Maker Pro
Maker Pro

Constant Voltage Transformer Operation?

A

amdx

Jan 1, 1970
0
I recently repaired a marine battery charger. It was set up to charge 3
separate 12v batteries. The repair was simple the charger had a *shorted
40uf 660v ac capacitor. The capacitor is connected across a winding on
the transformer. This is the usual constant voltage transformer
configuration. This is the extent of what I think I know.

Can someone describe the operation of a constant voltage transformer
that has 3 windings, input, output, and a third winding that is
connected across a capacitor.
I've seen these for decades but never have understood the operation.

Mikek

*bonus, why didn't the shorted capacitor (**measured 1.5 ohms) blow a
fuse or overheat the winding on the transformer.

**1.5 ohms plus 0.7 ohms lead resistance.
 
J

Jamie

Jan 1, 1970
0
Bill said:
http://en.wikipedia.org/wiki/Voltage_regulator#Constant-voltage_transformer

It seems to be be a ferro-resonant constant voltage transformer. The
third winding and the capacitor apparently set up a resonant tank
circuit.

The core is apparently intended to saturate, so the maximum flux in
the tank circuit is limited and more or less constant, and thus the
peak current and voltage. Because it is part of a a resonant tank, the
waveform on the output winding stays more or less sinusoidal.

Bill Sloman, Nijmegen
Apparently you're unsure of yourself, more or less!

Jamie
 
T

Tim Williams

Jan 1, 1970
0
The transformer has two sections: a core section around the primary winding,
and a section around the secondary. These are coupled somewhat looser than
a regular transformer, so (without the capacitor) the secondary voltage
would be lower and "squishier" than a regular transformer (i.e., higher
series inductance). Adding the correct value capacitor cancels the series
inductance, forming a resonant tank. But the increased voltage difference
across the barrier forces the transformer deeper into saturation, thus
limiting voltage.

The transformer needs to be designed so that, over the design range of input
voltage (usually +/-10%) and load (0-100% current), the transformer must
remain in saturation, without overheating. Finally, because the secondary
voltage is generally lower (in terms of volts/turn), a lot more copper is
needed. Iron operated in saturation also has high core losses. This makes
these transformers particularly large and low in effeciency.

One upside: the effective LC filter between primary and secondary isolates
harmonics and transients; although the input and output are still a bit
distorted, they make reasonable sine waves, so it doesn't matter much if
your primary waveform is part square wave, or that your load has similar
behavior (like a capacitor-input rectifier).

Tim
 
J

Jasen Betts

Jan 1, 1970
0
I recently repaired a marine battery charger. It was set up to charge 3
separate 12v batteries. The repair was simple the charger had a *shorted
40uf 660v ac capacitor. The capacitor is connected across a winding on
the transformer. This is the usual constant voltage transformer
configuration. This is the extent of what I think I know.

Can someone describe the operation of a constant voltage transformer
that has 3 windings, input, output, and a third winding that is
connected across a capacitor.
I've seen these for decades but never have understood the operation.

Current in the regulatiing winding saturates the core and decouples the
primary and secondary windings.
*bonus, why didn't the shorted capacitor (**measured 1.5 ohms) blow a
fuse or overheat the winding on the transformer.

the primary starts off poorly coupled and is rated for higher current
than in an ordinary transformer to survive the saturation.
**1.5 ohms plus 0.7 ohms lead resistance.

0.7 is the meter leads?
 
A

amdx

Jan 1, 1970
0
Current in the regulatiing winding saturates the core and decouples the
primary and secondary windings.


the primary starts off poorly coupled and is rated for higher current
than in an ordinary transformer to survive the saturation.


0.7 is the meter leads?

Ya, added that to make it *clear that the short was a little long!
I removed the 0.7 ohm meter lead resistance and still had a 1.5 ohm short.


Mikek

* apparently I didn't make it clear. :)
 

Similar threads

M
Replies
20
Views
4K
Mark & Mary Ann Weiss
M
J
Replies
84
Views
4K
Rich Grise
R
Top