# Why are there three wires in secondary coil of transformer?

Gold Member
This is a 12-0-12 transformer that I intend to use in my circuit:

Why are there three wires in the secondary? One is of course the neutral (in black). Why are there two live wires?

As an additional question, can I use four 1N5408 rectifier diodes to convert the output of this transformer to DC (full wave rectifier)?

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.Scott
Homework Helper
The center tap gives you the option of taking 6 volts and rectifying to 6VDC with two rectifiers instead of a full wave bridge.

Wrichik Basu
anorlunda
Staff Emeritus
Why are there three wires in the secondary? One is of course the neutral (in black). Why are there two live wires?
Probably there is more than one secondary winding, or more than one tap from a single winding, giving multiple output voltages.

As an additional question, can I use four 1N5408 rectifier diodes to convert the output of this transformer to DC (full wave rectifier)?

Why 4? You can't use diodes in parallel. Please show your proposed circuit schematic and tell us whay you are trying to accomplish.

Borek
Mentor
One winding and a four diode bridge, two windings and two diodes?

Gold Member
Why 4? You can't use diodes in parallel. Please show your proposed circuit schematic and tell us whay you are trying to accomplish.
Something like this:

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Gold Member
One winding and a four diode bridge, two windings and two diodes?
I can do that, but since I had already made a four-diode rectifier bridge last year (but never used it), I want to use it here so as to reduce some work.

Borek
Mentor
Check if you can treat both windings as one (ignore black wire).

Not sure how much sense it makes in terms of efficiency nor what voltage you will get (common reason suggests twice more than when using windings separately).

Gold Member
Check if you can treat both windings as one (ignore black wire).

Not sure how much sense it makes in terms of efficiency nor what voltage you will get (common reason suggests twice more than when using windings separately).
I haven't bought the equipment yet. So, I'll have to check after I buy it.

Anyways, thanks for your help. Problem solved (at least theoretically).

Gold Member
One last question, not worthy of posting in a separate thread: When I rectify AC, I will get the peak voltage as DC, right? (Irrespective of half-wave or full-wave rectification)

sophiecentaur
Borek
Mentor
No idea what you mean. You will get so called pulsed DC.

Wrichik Basu
Gold Member
No idea what you mean. You will get so called pulsed DC.
Leave it, I've understood.

.Scott
Homework Helper
One last question, not worthy of posting in a separate thread: When I rectify AC, I will get the peak voltage as DC, right? (Irrespective of half-wave or full-wave rectification)
So @berkeman showed you the difference between the full wave rectification and the half wave rectification. I'll warn you ahead of time that you need to take those voltage plots with a grain of salt. I will explain later.

Let me run through the possibilities:
If you don't use the black center tap, you will get the full rating of the transformer - 12V rms.
At 12Vrms, you can get full wave (4 diode) or half wave (2 diode) rectification and the result is what @berkeman showed.
If you use the center tap as the DC ground, then you will have the equivalent of 6Vrms. In that case, you only need two diodes to get full wave rectification.

Now let's talk about that "peak" issue. And for that we need to talk about capacitance and your load. In the simplest circuits (not ones that compensate for power factor), the next component after the full-wave or half-wave bridge is usually a good size capacitor (in the 100uF or mF range). With such a capacitor and no load, your voltage will hang out well above 12 volts. Basically, it will catch the peak voltage and never drop below that. So you will not see the scalloping that @berkman showed.

As your load increases, those sine wave peaks will become more and more evident. The voltage plot will become more and more lumpy. This is why power factor becomes an issue. You end up powering your device with only a portion of the waveform.

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Nik_2213, Wrichik Basu and berkeman
Gold Member
Thank you, @.Scott That was a very good explanation. I felt I had read about the peak voltage case somewhere, but I couldn't remember the capacitor. Later, I thought I was wrong. Your explanation was indeed very helpful, especially the centre tap topic, and the peak voltage issue.

Tom.G
Gold Member
Be aware:
That transformer is rated as 12-0-12, normally meaning that there is 12V each side of the center tap (Black wire). That gives a total voltage across the two Blue wires of 24V. When rectified, expect peak 32VDC output.

The datasheet is slightly ambiguous, so I suggest you measure all three secondary voltages before you connect any parts.

Cheers,
Tom

edit: fixed typo

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jim hardy, davenn and Wrichik Basu
rbelli1
Gold Member

In the future if you need to buy new parts anyway you can consider buying the bridge an one component. That saves more work.

BoB

davenn
davenn
Gold Member
2021 Award
If you don't use the black center tap, you will get the full rating of the transformer - 12V rms.

no, 24 VAC across the two blues

12 VAC is ONLY across the black and ONE of the blue wires

the next component after the full-wave or half-wave bridge is usually a good size capacitor (in the 100uF or mF range).

100uF isn't a good sized capacitor for this unless you were only drawing a few 10's of mA

1000uF minimum / 1 A of current is the normal rule of thumb
and 1000 uF would be the best place to start for any thing up to 1 A and go higher from there if the current is going to be higher
( of course that's got to be within the ratings of the transformer)

Dave

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berkeman and Wrichik Basu
Svein
Or you could get a "split supply" out of it:

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Wrichik Basu and Tom.G
Gold Member
no, 24 VAC across the two blues

12 VAC is ONLY across the black and ONE of the blue wires
Thanks for pointing that out.
100uF isn't a good sized capacitor for this unless you were only drawing a few 10's of mA

1000uF minimum / 1 A of current is the normal rule of thumb
and 1000 uF would be the best place to start for any thing up to 1 A and go higher from there if the current is going to be higher
( of course that's got to be within the ratings of the transformer)
Arduino's cannot it stand more than 40mA per I/O pin, otherwise the microcontroller might be crippled. I am designing my circuit such that it draws no more than 30mA from the pins. Since I am using two Analog pins and two Digital pins as of now, the current isn't supposed to exceed 120mA or 130mA at most.

I was thinking of using a large value of capacitor (say 6800μF, though the voltage won't be greater than 15V in my circuit under any circumstance). Will that do any harm to my circuit or the Arduino?

davenn
Borek
Mentor
No need for a large capacitor if Arduino is all that will be powered (especially as it has its own, additional powering circuit that will smooth the voltage further).

davenn's comment was of a more general nature, as he stated - it doesn't matter much when you draw low currents.

Wrichik Basu and davenn
gneill
Mentor
Or you could get a "split supply" out of it:
View attachment 232771
Something wonky with the diode orientations, methinks. I see direct paths through the diode shorting the secondary.

rbelli1
Gold Member
I was thinking of using a large value of capacitor (say 6800μF, though the voltage won't be greater than 15V in my circuit under any circumstance). Will that do any harm to my circuit or the Arduino?

With very large capacitors you need to be aware of the maximum inrush your diodes can handle.

BoB

Gold Member
With very large capacitors you need to be aware of the maximum inrush your diodes can handle.
You mean to say current, right? If that is so, then: my rectifiers (1N5408) can handle 3A current. But I'll take the advice and decrease the capacitance.

Actually initially I had thought of giving a 1F capacitor

jim hardy
Gold Member
Dearly Missed
Something wonky with the diode orientations, methinks. I see direct paths through the diode shorting the secondary.
You're right - Good Eye !
They swapped the AC and DC terminals of the bridge.
Should be Plus at the bottom. Minus at the top, and AC into the sides.

Going back to your OP's own sketch and making it split supply,

When looking at a full wave bridge rectifier for a power supply circuit...
DC side always faces two anodes or two cathodes (or three for three phase)
AC side always faces one of each

Someday you'll run across a "Ring Modulator" that violates the rule , but it's for a signal not a power supply circuit.

old jim

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davenn, Wrichik Basu, dlgoff and 1 other person
gneill
Mentor
You're right - Good Eye !
They swapped the AC and DC terminals of the bridge.
Should be Plus at the bottom. Minus at the top, and AC into the sides.
Thanks. Yup, over the (many) years I find I've become somewhat adept at spotting these little slip-ups in "pulp-published" circuits.

I just wanted to make sure that some newby didn't try to build the circuit as shown. Fried components can be quite discouraging.

Wrichik Basu, davenn, dlgoff and 1 other person
Rive
To the OP: please link here the source of the circuit you are trying to build. More than half of the posts (and half of your connected topics) are completely unnecessary or missing the point if applied to that specific circuit.

Gold Member
I think since my original question has been answered, so I will request moderators to close this thread.

To @Rive and others: I promise to post the project fully in the DIY forum when I finish with all the calculations.

Great advice was given in this thread, and I shall make good use of it. Thanks everyone for your input.

gneill, berkeman, jedishrfu and 1 other person
gneill
Mentor
Thread closed at the request of the Original Poster.

jim hardy