Re: Voltage selection options for dual-primary toroidal transformer?


<furtherside@xxxxxxxxx> wrote in message
news:1171841720.307217.123690@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
First, thanks to everyone who responded to my earlier question about
combining the secondaries on my toroidal transformer. That was the
thread from here: http://snipurl.com/1amow

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 ***,
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?


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.


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
transmission scheme, to get the best combination of power and speed.


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.

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?


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.)

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.

Well, I hope that info helps somewhat.

Jon


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