How to use a Half and Full Bridge Rectifier, in the Same Circuit? - Page 1
Forum section: electronics / beginners

author: t1d
author_rank: Super Contributor
timestamp: September 29, 2018, 01:27:24 am
content: Disclosure... I posted this question, on my e-load development thread, but I didn't get an answer. Instead of bumping it, as it is a topic unto itself, I thought I would post it here.https://www.eevblog.com/forum/projects/electronic-load-project-ltv171-irfp250-with-kicad-files/msg1842893/#msg1842893I would like to add a 12vdc output, to drive a cooling fan. But, I am not sure how to create the PSU, for it, because it needs a full bridge rectifier and there is already a half bridge rectifier on the board. Crossing up the ground with the negative source is the concern. See picture.I will continue researching this. But, I am not finding anything on using the two types of rectifiers, in one circuit.If you know the solution, please clue me in… Maybe, use the rectified +9vdc and -9vdc, being 18v and regulate that? Input = +9vdc, ground = -9vdc, output = 12vdc, with the fan input 12vdc and fan ground tied to circuit ground?I am a little upset, with myself. I should be able to figure this out... But, I just can't see it... Yet.

author: ArthurDent
author_rank: Super Contributor
timestamp: September 29, 2018, 02:41:48 am
content: In the upper part of your drawing you show that 12AC2 ic connected to ground. If you also draw that ground in the lower part of the drawing where you have 12AC2 to make it clearer you will see the diode I labeled ‘3’ has both sides grounded so it does nothing in the circuit.The diode I labeled ‘2’ is right across the 12VAC feeding the circuit so it is a short circuit. The diode I labeled ‘1’ (if ‘2’ was removed) would give a + rectified half wave at the #1 pin on the connector. Because this lead would only be positive, the diode I labeled ‘4’ would always be reverse biased and will do nothing.There is no reason to use a full wave bridge (the one drawn has '2' & '4' backwards as well) and the two supplies you show are only half-wave rectifiers (no bridge). If you remove diodes ‘2’,’3’,and ‘4’ and put another 220 uF cap in place of ‘4’ it will work. You might be able to connect the fan to the positive power flag but adding just one diode and one cap will give better isolation from noise.

author: t1d
author_rank: Super Contributor
timestamp: September 29, 2018, 03:24:20 am
content: Thank you, AD, for your help.As the diode and cap, that you suggest be added, already exist in the circuit, I have simply tapped the leg, after them. Please confirm that I have it correctly; yes?Will this cut the power in half, by only using one half, of the ac cycle?I will certainly have to scratch my head and suss it out. <grin>EDIT: Oops, I forgot to remove the 12AC2 leg, from 12VAC Jumper, Pin #2, but I know it should be deleted.

author: SeanB
author_rank: Super Contributor
timestamp: September 29, 2018, 10:50:41 am
content: Correct there, but increase the value of C10 to something like 470uF to 1000uF, 25V. you are now drawing more current through the positive half of the circuit, to power the fan, so the ripple voltage there will be higher. The voltage at the fan will be around 14-15V, so, depending on the fan and how much current it is using, a series resistor on the fan power might be needed to drop some voltage, so the fan is not going to overheat the internal coils, or be noisy. For a typical fan of 12V 100mA, around 10R 0.5W will do, and then you can add another capacitor (100-220uF 25V) after it to your GND point to reduce electrical noise from the fan being fed back to the rest of the power rails.Depends on the transformer and what you are powering as to how good this will be, but as this is going to power analogue circuits probably a good thing.Alternatively you place the bridge rectifier on the AC1 input and the GND, and treat the output as floating, and use a 1000uF capacitor to smooth it, and connect the fan across this supply directly, or via the second capacitor and 10R 0.5W resistor. This at least leads to the lowest DC component in the power transformer secondary, making it run cooler as it is not having a constant DC ( the half wave rectifier effectively means the transformer has a nett DC component applied to the core) offset pushing the core further into saturation on one half cycle.

author: 6PTsocket
author_rank: Regular Contributor
timestamp: September 30, 2018, 05:04:42 pm
content: Without any additional components, the bridge is wired wrong there should be no direct, common ground connection between the AC and the DC sides of the bridge A bridge has four connection points: two are AC in and the other two are + and - DC out. You are totally mis wired. If you are not using a marked, single bridge component, make sure the diodes are facing correctly and your AC and DC are connected to the right places on the bridge. Your circuit has numerous errors there. The bridge DC output should be floating, directly off the + and - if the bridge. You cannot tie the DC to one side of the AC you have marked as ground.You also need a filter cap across the bridge as thr fan may not work well on raw DC.I have no problem with your half wave and full wave outputs. See my crude drawing on hiw it should be wired.Sent from my SM-G900V using Tapatalk

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 10:33:06 pm
content: Oh, my gosh, Sean and 6PT, I am, just now, finding your solutions. I was notified of other posts, to this thread, but not yours??? Ian ended up getting me straightened out. Thank you, so much, for your posts.Sean, great suggestion, on adding a capacitorafterthe fan.Additional info on this subject can be found here:https://www.eevblog.com/forum/beginners/capacitor-calculations-correct/?action=post;num_replies=8