driving a CT from it's secondary?

I would like to induce a current of about 1ampere around a loop whose
impedance can vary from say 1mOhm to say 4 Ohms by using a CT and driving
it's secondary.
Now, transformers are not my strong point.
I'm suspecting that my idea is severely flawed due to the lack of sufficient
magnitising flux to achieve this end. Please put me out of my misery quickly
if this is the case.
So if I use a 100/5 CT (20:1 ratio) then I'd need to drive a 50mA current in
the secondary to cause my 1A in the primary. My 4 Ohm resistance would then
appear as 4*20^2 ohms to the secondary =1.6kOhms resulting in 1600*0.05=80V
driving voltage.
Question 1) am I correct so far?
Question 2) how do I make sure that it saturates at 4Ohms so that if
Rprimary reduces to say 100mOhms, I still only want to have about 1A in the
primary?
many thx
Danny
Reply to
Daniel Indyk
Loading thread data ...
On 3/30/06 4:16 AM, in article O_PWf.20246$ snipped-for-privacy@news-server.bigp> I would like to induce a current of about 1ampere around a loop whose
Obviously
All an ideal transformer does is for all the currents through their corresponding windings produce flux in the core adding up to zero. Because transformers are not ideal there is an additional magnetizing current required to support applied voltage.
To get constant current out of a winding, another winding has to be driven from a high impedance source.
If you want a constant current source, look up the Steinmetz monocyclic circuit that was once used for driving series strings of street lamps.
Bill -- Ferme le Bush
Reply to
Salmon Egg
You will have a very hard time finding a commercial 100:5 current transformer that will stand 80 secondary volts without saturating.
Think about it this way, if your 100:5 transformer supports 80 secondary volts, the transformer would be capable of handling a burden of 400 VA! Most small, inexpensive CTs have a rated burden of not more than 5 or 10 VA.
Or, calculate the flux needed to induce 4 volts in 1 turn. Then calculate the size of the core needed to support all that flux!
To your second question, a current transformer works on ratios just like any transformer. Saturation is the last thing you want. Within its operating range, the ratio of primary to secondary current remains constant.
I have the vague feeling that you are really trying to apply a regulated primary voltage and have the transformer supply a constant secondary current. Am I correct?
Reply to
BFoelsch
That is exactly what a monocylic circuit does.
Bill -- Ferme le Bush
Reply to
Salmon Egg
Yep. Salmon Egg introduced me to the Steinmetz network, which is intriguing. I am trying to analyse it with some difficulty. It seems to work when simulated, but the mesh equations don't seem to work out as 1 would expect. I will persevere as I'd like to know how to design for this circuit.
Thx anyway.
Danny
Reply to
Danny
er....does anyone no where i can find the analysis of this circuit. My kirchoffs law equations throw up some peculiar results.
thx
Danny
Reply to
Daniel Indyk
This is the circuit that i can't analyse.
-------r---------------------- | | | | L C | | | vsin(wt) |__ R __ | | | | | C L |----------------|------------|
w=1/(L*C)
i think this is the way it works: between R->0 and R->infinity the phase angle varies between 0 degrees and 180 degrees. this phase difference behaves in such a way as to change the voltage differential across the load R such that it stays constant. I feel after trying to solve this analytically for a couple of days now without success (kirchoff, thevenin/norton..etc) whether or not an analytical solution is possible but rather an iterative solution might be the go.
any suggestions gratefully considered. My only other thought is that as R varies the circuit is becoming slightly detuned such that it's usually working just off resonance at it's usual working point.
any thoughts (constructive 1's preferably) gratefully accepted.
many thx
Danny
Reply to
Danny
This is the circuit that i can't analyse.
-------r--------------- | | | | L C | | | vsin(wt) |---- R------| | | | | C L |--------|------------|
w=1/(L*C)
i think this is the way it works: between R->0 and R->infinity the phase angle varies between 0 degrees and 180 degrees. this phase difference behaves in such a way as to change the voltage differential across the load R such that it stays constant. I feel after trying to solve this analytically for a couple of days now without success (kirchoff, thevenin/norton..etc) whether or not an analytical solution is possible but rather an iterative solution might be the go.
any suggestions gratefully considered. My only other thought is that as R varies the circuit is becoming slightly detuned such that it's usually working just off resonance as it's usual working point.
any thoughts (constructive 1's preferably) gratefully accepted.
many thx
Danny
Reply to
Danny
----------------------------
Your diagram didn't come through clearly -try it with fixed text. Do you have any resistance in the left LC branch? As far as I can tell there is none so the voltage at the midpoint of this branch would be infinite at resonance. That can mess up analysis. With the inductance having resistance r (same as for the other branch, then you effectively have a bridge which should be solvable using any method. In any case you won't get a differential voltage for R =0 Is w=(L*C) correct?
Reply to
Don Kelly
W=1/sqrt(LC) r is the internal resistance of the supply and is thus common to both halves of the bridge. A finite R, I would have thought, would dampen any infinite voltage levels due to resonance. When R=0 -> no voltage differential: agreed....but when R is 'small' there will be a increasing differential as R increases.
Many thx 4 your interest, Don,
Danny
Ps how does one do fixed fonts on Google groups?
Reply to
Danny
On 4/3/06 4:34 AM, in article UK7Yf.22928$ snipped-for-privacy@news-server.bigp>
Reply to
Salmon Egg
If it's a homework problem then it's going to be handed in very late. I've been outside uni now for 6 years. Anyway, i think I've hit on a solution. I'm solving it slightly off resonance as in simulations the change in the current thru the load is negligable. I can then design for any current required. It's not too elegent...but i think it'll do wot i need.
thx
Danny
Reply to
Danny
If it's a homework problem then it's going to be handed in very late. I've been outside uni now for 6 years. Anyway, i think I've hit on a solution. I'm solving it slightly off resonance as in simulations the change in the current thru the load is negligable. I can then design for any current required. It's not too elegent...but i think it'll do wot i need.
thx
Danny
Reply to
Danny
If it's a homework problem then it's going to be handed in very late. I've been outside uni now for 6 years. Anyway, i think I've hit on a solution. I'm solving it slightly off resonance as in simulations the change in the current thru the load is negligable. I can then design for any current required. It's not too elegent...but i think it'll do wot i need.
thx
Danny
Reply to
Danny
OK.. sorry guys. I've solved it using plain old kirchoff. I was just cancelling out Xc'c and Xl's too early in the equations. Solving for all the currents, simplifying and then plugging in that w=1/sqrt(LC) seems to work out OK. I thru the load R= V/sqrt(L/C)+some small terms that depend on the load R.
Reply to
Danny
OK.. sorry guys. I've solved it using plain old kirchoff. I was just cancelling out Xc'c and Xl's too early in the equations. Solving for all the currents, simplifying and then plugging in that w=1/sqrt(LC) seems to work out OK. I thru the load R= V/sqrt(L/C)+some small terms that depend on the load R.
Reply to
Danny
OK managed to solve it with plain old kirchoff.
Current thru load resistor =Iload=V*sqrt(C/L) Supply current=Isupply=V*R*C/L where supply radial frequency=w=1/sqrt(LC)
it's an amazing circuit!
i thought it might be of use as a main safety earth check circuit. we use high voltage test trucks with a main earth lead and an auxilary earth lead. at the moment we just have a transformer driving current around the loop whereby if there is a high resistance in the loop the safety interlock is enabled thus disconnecting supply (as it's indicating that the safety earth may not be connected properly). I thought this might be a bit more elegent......not a homework problem at all.
thx anyways
Reply to
Danny
OK managed to solve it with plain old kirchoff.
Current thru load resistor =Iload=V*sqrt(C/L) Supply current=Isupply=V*R*C/L where supply radial frequency=w=1/sqrt(LC)
it's an amazing circuit!
i thought it might be of use as a main safety earth check circuit. we use high voltage test trucks with a main earth lead and an auxilary earth lead. at the moment we just have a transformer driving current around the loop whereby if there is a high resistance in the loop the safety interlock is enabled thus disconnecting supply (as it's indicating that the safety earth may not be connected properly). I thought this might be a bit more elegent......not a homework problem at all.
thx anyways
Reply to
Danny
On 4/3/06 11:28 PM, in article snipped-for-privacy@i39g2000cwa.googlegroups.com, "Danny" wrote:
I worked on this circuit many years ago. Even though it can, in principle, boost voltage to whatever is needed to break down gas lamps and the like without requiring a transformer, operation based upon resonance alone is not efficient. Ferromagnetic inductors saturate too easily to be of modest size. I have devised a transformer using a three-legged core similar to a three-phase core. It provides a turns ratio to match the design load. A winding on the third leg and an external capacitor enables the monocyclic behavior for a constant current output.
A transformer of this nature was used to charge a capacitor bank at close to unity power factor through a bridge rectifier. Because the fundamental secondary current was fixed, it was no trick to charge a completely discharged bank. By using a ferromagnetic core, output voltage would be limited by saturation. This would keep the constant current from continuing to charge the capacitors beyond the desired voltage. For a lossless and linear monocyclic circuit, the capacitors would continue to charge
Note that the primary current increased in proportion to the capacitor voltage.
Bill -- Ferme le Bush
Reply to
Salmon Egg

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