Making a Current Transformer

D

Dial-Up

Guest
I was reading another topic,
http://groups.google.ca/group/sci.electronics.design/browse_thread/thread/9f797bb57f86a2c/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?
 
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
Click to expand...
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?
Click to expand...
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.


--
Regards ..... Rheilly Phoull
--
Regards ..... Rheilly Phoull
 
Dial-Up wrote:

I was reading another topic,
http://groups.google.ca/group/sci.electronics.design/browse_thread/thread/9f797bb57f86a2c/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?

For a small CT, one could remove the secondary from a filament
Click to expand...
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.
 
Rheilly Phoull wrote:
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.


--
Regards ..... Rheilly Phoull
--
Regards ..... Rheilly Phoull
Click to expand...
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?
 
"Dial-Up" <foxryan@gmail.com> wrote in message
news:1123553734.761412.57660@g47g2000cwa.googlegroups.com...
I was reading another topic,
http://groups.google.ca/group/sci.electronics.design/browse_thread/thread/9f797bb57f86a2c/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?
Click to expand...


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!!
 
Dial-Up wrote:
Rheilly Phoull wrote:

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?
Click to expand...
Read the thread

http://groups-beta.google.com/group/sci.electronics.design/browse_thread/thread/a421ea66c8131080/75afe9b6a3477a58?lnk=st&q=%22current+transformer%22+group:sci.electronics.design&rnum=1&hl=en#75afe9b6a3477a58

from April 22 on. We had a decent discussion there on it. THen we can
fill in any details you don't understand.

John Perry
 
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