You are fine, the CT means center tap. Don't use it just use the two terminals (wires) that are marked 12v. The regulator will get a little warmer so you might have to use a heat sink.I don't know transformers very well, but I need to use one in a project soon. I am using it in a power supply.
I was looking at the details of a small RadioShack transformer (here ) and I was wondering exactly what the details mean. What does it mean when it says that the secondary voltage is 12V CT? Does that mean V continuous? Meaning DC? Or am I totally wrong?
I was planning to use it in a set-up in the like in the picture below, but with 120V AC instead of 240v AC. Would this work or will something terribly bad happen?
and I was wondering exactly what the details mean.
What does it mean when it says that the secondary voltage is 12v CT? Does that mean 12v continuous? Meaning DC? Or am I totally wrong?
I was planning to use it in a set-up in the like in the picture below, but with 120v AC instead of 240v AC. Would this work or will something terribly bad happen?
How does half the transformer output (6V) become 8.48 V DC? I know it has something to do with the fact that the 120V AC is actually the rms of the actual voltage right? I sort of understand that. But it seems like the diodes would drop the voltage down a bit.
I think I will stick with the 7805, because I have briefly looked into switching regulators, and it seems like I have a lot to learn before I start tinkering with those. Thanks for the advice, though.
One question just occurred to me. This will probably clear itself up once I have the transformer in front of to me reference, but where exactly would the ground go? For example, the diagram shows wires going to the ground, but I could not imagine where that would be when dealing with AC/DC conversions. I've never had to assemble anything with a transformer before, so I only have a vague idea of its leads. Will there be a lead on the transformer that serves as the ground?
Ok I understand the "center-tapped" thing now. But all this mentioning of heating up has got me concerned now. Would a 12V transformer generate that much more heat than a 9V one? I was trying to use whatever parts I could easily get to, i.e. without having to order them online, but if heat is a big problem, I might have to reconsider.
The project I'm working on now is a binary clock. Pretty simple, I know, but I'm just starting out with electronics. I figured that AC converted to DC power would be much more reliable and cheaper than batteries, but if it would not be feasible in this kind of project, it would be nice to know.
Thanks for all the help.
How does half the transformer output (6V) become 8.48 V DC? I know it has something to do with the fact that the 120V AC is actually the rms of the actual voltage right? I sort of understand that. But it seems like the diodes would drop the voltage down a bit.
So according to what everyone is telling me, I guess I will have to use a BA05CC0T instead of a 7805. Hopefully my local electronics store will have something like that.
Torben, you were right in assuming that I was not clear on that stuff. I have sort of taught myself electronics, so I missed out on the formal explanations of voltage, current and such.
Arhi, about the ground issue. So all those arrows on the diagram should be connected to the arrow in the bridge?
Also, what type of diodes would be best to use in the bridge?
Sorry I have so many questions about this. I don't start my EE courses until next year, and I am trying to learn as much as I can on my own and get some experience before I get graded on this stuff.
So according to what everyone is telling me, I guess I will have to use a BA05CC0T instead of a 7805. Hopefully my local electronics store will have something like that.
yup .. those "arrows" actually represent GNDArhi, about the ground issue. So all those arrows on the diagram should be connected to the arrow in the bridge?
Also, what type of diodes would be best to use in the bridge?
Sorry I have so many questions about this. I don't start my EE courses until next year, and I am trying to learn as much as I can on my own and get some experience before I get graded on this stuff.
well, you can get "rectifying bridge" .. it is a 4 pin element that has 4 diodes built in. If you want to make it with diodes any 1n400x would do for your 400mA max. I prefere to use Schottky diodes (as I have bunch of them) so I'd use 1n5822 but that's kinda overkilland I'd use those just because I have bunch of them, not because they are better for this particular case (they have lower voltage drop but not that it is important here)
The 7805's I have here have 1.7V drop ... most of those I saw are around 1.7 ... I think the 2V is the "worst case scenario" so 5.7V is min it *should* work with and 7V is kinda ideal ..
On the other hand, the 1N4007 that I often use (normal rectifiers) should have 1-1.1V drop right but that's @ 1A going trough it ... with 400mA the voltage drop is ~.5V so 6*sqrt(2)-(2*.5) = 7.48 ... thats much over 7V that is kinda ideal for 7805... now, that's my exp, I moved to BAxxCC0T as they are
- almost same price (78xx are 50% of price if you look at digikey but in local shop ba is ~10% more expensive then 78xx)
- come in TO220 that is isolated (so no need for mica's or similar isolators)
- .3V voltage drop typical (max is rated .5) ... every single one I measured had drop between .15 and .2V .. I love that
I also have some 10-20 78xx left ... I use them in devices that are powered from the network, but for batt operated devices is waste of power
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