transformer question

[soadrage7654]

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 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?

 

[Rolf]

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?

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.

 

[Torben]

Important: what kind of current capacity would you like the power supply to have? 450mA on the transformer is pretty small.

'CT' means 'centre-tapped', which means the secondary side will have 3 leads (the primary side is the side which connects to mains power; the secondary is the side which connects to the power supply circuit). On a mains transformer with a 120V primary, 12V CT means that when the primary is given 120VAC, the secondary will develop 6VAC between the centre lead and each of the outer leads, and 12VAC between the two outer leads. In your case, you'd use the outer two leads on the secondary and ignore the centre tap.

Hm. On second thought I think you'd want to use one outer lead and the centre tap and ignore the other outer lead. This would provide you with 6VAC out of the transformer, which after rectification and filtering will give ~8.5VDC. That should be enough to keep the 7805 happy, if you're drawing just small amounts of current from the power supply. I haven't worked out the ripple for this circuit yet (dinner is just about ready) but if nobody beats me to it, I'll take a look at that later tonight.


Cheers,

Torben

[Edit: Looks like Rolf beat me to the post and confirms my thinking that you'd be better off using just one half of the secondary.]

 

[soadrage7654]

Just to clarify, the transformer will put out 12v AC and the diodes are just for converting it to DC right?

Will selecting a transformer with a lower current output help to keep it cooler, or will I need a heat sink regardless? The project I am working only needs about 100 mA.

 

[arhi]

and I was wondering exactly what the details mean.

This is simple 120VAC to 12V AC
Maximum current you can draw (on 12V AC line) is 450mA

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?

Nope, the output is AC.

EDIT - IGNORE THIS: read: Current transformer - Wikipedia, the free encyclopedia

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?

should work,

note that 7805 will get hot and this will not be very efficient (as 7805 turn excess voltage into heat)...

some DC/DC or switching regulator might be much better AM1D-1212D-NZ for example (DC/DC) or LM2575T-5.0 (switching) or BA05CC0T (LDO)...

EDIT: for 5V DC output 9V AC transformer is most then adequate, 12V will just produce more heat

 

[soadrage7654]

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.

 

[arhi]

the 12V AC will enter the 7805 as 17V. 7805 will transfer difference from 17V to 5V into heat .. so, yeah, it will be hot ...

if you use DC/DC regulator or switching regulator (AM1D-1212D-NZ or LM2575T-5.0 for example) the regulation is bit different and you do not have that much heat emission.

If you insist on 7805, note that 7805 has voltage drop of 1.7V so it need 6.7V DC in order to produce 5V on the output and to emit minimum heat. If you use only "half" of the transformer 6V AC will be 8.48V DC and that's much closer to 6.7V then those 17V hence much less heat generated by 7805.

 

[soadrage7654]

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?

 

[arhi]

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.

After those rectifying diodes and capacitor it gets close to Vdc = Vac * sqrt(2)

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.

I understand but then at least use BA05CC0T if possible. It is similar as 7805 (TO220, 3 leads) only it has only 0.3-0.7V drop (7805 has 1.7V)

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?

nowhere
The ground goes from rectifying diodes, not from transformer. The leads from transformer go on one side to N/L 110V AC, and on another side to rectifying diodes. From rectifying diodes you get GND.

If you are asking about mains ground then it can go to the "case"

 

[Torben]

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.

No--as Rolf, arhi, and I suggested, just use half the transformer and you should be fine.

One thing which I'm not sure you understand (and forgive me if I'm wrong here) is that voltage is "pushed" into a circuit by its supply, and current is "pulled" out of the power supply by a circuit. So say you have a device which requires 5VDC at 100mA. You need to match the voltage requirement with the voltage rating of the power supply so that enough voltage is pushed in to run it but not so much voltage that the device burns out. However, the device will only pull as much current as it needs. So you could use a power supply rated at 5V/100mA, or 5V/10A, and either would work fine.

The problem with using the full 12VAC output from the transformer is that 12V will be pushed into the regulator, and it will have to burn off the excess as heat. Using only half the secondary means that there is much less excess voltage which the regulator would need to burn off.


Regards,

Torben

 

[Torben]

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.

Yes, you're right. And I was hungry and forgot to include the diode drops from the rectifier. You'll wind up with (6V * 1.414) - 1.4V (from the diode drops) which means that if everything is perfect you should have 7.084VDC going into the regulator--just barely enough to make the 7805 happy since it has a dropout voltage of 2V. I don't think that's enough headroom; I'd think a low dropout regulator would be a better choice. Or you could use Schottky diodes in the bridge, as they have a lower forward voltage drop than regular diodes. Or you could use a 9V transformer, as in the original schematic.


Regards,

Torben

 

[arhi]

the diodes will drop 1.4 max .. if you put standard rectifier you will get ~.8V dropout, even less...

BA05CC0T rulez go with it it is only 1$ (ok 7805 is .5 but ..)

 

[soadrage7654]

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.

 

[Torben]

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.

If you used Schottky diodes in the bridge, you could still use the same transformer and the 7805 because the voltage drop over the diodes would be much less. Of course, the problem with that is that I don't know whether Radio Shack carries Schottkies.


Torben

 

[arhi]

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.

donno where you are located but digikey.com / digikey.co.uk have them.

Arhi, about the ground issue. So all those arrows on the diagram should be connected to the arrow in the bridge?

yup .. those "arrows" actually represent GND

Also, what type of diodes would be best to use in the bridge?

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 overkill and 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)

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.

as long as you do not ask us to do your homework we are all happy to help. and if you do some research before you ask a question, as if I find answer to your question by pasting it to google and looking at the first few results - I'm gonna ignore the question

 

[soadrage7654]

Well, thanks again for all the help. I think ya'll have finally made the majority of this circuit clear to me. I can't wait to see how this turns out, too, because this is my first major electronics project.

 

[Torben]

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 overkill and 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)

Hi arhi,

Now I have a question for you. Are you saying the lower Schottky drop isn't important here just when using the BA05CC0T, or do you mean it's also not important in the case of the 7805? I know the regular diode drop gives just barely enough excess voltage to allow the 7805 to work within spec, but is there really enough headroom for it to be reliable? Just wondering. I don't like having to burn off more voltage in the regulator than I need to and I still have a bunch of 78xx around which I want to use up before I start buying better/more modern regulators.


Cheers,

Torben

 

[arhi]

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

 

[Torben]

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 ..

Ah, OK. I've never heard the 1.7V dropout figure for 78xx; the datasheet says 2V and that's what I've always heard of as the typical value so that's what I've gone with. I've also understood that since it's 2V 'typical', it could actually be slightly more than 2V for a given unit.

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

Thanks for the thorough answer. I also think regular 78xx aren't a great solution for battery-powered devices since they always waste the extra power. But for my purposes (which are usually powered from the mains) it's usually good enough. Most of my battery-powered designs so far haven't required that level of regulation. But then, I'm a weekend warrior, not a pro.


Cheers, and thanks again,

Torben

 

[arhi]

I measured 1.7, the 2.0V is in datasheet iirc, but last time I looked at 7805 datasheet was before internet ...

also, if you have voltage drop of 2.2V for e.g. and Vin of 7.0V the Vo will be 4.8 ... that;s tolerable for 99% of 5v powered chips

I'm weekend warrior too but I like making things that move so battery operation is kind of a must ... I got some "expensive ~20$" dc/dc hybrid 5V and 3V regulators (in sip package covered in black gue) by mistake (I was ordering bunch of stuff and ordered them by mistake, 5 of each) so I use them (move them from project to project ) for power critical stuff .. Vdrop is ~0 the efficiency is near 1 (IIRC the datasheet showed something like .98) ...

I'm moving to msp430 and that's another philosophy now ... and I'm still searching for "proper BJC's", have not found the proper (MOS)FET that can be acquired in local store that can be opened with 3V3 .. but darn uC uses less power then battery is leaking on itself ... very very nice