beananimal said:
John Popelish wrote:
Well actually yes, but my comments were more of a rhetorical question
regarding what I though I understood about the pros and cons of each of
the basic methods described to me.
To be frank I am having trouble wraping my head around all of this. I
get part of it, but get lost in some of the details.
A transformer is a transformer. That is, a transformer has a fixed
(determined by turns ratio) ratio of input to output voltage and a
fixed ratio of input to output current, neglecting losses and leakage
flux between windings. If the primary is fed with a stiff voltage
source (like being connected across the power lines) then the output
also acts like a stiff voltage source, that varies its voltage only
slightly as the load current changes.
But if the primary is fed with a nearly constant current (as when a
current transformer is connected in series with a load, which is
dropping almost all the line voltage, and the current transformer
primary is dropping only a tiny fraction of the line voltage, so any
voltage variation reflected back to it from its secondary voltage will
have almost no effect on the load current) then the output of that
transformer is a current, related to the load current by the turns
ratio, regardless of how much output voltage the CT produces (as long
as its core doesn't saturate from producing that voltage). Any
inductor (including transformer windings) produce voltage in
proportion to the rate of change of the flux surrounded by their
windings. Larger voltages require larger flux swings. This is why
CTs have a maximum burden resistor rating. Multiply that rated
resistance by the rated output current (rated input current divided by
the turns ratio) and you have the maximum output voltage the core can
produce without saturating.
So, adding rectifier drop between a current transformer and its burden
resistor forces the current transformer to produce extra voltage, to
cover the diode drops, so that the total current through the burden
resistor is essentially unchanged, except for the direction of
alternate half cycles. You get very little nonlinearity added by the
rectification process. It is very close to ideal rectification, as
long as the desired output is a current.
So having the diodes before the "burden resistor" takes their voltage
drops out of the picture.
Yes. Almost perfectly.
But placing the burden before the diodes is
safer (in case of diode failure?) but the voltage drop of the diodes
become a problem?
Yes. In that case, the voltage drop across the burden resistor
becomes a signal source that is subsequently rectified, with the
normal diode losses. The additional failure rate does not go up very
much if the rectifiers and burden resistor are well over rated, to
include the expected source current peaks any start up. Or you add
peak limiting components, but those also have to handle the peak start
up current, but at an even higher wattage, because of the higher voltage.
This is just a wild guess, and at this point it
should be obvious that the theory part of this is a bit over my head.
I am 100% self taught... and learning every day. I am trying to sort
all of this out so that I can at least get one or two basic design on
hte breadboard.
Most signals we deal with are treated as voltages, so it takes a
slight adjustment to think about a signal that is inherently a current
(and the voltage can be anything, over some range).
If it appears that I am somewhat dazed in a sea of informed people...
well I guess that would be a fair observation. However, this is the
only way I can learn (by experience and asking questions).
You seem to be doing just fine from where I am watching.
))
