Ac to dc conversion after rectification - confused! - Page 1
Forum section: electronics / beginners

author: ScottGAndersen
author_rank: Newbie
timestamp: October 16, 2021, 11:57:28 pm
content: How I calculate what AC voltage I need for a specific DC voltage I want after rectification? .......I want 25 V DC so divide by 0.636 to get PEAK AC of 39V then divide by 1.41 to get AC RMS of 27V??. I have read though also that DC voltage is AC x 1.41 to give 17.7V AC RMS being the AC voltage I should use??, why is there different advice and different answers which is correct? I have googled everywhere and just confused. Many thanks!

author: bob91343
author_rank: Super Contributor
timestamp: October 17, 2021, 12:46:41 am
content: I understand the confusion. Unfortunately, there is no simple answer. When rectifying a sine wave, the unloaded output of the rectifier is approximately the peak value. As the load increases, the output voltage drops because the current from the input wave flows and a portion of the input voltage is replicated in the output. So the output is no longer the peak input, but ripples from peak to a lower value. As load increases further, the portion of the sine input increases and again is replicated in the output.Since the output contains ripple, the average output is lower than the peak. Many efforts have been put forth to reduce all of this to mathematics but there are other factors, such as source resistance, rectifier drop, and is further confused by filter capacitors.

author: ledtester
author_rank: Super Contributor
timestamp: October 17, 2021, 01:31:08 am
content: Quote from: ScottGAndersen on October 16, 2021, 11:57:28 pmI would go with that one, with the following adjustment:DC peak = AC peak - rectifier diode dropVrms = 0.707 * AC PeakThe drop due to the rectifier will be one or two diode drops (i.e. either 0.7 or 1.4 V) depending on the rectifier topology.Some more formulas for different rectifier and filtering configurations:http://www.hammondmfg.com/pdf/5c007.pdfIn this PDF any AC measurement (i.e. V A.C. or I A.C.) is RMS.

author: GlennSprigg
author_rank: Super Contributor
timestamp: October 17, 2021, 11:21:56 am
content: You should follow what 'Ledtester' said above, but with the following addendum...Make the output voltage a bit greater than 25v, (say 30v), and use a Regulator chip!! I don't know what current you want,but if you use an LM138 with a TO-3 Can, or LM338 with a TO-3 Can or TO-220 Package, that will handle 5-Amps.Only 2 resistors are needed, to get any voltage you want, including your 25-Volts. And use an ample filtering Capacitor...Full details here....https://www.ti.com/lit/ds/symlink/lm338.pdf

author: BeBuLamar
author_rank: Super Contributor
timestamp: October 17, 2021, 01:50:36 pm
content: First of all if you put a full wave rectifier at the output of the 17V transformer you get a little less than 17VDC without load due to the diode forward voltage drop. The voltage would jump to 25V only after you put in a capacitor. And that is with no load. With load the voltage will drop but how much it depends on a lot of things like the current capacity of the transformer, the capacitor value etc..

author: TimFox
author_rank: Super Contributor
timestamp: October 17, 2021, 01:56:20 pm
content: For conventional rectifier/filter circuits, I recommend using "PSUD II" freeware from Duncan Labs.https://psu-designer-ii.software.informer.com/2.2/Note that when using this software, you must specify the transformer in terms of no-load voltage and equivalent resistance at the secondary. The information from Duncan includes how to estimate these parameters from the usual specification of voltage with specific load current.