# Voltage selection options for dual-primary toroidal transformer?

Discussion in 'Electronic Basics' started by [email protected], Feb 18, 2007.

1. ### Guest

First, thanks to everyone who responded to my earlier question about
combining the secondaries on my toroidal transformer. That was the

I like the voltmeter test idea...I will try that method...thanks!

I'm preparing to build a power supply for a set of servo drivers for a
home workshop CNC router I designed. I can run these servos from
between about 17VDC up to 56VDC. According to the servo data sheet,
peak power is achieved at approx 41VDC, 3200RPM. However, due to some
of the other mechanical aspects of the machine, I'd may like to
perhaps run the servos at a lower voltage and speed.

The transformer has two 115V primaries, and two 33V secondaries, each
rated at 7A.

Q1:

Say I wanted to have the option of either "full' voltage (33V) or
"half" voltage (16.5V). Is there a way to wire the primaries of the
transformer (in series?) to produce half the rated output voltage,
across the secondaries? If this is possible, do I get 14A out of it?

It would be very cool if I could use this transformer and put a simple
switch on the front panel of my control box that says "full voltage or
half voltage". Then, I can experiment with the gearing in my power

Q2:

My household service is usually around 120VAC, so I think my 115V /
33V transformer is probably going to produce around 34 or 35 volts. I
realize that creating a truly regulated PS at some lower voltage is
way beyond the scope of what I'm trying to do here. So, instead of
regulating to lower than the 42VDC @ 14A expected final output of my
power supply, is there a simple way to induce an "extra" voltage drop
in the circuit? I know that the rectifier diodes in a full-wave
configuration are good for around a 2.4V drop -- what if I wanted to
create (for example) a 10V drop? Can it be (safely) done?

If this is possible, and my Q1 is possible, that would give me a lot
of options of "final" voltage out of my unregulated supply, at the
amperage I need.

-Chris

2. ### John PopelishGuest

Sure. Wire the two primaries in series and the volts per
turn for all windings falls by half. The current still has
to pass through both windings in the same direction it did
when they were in parallel, so you have to disconnect both
ends and reconnect one end of one winding to the end it
wasn't connected to of the other one. The reconnection can
be wired to a double pole double throw switch, so you can do
it on the fly.
If each of the 33 volt windings is rated for 7 amperes, and
you keep them in parallel, that makes a 33 volt winding
rated at 14 amperes. If you change the primaries from
parallel to series, you retain that amp rating, though the
total power output falls to half, because of the lower voltage.
You permanently connect the hot (after an on off switch and
a fuse) to (for instance) end 2, and the neutral permanently
to start 1. End 1 goes to the center contact of one of the
poles, with high voltage position to end 2 and the low
voltage position going to start 2. The center contact of
the other pole goes to start 1 and the high voltage position
goes to start 2. Nothing connects to the low voltage
position of that pole.
It would create a lot of heat if it was done in any way that
wastes the voltage. At this current level, I would be
thinking in terms of an inductor input filter, instead of
connecting the rectifier output directly to the filter
capacitor. This will reduce the voltage to the average of
the rectified waveform, rather than try to pump the caps up
to the peak voltage of the waveform. It also greatly lowers
the ripple current heating in the capacitors and slightly
raises the current rating of the DC supply because the
current is not concentrated in brief pulses near the peak of
the wave. The inductor will have to be about as big as the
power transformer, to be effective in this service.

3. ### Jon SlaughterGuest

Yes, Essentially you have 2 identical transformers(hopefully) and you can
combine them in 8 possible ways. Your voltages will be either 66V, 33V, or
16V depending on the turns ratio(series or parallel of each side).

VP/VS = NP/NS = IS/IP

where the P represents the primary side and the S the secondary side.

You have said that VP is 115 for one primary and 33v for one secondary. This
implies that NP/NS ~= 3.5. We can take NS = 1 and NP = 3.5 for conviences as
long as the two primaries are identical as also the two secondaries.

In that case can see that if we hook the primary side up in parallel we do
not increase the turns on that side but allow for double the current. If
they are in series we double the voltage but cut the current(since
resistance doubles). If we only use one primary then we do nothing to that
side.

The same logic holds for the secondary side. So basically you have

a/b*NS/NP*VP = a/b*33V = VS
b/a*NP/NS*IP = b/a*7A = IS (I'm assuming 7A is measured from the secondary)

a and b are indicators that tell you which side and which configuration the
inductors are in. a is for the secondary and b for the primary. If a = 1
then its parallel and a = 2 means series. If you only use one primary or
secondary then the formulas won't work but you can use the same logic to get
the voltages.

If you have the primary configured as parallel and the secondary as series
then a = 2 and b = 1. This gives your secondary voltage as VS = 2/1*33V =
66V and IS = 3.5A. That is, you have sacrificed voltage for current. (the
power is still the same)

If you configure the transformers with primary in series and secondary in
parallel then then b = 2 and and a = 1 so that VS = 16V and IS = 14A.

So if you want to drop the voltage you can increase the mas current rating.
This isn't entirely true and I have no idea how well it works in the real
world as you can saturate your core which makes the transformer less
efficient and it can burn up.

The main thing to note is that when you hook a side up in series you
increase its voltage that it supplies or that it takes(depending on which
side). If you hook them up in parallel then you increase the current that it
can take but decrease the voltage.

Yeah, you can do this many ways and a switch would be easy. Just remember
that you could ruin the switch if you don't account for inductor kickback.
Maybe just use a resistor. If you need to drop a few volts, say 10v at 14
amps then thats 140W. (so a lightbulb would do. A bit hot though.)
Well, I hope that info helps somewhat.

Jon

4. ### Paul E. SchoenGuest

Toroidal transformers have the advantage that you can usually add a few
extra turns to either the primary or secondary to tweak the voltage a bit.
A 600 VA transformer like this probably has about 0.5 volts per turn, so
ten turns of #16 to #14 wire can give you an additional 5 VAC that can be
added or subtracted from the output. This would be a lot more efficient
than burning up the extra voltage in a resistor.

Remember that most transformers are specified at the rated current, at
which point the output voltage will be 2% to 10% lower than at open
circuit. Also, at the rated line voltage, the primary will be saturating
just a bit at the voltage peaks, whereas this will not happen with the
primaries in series. So if you read 33 VAC at 120 VAC with parallel
primaries, you will probably read 17 or 18 VAC with a series connection,
and the waveform will have less distortion (no flattened peaks).

It is possible to drive a transformer with a properly designed triac or SCR
circuit, which can even be controlled in a feedback loop to produce a very
well regulated DC voltage. This is a very efficient method that is used in
some very high current supplies, but it is important to have a solid low
impedance source and a well balanced phase firing circuit. Any imbalance in
the phase angle will cause a net DC current and can lead to instability.

The most efficient safe way to adjust the output is by means of a variable
autotransformer (Variac or Powerstat).

Paul

5. ### Guest

Paul, and everyone who responded (to both my recent threads) -- thank
you. This information helps a lot.

Best regards,
Chris

6. ### The PhantomGuest

If the secondaries are in series, the output will be 66 volts with a
current capability of 7 amps, not 3.5 amps.
How would the core become saturated?

7. ### Guest

The tranny has two independent primaries and two independent
secondaries. So if each primary seeing 115V causes each secondary to
output 33V, then wiring the primaries in series should cause each
secondary to output 16V, right? It's like wiring the transformer to
have a center-tap.

I'm the newbie here, so may be wrong...but that's the way I was

-Chris

8. ### Rich GriseGuest

16.5V. ;-) Otherwise, yes. The voltage ratio is the same as the turns
ratio.

And they do wire them like that for a center-tap; it's quite common.

Cheers!
Rich