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Making a Current Transformer

R

Rheilly Phoull

Jan 1, 1970
0
One day Dial-Up got dressed and committed to text
I was reading another topic,
http://groups.google.ca/group/sci.electronics.design/browse_thread/thread/9f
797bb57f86a2c/010987b260d96bdb?q=make+an+ammeter&rnum=37#010987b260d96bdb
and was wondering: How exactly do you make the CT described?

I understand that there'll be your two coils, one that the wire you
want the measure the current of goes through, and then the other
which, I'm assuming, goes to your load. What I don't understand is
what to do with the leads of the first coil. Do I just connect them
together?

Think a bit :)
How can the current being measured pass through two wire that are joined
together ??
Mebbe you should connect one of the heavy wires to the supply and the other
end to the load ??
DONT ever forget to short the output winding if the ammeter or whatever is
removed, since very high and lethal voltages can be developed if that
winding is open circuit.
 
R

Robert Baer

Jan 1, 1970
0
Dial-Up said:
I was reading another topic,
http://groups.google.ca/group/sci.e...db?q=make+an+ammeter&rnum=37#010987b260d96bdb
and was wondering: How exactly do you make the CT described?

I understand that there'll be your two coils, one that the wire you
want the measure the current of goes through, and then the other which,
I'm assuming, goes to your load. What I don't understand is what to do
with the leads of the first coil. Do I just connect them together?
For a small CT, one could remove the secondary from a filament
transformer from RS - leaving some space for an insulated one turn high
current sensing primary.
The 110V winding left on becomes the secondary and needs to be
loaded; ideally shorted.
The secondary current times the turns ratio (secondary turns divided
by primary turns) equals the primary (input) current.
 
D

Dial-Up

Jan 1, 1970
0
Rheilly said:
One day Dial-Up got dressed and committed to text


Think a bit :)
How can the current being measured pass through two wire that are joined
together ??
Mebbe you should connect one of the heavy wires to the supply and the other
end to the load ??
DONT ever forget to short the output winding if the ammeter or whatever is
removed, since very high and lethal voltages can be developed if that
winding is open circuit.

Heh, now I'm slightly more confused... I want to measure the current
drawn by the load, but what I gather from your explaination, I'm
supposed to put the coil in series with it?

Perhaps there's more than one kind of CT? The kind I am familiar with
is what I use at work: they look like black doughnuts with a pair of
twisted wires and a ground coming off of them. Through the hole, we put
our big 4/0 wire. The wires coming off of the CT are only 14 gauge, so
I'm assuming it's relatively low current. It goes off to a low-voltage
panel to monitor something-or-other. I don't know much about the
low-voltage panel, and its doings.

So in my idea of a CT, it's not powered.

If I'm thinking of the wrong thing, then perhaps, could you draw a
small diagram?
 
D

DaveM

Jan 1, 1970
0
Dial-Up said:
I was reading another topic,
http://groups.google.ca/group/sci.e...db?q=make+an+ammeter&rnum=37#010987b260d96bdb
and was wondering: How exactly do you make the CT described?

I understand that there'll be your two coils, one that the wire you
want the measure the current of goes through, and then the other which,
I'm assuming, goes to your load. What I don't understand is what to do
with the leads of the first coil. Do I just connect them together?



You make a CT by winding a large number of turns of wire onto a core,
usually a toroid. But in your case, to make things simple, a small filament
transformer would do just fine.
Remove the secondary winding from the transformer. The transformer's
primary winding becomes the secondary of the CT.
In the space left by the winding that you removed, wind a single turn of
wire. This becomes the CT's primary.
Connect a resistor of 10 - 100 ohms across the CT secondary.
Connect one end of the CT primary to one side of the AC source.
Connect the other end of the CT primary to the load. This load will be the
device that you want to measure the current that is passing through it.
Connect the other end of the load to the other side of the AC source.

When you apply power from the AC source and turn the load device on, the
current that is passing through the load will be transformed by the CT to a
voltage impressed across the 10 - 100 ohm resistor in the proportion of E =
(I(load) / N) * R
where
N = turns ratio of the CT, which will be the number of turns on the
transformer's secondary (the original primary), assuming a single primary
turn
R = specific value of the resistor across the CT secondary

The circuit will look like this (viewed with a fixed pitch font such as
Courier):

CT
AC --------+ +--------+--------+
Source ) || ( | |
) || ( / |
) || ( \ |
1 turn ) || ( N / R AC Voltmeter
) || ( turns \ |
) || ( | |
+---+ +--------+--------+
|
|
|
Load
|
|
|
AC -----+
Source



--
Dave M
MasonDG44 at comcast dot net (Just subsitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!
 
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