Calculating smoothing capacitors - Page 1
Forum section: electronics / beginners

author: AtomAmp
author_rank: Newbie
timestamp: June 19, 2014, 04:56:02 am
content: Hi, new person here and somewhat confused about calculating smoothing capacitors. I'm in the US so my values won't be the same as most others. I've done some searching on calculating smoothing capacitors and I found a tutorial that I like and is easy to follow however, I have a question at the end. It goes like this:1. Determine the output voltage of the secondary winding. ( my example, 30v ac )2. Subtract 1.4v from this value to determine the output voltage of the bridge rectifier. ( 30v - 1.4v = 28.6v )3. Multiply step two's value by the frequency of the secondary winding. ( 60hz * 28.6v = 1716 )4. Multiply the total current of your load by 5. ( I will be drawing 4a so, 4 * 5 = 20 )5. Divide the value of step four by the value of step three. ( 20 / 1716 = 0.011655 )6. Express this value in microfarads by multiplying by 1000000. ( 0.011655 * 1000000 = 11655uF )Pretty simple. here's my question, would the frequency be doubled because I'm using a full wave rectifier?Hence making the capacitor 5800uF instead.Thanks for any help...

author: Shredhead
author_rank: Regular Contributor
timestamp: June 19, 2014, 07:09:35 am
content: I am dumb so don't take this the wrong way but is there a way to factor transient burst power into the formula for the smoothers too? Like an amplifier that has a high burst wattage for a period of time before it sags?

author: PChi
author_rank: Frequent Contributor
timestamp: June 19, 2014, 10:16:05 am
content: I am guessing that the formula assumes that it is for a full wave rectifier circuit. Diodes are cheap and capacitors expensive so half wave rectification is rare.I wouldn't get too hung up on the formula becuase it is a simplification and is ignoring transformer output impedance (regulation). It is also ignoring what is an acceptable ripple current through the capacitor.A rule of thumb that I heard about a long time ago is 1,000 uF per amp was just OK but that 2,000 uF per amp is better. It's more complicated than that though. Either simulation or build and measurement are required.

author: mariush
author_rank: Super Contributor
timestamp: June 19, 2014, 11:07:15 am
content: for 30v AC and 60Hz (assuming US)A full wave rectifier converts AC Rms voltage to DC, the peak dc voltage (before the losses in the diodes of the bridge rectifier) being Vpeak = sqrt(2) x Vac = 1.414 x 30v = ~ 42.5vThe current output of a transformer is aproximated with the formula Idc = 0.62 x Iac ... let's say you have a 30Vac 100VA transformer, then the maximum dc current would be Idc = 0.62 x (100va/30vac) = 2.06ANow note that transformers are designed like this and usually output a higher voltage at low currents. This is especially the case with small transformers (let's say under 50-75VA) , the ac voltage coming out of the transformer at low currents can be 5-10% higher... so your peak DC voltage (before rectifier losses) can be as high as 1.414 x 33-35v = ~ 49.5v max. Very important to known when choosing voltage rating of capacitors. For this example. you wouldn't choose 50v rated capacitors, you would go with 63v or even 80v.. or even higher, depending on your current ripple requirements and.. well, sometimes you may want bigger rated capacitors simply for their better temperature rating, for longevity etc etcAs the AC voltage is rectified, there is some loss in the rectifier, the current always moves through two out of four diodes in the rectifier. The voltage drop varies with the current, for example a 10A rectifier could have 0.6v drop per diode at 100-250mA but could go as high as 1v per diode at 8-10A.Going back to our example and ignoring variation of voltage at low currents, let's just assume the bridge rectifier will have 0.8v drop per diode, so we'll have 42.5v - 1.6v = ~ 41v. Remember, this is a peak DC voltage, you'll still have 120 pulses of energy a second, with valleys and peaks. This is where smoothing capacitors come in.A formula that approximates the capacitance is C = Current / ( 2 x ACfrequency x Vripple) where Vripple is how much you're willing to let the voltage sag from the peak dc voltage.With our figures (41v peak dc after rectification, 2.06A, 60Hz AC mains frequency) let's say we're fine with voltage as low as 38v. In this case, the capacitance can be approximated to : 2.06A / [2 x 60 x (41v-38v)] = 2.06 / 360 = 0.0057222[..]222 Farads = ~ 5722 uFNow of course, you can go higher than this for safety, or for convenience, nobody stops you. Just keep in mind that higher capacitance means when you plug a transformer in there's going to be a surge of current going into the capacitors so especially with big transformers you have to be careful not to pull more than the fuse rating of your mains socket.It's possible to add a thermistor in series with the transformer - as current flows through it it heats up and its resistance changes so the current is limited initially for a few seconds and then as it gets warm it lets more current flow.

author: dentaku
author_rank: Frequent Contributor
timestamp: June 19, 2014, 07:52:46 pm
content: That was very educational mariush

author: Lightages
author_rank: Unknown
timestamp: June 19, 2014, 08:45:11 pm
content: A quick method I learned from reading Don Lancaster in Radio Electronics:"In an 8300 Microfarad Capacitor, the VOLTS of ripple will equal the AMPS of load current."Of course this only applies to 60Hz and full wave recitifcation.

author: AtomAmp
author_rank: Newbie
timestamp: June 19, 2014, 10:10:08 pm
content: Shredhead,Apparently I'm dumber because I don't know what you're talking about.PCHi,Thanks for your input.Mariush,Thanks for the lesson but you're over my head. The results that you came up with however, answered my question.I do need to double the Hertz. Again thanks for the voltage correction. I will change that.Dentaku,That was quite educational wasn't it? I'm almost 60 years old and still learning while I tinker. An O-scope sure would come in handy. I have to see it to get it.

author: AtomAmp
author_rank: Newbie
timestamp: June 19, 2014, 10:14:58 pm
content: Lightages, What?

author: bingo600
author_rank: Super Contributor
timestamp: June 20, 2014, 02:34:39 pm
content: @MariushGreat explanation , but ..... I'm in 50HZ landWhat would be the difference in the above calculations ?/Bingo

author: mariush
author_rank: Super Contributor
timestamp: June 20, 2014, 03:07:11 pm
content: You have in the formula C = Current / (2 x AC Frequency x Vripple) where AC Frequency is your mains frequency, 50 Hz....In the post above, i used 60 Hz because OP is in US, hence my first line "for 30v AC and 60Hz (assuming US)"So instead of C = 2.06/(2*60*3), you would have 2.06/(2*50*3)

author: Lightages
author_rank: Unknown
timestamp: June 20, 2014, 03:07:27 pm
content: Quote from: AtomAmp on June 19, 2014, 10:14:58 pmSorry if that was too simplified. The idea is that if you want to know what ripple you will have, then if you are going to draw 0.1 amp from the power supply and you use a 8300 microfarad cap, then you will get around 0.1 volt of ripple. If that much ripple is not to your liking, then double the capacitance and you will get 0.05 volt ripple. It is an easy way to remember the formula.For 50Hz, the capacitor needed for the same ripple will just be the ration of the frequencies, ie: 60/50 x 8300, or closer to 10000 microfarad.

author: bingo600
author_rank: Super Contributor
timestamp: June 21, 2014, 09:46:26 am
content: Quote from: mariush on June 20, 2014, 03:07:11 pmThanx mariush , wonder how i missed that in your post/Bingo

author: David Hess
author_rank: Super Contributor
timestamp: June 21, 2014, 12:35:10 pm
content: I use the same rule although I did not know Don Lancaster came up with it. In factor-label notation:8200uF V / Aor8200 microfarad volts per amp