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

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

Thanks kindly for your support ;)

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 :)

Thanks mate :)

Old info, bin there, done that 25 years ago.

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.

Thanks ;)

You've made my day ;)

:)

are you sure its cap....could be a MOV such a cap can reduce line noise/spikes

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

Zener diode is used when you need a specific vpltage , you now get a zener diode of dat rating connect as required

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

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 ;-)

yes