fluke Posted February 10, 2009 Report Share Posted February 10, 2009 Without any rectification, the capacitors on the input side of the 7805 are simply acting as resistors to ground, with a resistance of -j/2?fC (ie 1/2?fC with 90° phase lag). So all you've got there is effectively 0.4? resistor to ground, which would explain the ripples. Quote Link to comment Share on other sites More sharing options...
/tilted/ Posted February 10, 2009 Report Share Posted February 10, 2009 The waveform doesn't look like clean AC.It looks more to me like the AC is being shunted by the bridge, when it gets above or below certain points.These points are around +/-10mV from the baseline voltage (I had the oscope set to AC coupling, because I don't seem to have an offset, so this is the only way to look up close).And the transformer output is rectified. it's just sort of done weirdly. I have finally figured it out.It's kind of like a bipolar suppy, where the reference is 5V, the negative is 0V, and the positive is 9V. this is essentially the same as a +4V,0,-5V supply.Which is something that occurs to me. m00dawg, when your supply bogs down to ca 4.5V, does your 9V rail stay dead-on stable? Quote Link to comment Share on other sites More sharing options...
seppoman Posted February 10, 2009 Report Share Posted February 10, 2009 Hi,I don't have much time to answer right now, just one comment:"truncated AC" - you did notice that m00dawg has drawn the transformer input terminals in a way that doesn't make reading easier, i.e. the wire going to the 7805 input should be connected to the center tap?Sedit: 808 posts, cool ;) Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 10, 2009 Author Report Share Posted February 10, 2009 @/tilted/Actually the 9V bogs down too I believe, just not near as much. So if I'm doing 4.5V, the 9V is around 8.98V or something around that. It's definitely more stable than 5V. I can hook the board back up and run another test to see if you need concrete numbers.@seppomanI did that so to make the traces easier on the board. My apologies for the confusion there. Center-tap is X1-3 in the schematic I posted last. Quote Link to comment Share on other sites More sharing options...
/tilted/ Posted February 11, 2009 Report Share Posted February 11, 2009 It's alright, I knew that.In my hook-up, the taps are: 0 - 6V3 - 9V40V and 9V4 are connected to the bridge, with the 6V3 tap to the 7805.Think hard before you jump on me for this, remember, I'm not using a centre-tap tranformer.Another way to say the same thing would be that the "0 tap" is to the 7805, while the "6v3 tap" and the "3.1 tap" are to the bridge.This may explain the excess of AC in my proto, I'll switch to a 0-6V3-12V6 (ie 12V6 CT) when I get home from work tonight. Quote Link to comment Share on other sites More sharing options...
seppoman Posted February 11, 2009 Report Share Posted February 11, 2009 Another idea: maybe 6 or 6.3V is not enough to supply the 7805 under load condition?S Quote Link to comment Share on other sites More sharing options...
stryd_one Posted February 11, 2009 Report Share Posted February 11, 2009 6V3DC is wishful thinking. AC, maybe. Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 11, 2009 Author Report Share Posted February 11, 2009 I know this is a silly, and probably basic, question, but how do I read "6V3DC", "12V6", etc.? I assumed it had something to do with the center-tap, but I'm especially confused by "0-6V3-12V6" ?As for the regulator dropping out, I don't think that's my issue - or, if it is, it's dropping out and coming back in rapidly, which may be produce the lower voltage (someone mentioned that in an earlier post in this thread, but I can't seem to dig it up). Otherwise, I should be getting close to 0V off the regulator, not 4.3V right? Quote Link to comment Share on other sites More sharing options...
stryd_one Posted February 11, 2009 Report Share Posted February 11, 2009 6V3 is another way of writing 6.3V, so 6V3DC is like 6.3VDC (as opposed to 6.3VAC). Sorry about that, was just following the conventions I saw in the few previous posts. Perhaps I should be following this thread more closely, if I get a chance to finish the code I'm working on today I'll read the whole thing.I think that what seppo is suggesting is that you are approaching/traversing the dropout curve on the regulator. It's a pretty steep drop, so usually you'll get either 5V or nothing, but there *is* a sweet spot where your input can be too low to give the desired 5V output, but not quite low enough to cause the output to drop out entirely. Check out the 7805 datasheet to find the graph. It is very rare to hit this sweet spot, but I've done it a few times. Guess it depends how unlucky you are ;) Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 12, 2009 Author Report Share Posted February 12, 2009 Thanks for the explanation on the notation, stryd!The sweet spot could be a possibility. I'll need to see what voltage I'm feeding it, although /tilted/'s findings also explain things. I probably am feeding it more AC than DC given that I have no diode off the center-tap. It's also weird that the voltage continues to drop in proportion to the load if I found the "sweet spot"?I'm going to be bringing this issue up at my local hardware meeting next Monday so maybe something will come of that since I may have access to an O-scope. *shrug*Either way, this is a really fun exercise in basic power supply theory :) At least I'm learning something from all of it :) Quote Link to comment Share on other sites More sharing options...
/tilted/ Posted February 12, 2009 Report Share Posted February 12, 2009 Yeah, sorry about the 6.3V / 6V3 thing. It's an old convention which probably doesn't need to be followed any more. I think the original idea was that if you (for example) make repeated photocopies of something saying 6.3V, or write it with a bad pen, the decimal point might get hard to read, and the figure becomes 63V, and errors creep in. If you write 6V3, you'll lose the letters before you lose the decimal place. This has been taken for all sorts of measurements, eg 1R5 = 1.5 ohms, 1k8R = 1800 ohms (1.8k ohms)Really not an issue in an online forum though...Anyway. I wonder if the problem here might just be one of pure circuit topology. By this I mean that the rectification and primary DC smoothing seems similar to that used in a bipolar supply, but the regulation is taken more from a dual rail configuration.Mostly these differences seem cosmetic, but if you think about it for a second... in a bipolar supply, maybe the 0V rail would always wobble around like this, and have this amount of AC on it, but since the regulators are referenced to the 0V rail, it isn't an issue. This brings us back to the flapping bird analogy. Perhaps the issue here is that the 5V rail is the one that is wobbling the most, and the 9V is not referenced to the 5V, but to the 0V rail. Perhaps the bird's two wingtips are nailed (referenced to each other), and the bird's body is moving. -So the 5v is swinging around in the middle somewhere. I wonder if the situation would be changed if it were a -5V - 0 - +4V supply, with a negative 5volt reg, and a positive 4volt reg? Quote Link to comment Share on other sites More sharing options...
stryd_one Posted February 12, 2009 Report Share Posted February 12, 2009 Yeah, sorry about the 6.3V / 6V3 thing. It's an old convention which probably doesn't need to be followed any more.Doesn't hurt. Yeh maybe it's obsolete on a forum, but It's easier to read/harder to 'miss' the dot too. I learned this crap from the military so maybe I'm biased, but I like it. :) Anyway back to our scheduled programming ;) Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 14, 2009 Author Report Share Posted February 14, 2009 -So the 5v is swinging around in the middle somewhere. I wonder if the situation would be changed if it were a -5V - 0 - +4V supply, with a negative 5volt reg, and a positive 4volt reg?Can you elaborate on that a bit? Are you saying that can be somehow used to produce 9V as well as 5V? You'll have to forgive me, here, bipolar supplies are something I know little about. Speaking of, if someone knows of a good source for reading up on this that would be awesome! I know there's good Wiki articles on how to build one (they've already been sited in this thread) but I don't really understand the difference between positive and negative DC voltages, other than the absolute basics. Quote Link to comment Share on other sites More sharing options...
/tilted/ Posted February 16, 2009 Report Share Posted February 16, 2009 Well, technically. yes. It could be used to create a +5v and +9v supply rail. It would be a really weird way to do it though, and you'd be leaving yourself open to really cocking things up if you made a slip.Given that in said configuration, you'd also be running the "GND" through a regulator, I don't think it's really advisable.Personally, I've never had any problems with the old 'multitap transformer, several rectifiers and several regulators' approach.Maybe there was some compelling reason why this approach wasn't used here. Guess I'll have to re-read the thread. Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 16, 2009 Author Report Share Posted February 16, 2009 Personally, I've never had any problems with the old 'multitap transformer, several rectifiers and several regulators' approach.Maybe there was some compelling reason why this approach wasn't used here. Guess I'll have to re-read the thread.That was discussed, but not in great detail. I wasn't sure how to derive GND, err, unless you're talking about a true multi-tap transformer (as opposed to a center-tapped one). I've been having a hellofa time tying to find a multi-tap that can output multiple voltages and don't see a huge benefit over, say, a dual secondary transformer over a single (as long as it is rated high enough).My current thought is to just use a regular 9VAC transformer, rectify and regulate that down to 9VDC and then hook another regulator up to that to step down to 5VDC. I've done that before with success and, apart from heat, it seems to work reasonably well (though I haven't used it to power a full control surface yet). Quote Link to comment Share on other sites More sharing options...
/tilted/ Posted February 17, 2009 Report Share Posted February 17, 2009 OK, well for the dual-rectifier design, there's no reason you can't use your current transformer. It is essentially a multi-tap transformer with two taps, providing symmetrical voltages.The other good news is that you can use the board design you have currently, sort of.At the moment you have a CT transformer with the outer taps connected to the AC (^v) points of a rectifier, the (-) becomes your 0 volt, the (+) feeds some caps, a regulator, more caps, etc. (Pardon the appalling ASCII art)..."All you need to do" is connect another bridge rectifier's AC points to:(1) the CT of your transformer, and (2) one (NOT BOTH) of the other taps.then:the (-) of this rectifier is tied to the (-) of the other rectifier (as these are both for your 0 volt), andthe (+) of this rectifier is fed to the filter caps, 5v regulator etc of your board.This would have a similar effect to simply adding a single diode between the CT and the regulator, but instead of being half-wave, it is full-wave (which is much more efficient, and will allow for more current, with less loading on your filter caps).This is of course only one of many ways to do it, but it's the way I've done it, and tend to do it. Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 17, 2009 Author Report Share Posted February 17, 2009 Ah well I actually tried doing that very thing in an earlier design and somehow ended up with the same voltage after both rectifiers. I did run that design through the simulator and ended up with weirdness too. Specifically when tying the negatives together. I can have another go at it though. Quote Link to comment Share on other sites More sharing options...
m00dawg Posted February 20, 2009 Author Report Share Posted February 20, 2009 Brief update - I received the printed boards from BatchPCB and thought I'd see if I could at least replicate some of /tilted/'s findings. Sure enough, I end up with around 3VAC off the center tap to the GND (the - on the rectifier) which seems to indicate it is quarter wave.As I mentioned before, while I still want to eventually tackle this, I think I'm going to work on a more standard setup so I can power my MB-SID and finish that up. After that I might circle back around and see if I can make this design work. Quote Link to comment Share on other sites More sharing options...
mescalinum Posted January 28, 2010 Report Share Posted January 28, 2010 Personally, I've never had any problems with the old 'multitap transformer, several rectifiers and several regulators' approach. what's this approach are you talking about? I ask because I want to use a multitap transfo (0 - 4.5 - 6 - 7.5 - 9 - 12 - 15 - 18 - 24) to power a SID module and a CORE module, but I've read bad things on this forum thread (like the low voltage "shifting" up a lot). I was thinking about two options: 1) use 0 - 7.5 for sending it to CORE module, getting stable 5 VDC, and use 0 - 15 (or eventually 9 - 24) for sending it to SID module, getting stable 12 VDC. this basically puts the two loads on the same windings, right? 2) cut the secondary winding between 7.5 and 9, so that I get two secondary windings. (this isn't multitapped anymore, right?) that way I have 7.5 VAC on 0 - 7.5 pins, and 15 VAC on 9 - 24 pins what do you think about that? any counter-indication? I'd prefer option 1, cause I don't know which is the right wire to cut: I should cut the winding between 7.5 and 9, but two wires are attached to pin 7.5 and two wires are attached to pin 9. If I do the wrong cut I trash the transformer. (moreover, I'm not even sure 2 is completely correct) Quote Link to comment Share on other sites More sharing options...
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