start from the beginning. :-)
Very useful, thanks. Subbed.
Great video. To the point and well organized. Well done. Count me in as a subscriber.
Interesting, very well explained
:)
Awesome video!!!
Very useful.....thanks mate!
Great video!
By any chance have you done a full video of DC power supply using transformer and capacitors? Great video by the way. Nicely done!
God bless u, thanks for a very nice job
Nice vid series thanks.
liked and enjoyed .. thank you for sharing.
Thank you for that great explnation. I have ? if i have only normal diode & i have to make BR from it with Capacitor. if I want 18V & 15A or higher, from 220 transformer. How much should be the diode & capacitor? Thank u for answering if you can.
i want more videos on your channel, related to electrical engineering.
Excellent video & tutorial. I am in the process of replacing the Bridge Rectifier in a Daiwa PS-30XMII. When I removed the faulty BR it had a 14mm brown ceramic disc capacitor across the AC terminals. I can find nothing about this on the schematic. Is the capacitor of any real value in this situation or would you recommend
fitting the new one without it please?
are you sure its cap....could be a MOV
such a cap can reduce line noise/spikes
Thank you , really helpful
Hi, any advice on the brand to choose, there are so many ! Thank you
Your videos are great! You've got yourself a subscriber
Thanks ;)
You've made my day ;)
I have seen videos using a center-tap transformer with 2 diodes for full-wave rectifier and a single smoothing capacitor and I have seen them with the standard full-wave rectifier with the center-tap tied between 2 smoothing capacitors. Transformers are rated in power as either Watts or Volt-Amp rating which is still the same thing. So is there any one that is better for a power supply. If you have 30 volts AC at 4 Amps and after DC Rectification and Regulation - is that going to provide 15 Volts at 8 Amps?
Creative video, thanks :)
well done !
Thanks. Joe
Perfect, thanks.
Great mate.
I was gonna build a dc supply with a 1000w microwave transformer. gonna rewind it to put out around 50v I have a couple 1000v 50a bridge rectifier laying around could I just use them or would that be way overkill and have negative effects?
Can you build a simple power supply for testing a inverter compressor. Where it is not known if the compressor or inverter or both are bad.
Thanks, really enjoying your vids. Building a twin power supply (2x55 watts) at the moment and bought some (probably) oversized caps because I found no suitable formula on the internet. Now I can calculate if they are sufficient.
Btw, where is the hawaii shirt from the early vids? ;-)
Lejf Diecks. Thanks for the feedback. At the moment I'm building a 900watt programmable bench power supply (Build video coming) it was a little chilly for a tropical themed shirt ;-)
This was so helpful. I went down to subscribe, but saw i already was. Rad
excellent videos keep them coming. im interested to find out how to pair a mosfet and a protection diode cause im sick of destroying mosfets lol. im running a 240vdc treadmill motor off of a 12v battery to 240vac inverter to a bridge rectifier (now i know how to match a cap for the job) to a pwm circuit with mosfet then to the motor. the motor is 100% dc not ac but the mosfets don't last long. any help would be awesome
Thanks! Plenty more videos coming! My first thought is the mosfet being adequately cooled with a heatsink? If the mosfet is exceeding the rated operation temp it's life span will be significantly reduced. Make sure you have a mosfet rated for at least 20% more current & higher voltage than what you require
at 5:57 bridge generate 2 x VF or 33 Watts . great vid
Great video thanks, the only slight issue is that the formula states that it equals ųF but it actually equals mF. Though in your example you had already converted it (15000 vs 15)
Nice one, I just made a comment on this too haha
uF (microfarrad) IS mF (microfarrad).
They represent the same thing.
They are the same thing.
@@googleedwardbernays6455
uF is microFarad (x10-6)
mF is milliFarad (x10-3)
@Nohr Tillman
Aaaahhhh. You are correct!! I always forget about the milliFarrad due to it not being used as the other three (u, P, N) in reference to electrolytics or in reference to capacitors. And since i am unaware of a single instance where milli has been used in terms of capacitor. My bad, homie. Thanks for correcting me
Can a full bridge rectifier be used as a choke in place of an inductor between a control board and a DC treadmill motor? The inductor that came with the treadmill had a bullet hole in it and was beyond repair. Also, don't know how to use a transformer as an inductor because I don't know what values would be needed. You probably have a video on inductors that I should watch. Cheers
Thank you sir...!!!!
good video
So I guess my questioning would be can you use multiple compasitors to reach the compasitance you require? Like idk if that would even work in series or parallel or dose it have to be one compassitor i guess depending on the application?
I wanted to know when to use a Zener Diode and where to put in the set-up. Please let me know.
Zener diode is used when you need a specific vpltage , you now get a zener diode of dat rating connect as required
How about a 3 phase windgenerator... how can I know what capasitor to use? Since its up and down on the power all the time?
Regards.
Hello, I have an ATV with no battery, just pull start which outputs AC power and I want to run a 12 V DC fan for the radiator (no other powered accessories such as lights etc.). The stator has a lighting coil that outputs 150 watts. I cannot find the wattage of the fan but I did measure the resistance and got 3.8 ohms, so I believe the wattage is about 38 through calculation. The engine manual says the lighting coil produces 32V at 2000 rpm, 50V at 3000 rpm and 67V at 4000 rpm. The motor is rated at a peak rpm of 7200 (not sure if all of this info is needed). This is a retrofit so I have not added the voltage regulator yet but I assume it will bring the voltage to 12V? With that information, can you provide guidance of which rectifier and capacitor to use? I am somewhat electrically helpless, so any assistance would be appreciated. Thanks!
How do i calculate the capacitor value if the ac source is from a bicycle dynamo? Any way to make an estimate? Also, does the ready-made bridge rectifier comes with capacitor??
A full bridge rectifier is basically four hardwired diodes .... great piece of electronic hardware ... they commonly have a hole in the middle to allow mounting on a heatsink if you are going to push its limits ....
Thanks man...🙏🙏
In my case adding a smoothing capacitor leads to an increase in power consumption from the grid (x3 times more). Why is this happening and what should be done to eliminate this effect?
Hey Schematix where did you go dude... its 2019... are you still alive out there ?
RaithUK I was thinking that he died
Can someone explain? At 3:38 we are told that _"To calculate the volts after rectification multiply AC input voltage by 1.414... "_ . Why then does the calculation at 8:27 only use 70.5V for the capacitor calculation? Both the 100V (approx.) *and* the 70.5V are referred to as DC voltage "after rectification". Is this because the 100V is only a peak (as referenced in the RMS calculation), and the capacitor should be calculated based on the nominal 70.5V? Thanks for any explanation.
intuitive, so if I want 50amps and 16 volts what type and how many capacitors would I need?
Nice tutorial very educatioal
10:19 - Please Sir, is an "electrolic" capacitor the same as an electrolytic one?
I mean, did you mean to say "electrolytic" rather than "electrolic" ?
You clarification would be appreciated.
Thanks.
Great video by the way.
its just a dialect. or a short form for the all too lengthy word. the electrolytic type is usually choosen in power supplies, as it is well predictable which side is always positive. simply because they are cheaper.
Thanks
Liked & subscribed
Well presented thanks..I have one question regarding my DIY project/experiment..I'm trying to rectify the HV/HF output from the Secondary of my mini Slayer Exciter tesla coil and not sure about the type and the current rating of the rectifier I require..any input appreciated thanks
While you could purchase an ultra-fast recovery diode, the problem is the voltage. A quick browse online shows a mere 40kv or higher diode is in $$$$ range. Even though the humble slayer exciter is a baby tesla coil in the grand scheme of things. It's very capable of outputting over 40kv. I've never heard of anyone rectifying such a high voltage/frequency before.
thanks
Hey raydan18, does your project works? I am also trying to rectify a HV HF voltage with f=49khz. May I know what are the specs of ur rectifier?
Hello, thanks. In conjuntion with you video on modifying a microwave transformer, if I wanted 15VDC, would the following be correct: 11.5VAC (Approx from secondary winding) --> 16.5VCD --> 15.4VCD after bridge (is it a problem if the bridge is rated at 1000V but the real voltage is 15?). For capacitor: 10A*10ms*0.5V (acceptable voltage drop)= 5000 microfarads. Could I get away with a 25V 4700uF? Or a 25V 22000uF? (that is what I have at hand at present), thanks.
His formula in the video is a little sketchy. First, you don't multiply by 10 for 10 milliseconds (milli = 10^-3), you multiply by 0.01 (or by 0.0083 for 8.3ms). Second, the final answer is in farads, not microfarads (micro = 10^-6). For your application, it works like this:
(11.5VAC x 1.414) - 1.1V = 15.2VDC
(10A x 0.01s) / 0.5V = 0.2F
That final answer is 200,000 microfarads. At high currents, it takes a *lot* of capacitance to produce very low voltage drop (ripple). The 26,700uF of capacitors that you have on hand will result in about 3.75V of drop at 10A:
(10A x 10ms) / 0.0267F = 3.75V
A few more things to consider...
If you're measuring 11.5VAC from your transformer with no load, that's *not* what it will produce under a 10A load. It will be significantly less (as much as 20% less, in some cases). If you're getting the 11.5V number from the full-load rating on the transformer's label or data sheet, then you're in good shape.
You don't usually use the transformer-rectifier-filter circuit to achieve your desired final voltage tolerance (0.5V, in your case). Instead, you should design it to supply a *minimum* voltage to your regulator that never goes below its dropout voltage. It's the regulator's job to eliminate gross ripple and provide the stable voltage you need. My point is that if you're making a regulated DC supply, you probably don't need to get 0.5V ripple from your filter capacitor.
For example, let's say you want 12VDC at 1A and you decide to use an LM7812 linear voltage regulator. The LM7812 has a typical dropout voltage of 2V, so the output from your filter capacitor(s) must never fall below 14V. Any ripple above that value (within reason) is handled by the regulator.
Tip: if your design is really tight on dropout margin, you can "save" half a volt or so by making the rectifier out of discrete Schottky diodes.
Yes, a 1,000V rectifier will work fine. Its rating just has to exceed the peak AC voltage from the transformer's secondary winding.
Hello, thank you for the detailed response, appologies for not responding sooner. The application is to give 15V to a car battery at around 5Amps without forcing the transformer/capacitor Etc (for car battery recovery for a solar set-up). I know I could buy this from ebay, just trying to play around (with care). Once I manage to put something together I will share results!
+FlyingShotsman why did you substract 1.1v while converting vac to vac max (vrms * 1.4142)?