is my thinkng right? - Page 1
Forum section: electronics / beginners

author: p.larner
author_rank: Frequent Contributor
timestamp: April 12, 2024, 07:03:48 am
content: I have 19volts raw ac input going into a full wave rectifier+ resevoir cap so 19x 1.14=21.66 rms +19v=40.66v,is that how peak to peak to rms pans out after rectification and smoothing at no load?.

author: Damperhead
author_rank: Contributor
timestamp: April 12, 2024, 07:08:15 am
content: To calculate the RMS voltage after rectification and smoothing, you need to consider the peak voltage of the AC input and the characteristics of the rectifier and smoothing capacitor. Given that you have a 19 volts RMS AC input, after full-wave rectification, the peak voltage will be √2 times the RMS voltage, which is approximately 1.414 times the RMS voltage. So, for a 19 volts RMS input, the peak voltage would be approximately 19 * 1.414 = 26.866 volts.After rectification, you would have pulsating DC with peaks at this voltage. Then, after smoothing using a reservoir capacitor, the voltage will increase slightly above the peak voltage, depending on the capacitor's value and the load. But, generally, for no load conditions, it's safe to approximate that the output voltage will be close to the peak voltage. So, in your case, with a 19 volts RMS AC input, after rectification and smoothing at no load, you can expect an output voltage close to 26.866 volts DC.

author: BennoG
author_rank: Frequent Contributor
timestamp: April 12, 2024, 07:08:31 am
content: no it should be 19 x 2 squared so 19 x 1.414 = 26.8 Vso you are doing something terribly wrong.Benno

author: Gyro
author_rank: Super Contributor
timestamp: April 12, 2024, 08:57:30 am
content: Do you really have to start yet another thread on this?https://www.eevblog.com/forum/beginners/are-these-readings-correct/

author: tggzzz
author_rank: Super Contributor
timestamp: April 12, 2024, 09:19:07 am
content: Quote from: Gyro on April 12, 2024, 08:57:30 amThat's the OP's SOP, unfortunately.However, we should give the OP credit for taking account of this thread - which has, remarkably, become "sticky" in less than a dayhttps://www.eevblog.com/forum/beginners/why-writing-style-and-grammar-matters-in-posts/msg5444555/#msg5444555

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 12, 2024, 12:30:21 pm
content: Quote from: p.larner on April 12, 2024, 07:03:48 am19volt what? peak to peak? rms? we dont compute what "raw" is...Quote from: p.larner on April 12, 2024, 07:03:48 amhavent you learnt the right notation? 2 algebraic terms with different algebraic variables cannot be added together. rms is not the same as v... and thats not how math workflow is written (suddenly add 19v out of nowhere) if you cant organize it in a manner that hopefully your audience can understand. how can you expect an answer that you can understand?

author: BeBuLamar
author_rank: Super Contributor
timestamp: April 12, 2024, 12:36:09 pm
content: 19VAC into the full wave rectifier without the cap the output would be about 19VDC. Some different because the DC is average and the AC is RMS. Only with the cap that it would be about 26VDC because the cap stores the peak value (there is no load to drain the cap).Just did a calculation and without the cap the DC voltage is 17.11VDC.

author: CaptDon
author_rank: Super Contributor
timestamp: April 12, 2024, 12:42:06 pm
content: P.L, I couldn't even guess where you came up with the formula you used, and it is 1.41 not 1.14. At any rate just google up linear power supply design. All 50 of your current questions plus 50 of your next questions will be covered in detail. Look at the data sheets for the linear I.C.'s, they will tell you the max +/- rail voltages. You don't need to ask here. It is all covered in great detail in datasheets, online tutorials and manufacturers application notes. As noted in one of my prior replies, 19 X 1.41 = 26.8 (and probably more with the transformer unloaded) so figure around 28 for the positive rail REFERENCED TO THE GROUND RETURN RAIL, and your schematic shows an auxilliary rectifier for the negative rail which may provide another -25 volts. Those voltages appear to go through a regulation circuit so maybe + and - 20 volts or so regulated. Read and learn, then you can put the kludge together on your own and get predictable results without questioning every power supply design parameter since the dinosaurs.

author: Zenith
author_rank: Frequent Contributor
timestamp: April 12, 2024, 12:58:54 pm
content: There's even a handy bit of software that runs on Microsoft operating systems, and lets you play with a range of transformers, rectifiers, rectifier configurations, filters and loads. It produces graphs. I've used it a few times in the past.https://www.duncanamps.com/psud2/You might have to join the group to download it these days.There are no doubt others.

author: CaptDon
author_rank: Super Contributor
timestamp: April 12, 2024, 01:14:33 pm
content: Zenith, that is indeed an excellent bit of software! I have used it when designing several high voltage supplies for RF amplifiers. The more details you enter, the closer the results are to the real world tests. I have used it to work out different bleeder resistors vs. voltage regulation for power supplies used with Single Sideband amplifiers where the load current varies greatly. Slightly heavier bleed current gives better regulation with slightly better IMD and THD values coming out of the amplifier. I think the rules read roughly like "Everything coming out of the transmitter that isn't the primary signal must be collectively at least 30dB below the primary signal". That would imply 'all the junk added together'. Cheers mate! Thumbs up to that Duncan site!!