# Center Tapped and Regular Transformer

• posted

Hi,

Could you please explain the following points regarding a center tapped transformer (some questions apply to a non-center tapped "regular" transformer as well):

a. Is the center tapped transformer wound differently than a non-center tapped transformer? Or is it just a regular transformer for which the center point of the secondary winding is "brought outside".

b. Considering the secondary voltage of a transformer is Vs, the two terminals of the secondary are at +Vs/2 and -Vs/2. This implies a voltage gradient across the secondary. The gradient passes through a zero point which we "tap". What causes this voltage gradient?

c. Can we say that all the turns in the secondary winding of a transformer have the same amount of flux passing through them at a given instant or do they have a different amount of flux (with the flux depending on the position of the turn)?

Thanks, Anand

• posted

I don't know the answer to these, and I'm interested to know, but they are sound very much like homework questions and I wouldn't be surprised if the answers you get are along the lines of "do your own homework".

• posted

For power and low frequuency transformers, they just bring out the midpoint of a consecutive, layered winding. For high frequency stuff, the winding tends to be bifalar, with the wires of the two windings side-by-side. You can tell which is which by measuring the DC resistances. For bifalar, the resistance from CT to the two ends will be very close. For a tapped layered consecutive winding, the outer winding has a bigger radius so the wire is longer and has more resistance than the inner one.

Each turn of the secondary has a time-varying magnetic field pass through it, so each turn has a voltage induced into it. The magnitude is typically on the order of 1 volt RMS per turn, for power transformers at least.

Pretty much the same. The flux is concentrated in the iron core, and all turns loop that core. There is a small amount of magnetic flux that cheats and flows through space, not iron, that can make small variations in the flux on different turns. This causes "leakage inductance", usually under 1% difference between different turns.

```--
John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com```
• posted

"John Larkin"

** That would need to be a rather large ( multi kVA) transformer.

The "rule of thumb" for a laminated iron E-core is 5 to 6 turns per volt, for 1 square inch core cross section.

( 5 turns if it is for 60 Hz only and 6 turns if it is for 50Hz )

If the core is a strip wound toroidal, then the numbers is reduce to 4 and 5 turns.

So, for 1 turn per volt, the cross section needs to be around 5 square inches.

... Phil

• posted

Is a non-center tapped transformer wound the same as the non-center tapped transformer sitting next to it?

Maybe.

Like Larkin said, line power and audio transformers are wound in a fairly straightforward way. But as the frequency goes up the inter-winding capacitances start to get important, and all sorts of winding schemes (of which bifilar is but one) come into play.

Considering that the secondary voltage of a non-center tapped transformer is non-zero, this implies a voltage gradient across the secondary. What causes this voltage gradient?

Yes and no. It depends on the transformer, and on what your threshold is for considering an amount of flux to be "different".

For nearly all line voltage or audio, iron core transformers, the amount of flux per turn is practically the same. But there are exceptions.

Power transformers for microwave ovens, OTOH, have intentionally high leakage inductance to limit the current to the magnetron. They keep the flux/turn constant (or nearly so) for the HV secondary, though. Given that power transformers have been with us for over a century, I wouldn't be at all surprised at applications (particularly 1950's-era mil-aero) that intentionally split the magnetic flux between two or more secondaries.

More prosaically, any transformer that is air-wound or has a large gap is going to have different flux densities in different windings. I doubt that there are many such in current production electronics, but again, I wouldn't be surprised at any weird thing found in consumer kit.

```--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.```
• posted

Daniel, John, Phil and Tim, thanks to each one of you for your reply.

Let me assure you, this isn't a homework question that I am trying to push on to the newsgroup. (If it was homework, I wouldn't do it. :) )

I stumbled on these doubts while trying to develop a negative DC voltage source. Amongst the other features in the requirement spec., one requirement is that we would like to make this negative voltage source as far as possible without using a center tapped transformer. As it so "turns" out - its difficult.

What was irritating is that I could not find (printed or online) any standard text which discusses these points:

a. A transformer with its primary terminals connected to Vp and ground (zero volts) will have its secondary terminals as +Vs/2 and

-Vs/2. (For a small bit of time, I continued to think that the secondary terminals would be: +Vs and 0V, mimicking the single ended primary.)

b. If the flux crossing all the coils is the same (as in same density), what causes the gradient in the voltage across the coil? You see, the gradient on the coil is a change in the voltage magnitude and polarity from one end to the other! I have concluded that this happens because at one end, the flux is *entering* the coil and *leaving* at the other. This *entering* and *leaving* causes the opposite polarity (some "thumb rule"), but what about the drop in magnitude? Then again I "imagine" that the voltage generated all across the coil is such that the positive voltage is completely worn out by the time it reaches the other end and vice versa for the negative voltage. The positive and negative voltages meet at the mid-point, cancel each other and give us a beautiful zero! But why?

I will continue to be worried that my understanding is a figment of imagination until I see some text/formal/mathematical treatment which confirms/denies all this.

I will be embarrassed to see any URLs which discuss these points straight away. But then they are welcome anyway.

Thanks once again, Anand

• posted

And why is producing a negative voltage with a non-center tapped transformer a problem? Use a couple of bridge rectifiers (fuse AC side please). Tie one of the bridges negative terminal to common, the second bridge with the positive to common. Now you have an equal positive and negative supply that you can filter and regulate as required.

If you only need negative then just use one bridge and ties its' positive terminal to common...

John :-#)#

```--
(Please post followups or tech enquiries to the newsgroup)
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9```
• posted

In a normal transformer, the primary and secondary windings are electrically isolated from each other. If you connect the primary winding to Vp and ground, then try measuring the voltage between ground and either end of the secondary, you will not measure any voltage, as there is no current path between the secondary and primary. You will, however, measure a voltage between the ends of the secondary winding.

You are over-analyzing this, and are being misled by your assumption stated above that the voltage on the secondary is galvanically related to the voltage on the primary.

To make your negative voltage source, you can simply make a positive supply, and call the "+" terminal "ground", and use the terminal you would normally connect to ground as the negative terminal. Alternatively, make a normal half-wave rectifier circuit, but reverse the polarity of the diode and capacitor - then the junction of diode anode and capacitor "-" will be your negative terminal, and the capacitor "+" terminal will be ground.

```--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca  ```
• posted

To "use a couple of bridge rectifiers.....", you are assuming that the bridge rectifiers are galvanically isolated from each other. That is, the source for positive voltage regulator and the source for the negative voltage regulator should not come from the secondary of the same transformer.

If you do "tie one of the bridges negative terminal...." when the source of the positive and negative regulators are the same there is a problem. For a full wave bridge rectifier, the capacitor charges to 2Vpeak (no load condition) as in: +Vpeak on its positive terminal and -Vpeak on its negative terminal. So connecting the positive of one bridge to the negative of one bridge is not a solution.

You could simulate this or build this. Just measure the voltage between the two points you want to tie as common before connecting them. You will see a 2Vpeak voltage between them. (Obviously, you don't go ahead and connect them.)

If using two separate windings or separate transformers is your solution, then I agree. It is easy.

Thanks, Anand

• posted

I don't see where I have assumed or stated that the secondary is galvanically related to the primary. All that I have said is valid with or because of the galvanic isolation.

Sorry I wasn't clear about this, the idea is to make a positive *and* negative supply from the same source (as in from the same secondary winding) without using a center tap. (In which case "calling the "+" terminal ground doesn't help. And flipping the diode's orientation just moves which side of the bridge the positive and negative voltage appears.)

Thanks, Anand

• posted

I am sorry, I haven't been clear. We need to generate a positive *and* negative voltage using the same transformer secondary without a center tap.

(I kind of started writing where I was irritated thinking about the negative voltage.....)

• posted

shows how (see the first example).

Basically: using a single-winding (non-centertapped) secondary, you just ground one end of the secondary, and use two half-wave rectifying circuits (one diode and one capacitor each) to generate the positive and negative DC supplies.

I believe this approach will have somewhat higher amounts of ripple on the unregulated DC supplies, because each capacitor will be recharged during only half of the powerline cycle.

```--
Dave Platt                                    AE6EO
• posted

But you have to examine the whole cost, and it may be simpler to go with the centre tapped transformer.

You could use a second transformer. This works especially well if the second voltage is relativley low current, you can use a much smaller transformer.

Got to a switching supply, more complicated but with multiple voltages it may work out cheaper.

Especially if the second voltag is low current, use a dc to dc converter, ie connect to the positive supply an oscillator and rectify the output for the lower current negative voltage.

Use a higher voltage supply and split the DC voltage in half. This works well if the two voltages are the same but of different polarity, and works best if the current is relatively low. You in effect are setting an artificial ground at the output of the rectifier.

Michael

• posted

h
t
)

OK can I add more components to the none ceter tapped secondary?

Then make yourself (say) a 30V supply. Add a power opamp rail splitter*, and tie the center point to ground. Viola... +/-15 Volt supply.

George H.

*in it's simplist opamp is a buffer with non inverting input driven from two resistors one tied to each rail of the power supply.
• posted

Don't forget interwinding capacitance. A 10 meg DMM will almost always see something.

```--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."```
• posted

"Anand P. Paralkar"

** It is ridiculously simple to make equal "+" and "-" voltages from a single winding.

The circuit is called a "full wave voltage doubler". Two diodes and two filter caps, each pair wired as a half wave rectifier gives two DC rails of opposite polarity from one winding.

The centre of the two caps is the zero volts point.

.... Phil

• posted

-1.jpg

Agreed. Here is a simulation of a circuit that I actually built:

Here's the ASCII file if you want to play with it:

Paul

• posted

Ground one end of the secondary. Note that "ground" might be a chassis, might be a water pipe, or might be anything else that we care to use as a reference level. It's where we're going to clip the black wire of our voltmeter, not much more.

Connect the other end of the secondary through a diode, to a capacitor, and call that junction "DC+". Ground the other side of the capacitor. This diode has anode on the transformer, cathode on the cap.

Now the diode rectifies the AC from the transformer, dumping the AC peaks into the capacitor. You have DC voltage across the cap, with a bit of ripple.

If you reverse the diode, you'd get negative DC at the cap.

With two diodes and two capacitors, you can do both, namely get DC+ and DC- simultaneously, both relative to ground.

(In that last link, he has the current arrows backwards. The Internet is a mess.)

```--
John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com```
• posted

sure that's easy..

use one leg of the secondary as the common and the other leg will have

2 diodes branching from it. Each diode will be such that you'll get a
• and a - source...

THere is one big problem with this how ever, you need to use a cap double or more than what you'd normally would to filter it on the DC side. This is because you'll only get 30 hz per diode. Basically, you'll have a gap on each side when the opposite side is conducting.

Of course, if your requirement for the - source is a low current type that could be done with a added circuit that operates from a single power source.

Jamie

• posted

Another way, that I saw in an Elektor project, is to use a full wave rectifier for the main supply and use two half wave doublers hooked to each end of the secondary. (Or you could call it a capacitivly coupled full wave rectifier).

------|(-----+----||--gnd +--|(--gnd | +-->|--gnd | | | ------|(-----+----|

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.