PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
OverUnity Research > Electrical / Electronic Theory and Learning Center > Inductors > Saturable reactors.
2024-11-26, 22:21:05
News: If you have a suggestion or need for a new board title, please PM the Admins.
Please remember to keep topics and posts of the FE or casual nature. :)

Pages: [1]
Author Topic: Saturable reactors.  (Read 9729 times)
Topic Options img
img  szaxx
Saturable reactors. « on: 2015-11-20, 18:50:40 »
Sr. Member
****

Posts: 276
If we wind 1000 turns on a magnetically closed loop sat reactor and wind another separate 10 turns of thicker wire as a control input. So far a typical setup for AC control issues. However, if we were to discharge a capacitor into a load through the 1000 turns and when the core is only half saturated, we then applied a DC Pulse to the control winding that saturated the core.
What would occur to the stored magnetic charge in the reactor referencing the current already flowing through ?
   
img  Smudge
Re: Saturable reactors. « Reply #1 on: 2015-11-20, 20:23:17 »

Group: Professor
Hero Member
*****

Posts: 1940
When the core saturates the voltage induced into the coil from the rising current (that is the voltage that is controlling the rate of rise) suddenly increases so as to be above the capacitor voltage, so now current flows backwards into the capacitor.  Some of the energy previously stored in the high value inductance now gets quickly fed back to the capacitor until some equilibrium is reached.  Then the capacitor again adds charge to the inductor but now at a faster rate because of the lower inductance until eventually the current rises to a much higher value at zero capacitor voltage when all the original energy is now stored in the low value inductance.  That's my analysis, assuming all this takes place without significant resistive loss, i.e. the timings are all faster than the L/R time constant.

Smudge
   
img  BEP
Re: Saturable reactors. « Reply #2 on: 2015-11-20, 21:15:40 »
Group: Guest
Quote from: Smudge on 2015-11-20, 20:23:17
When the core saturates the voltage induced into the coil from the rising current (that is the voltage that is controlling the rate of rise) suddenly increases so as to be above the capacitor voltage, so now current flows backwards into the capacitor.  Some of the energy previously stored in the high value inductance now gets quickly fed back to the capacitor until some equilibrium is reached.  Then the capacitor again adds charge to the inductor but now at a faster rate because of the lower inductance until eventually the current rises to a much higher value at zero capacitor voltage when all the original energy is now stored in the low value inductance.  That's my analysis, assuming all this takes place without significant resistive loss, i.e. the timings are all faster than the L/R time constant.

Smudge

I believe your explanation is correct. What energy isn't used by the load will be radiated as heat from the core because of the reactive nature of capacitors and inductors.

I remember a circuit for a VFO (variable frequency oscillator) that used the same principle to control the frequency. i.e. frequency control was due to controlled saturation of the tank coil around two toroidal ferrous cores. A DC current was used to control saturation of the cores. Two cores were used with saturation windings wound in an opposing fashion so the DC source didn't see induced current from the tank windings.

It was a very old patent, if I recall correctly.
   
img  szaxx
Re: Saturable reactors. « Reply #3 on: 2015-11-20, 21:53:57 »
Sr. Member
****

Posts: 276
Thank you Smudge.
That's what I was thinking too.
If we cascaded a chain of sat reactors I'd assume the produced voltage to be far in excess of that supplied by the discharge capacitor.
Following this, a transient suppressor across the cap cct would therefore send the excess voltage in the direction of the load. (Assuming we had a resistance after the cap).
Regarding the back EMF constant current rule, is this applicable as the core becomes invisible magnetically or is there something else working here creating the voltage pulse.

Hi WW, long time.
I remember them in the old valve tv's too for a similar purpose.
I'm chasing the controlled pulse sharpening idea I had a while ago and the inner workings were never satisfactorily found.
It's part of an old thought train I had 10 years ago that seemingly has resurrected itself.
New info adding to the jigsaw.
   
img  Smudge
Re: Saturable reactors. « Reply #4 on: 2015-11-21, 09:22:02 »

Group: Professor
Hero Member
*****

Posts: 1940
Quote from: szaxx on 2015-11-20, 21:53:57
Regarding the back EMF constant current rule, is this applicable as the core becomes invisible magnetically or is there something else working here creating the voltage pulse.

What is the back EMF constant current rule?

Quote
Hi WW, long time.
I'm chasing the controlled pulse sharpening idea I had a while ago and the inner workings were never satisfactorily found.

Using a saturating ferrite toroidal core is a classical way of sharpening up the leading edge of a pulse.  Essentially with the ferrite unsaturated the inductance is high so little current flows through the inductor into the load, but when saturation is reached the inductance switches suddenly to a low value whence maximum current now flows into the load.  The effect is to hold off the voltage connection between load and pulse source until the slow voltage rise has built up to its maximum value.  The defining equation is int(V.dt)=NAB where N is number of turns, A is area of core and B is the flux density.  You arrange that int(V.dt) over the slow rise from zero V to maximum V takes B from zero to its saturation value.  For some control of that saturation point you can DC bias the core so that B goes from some set value up to saturation.

Smudge
   
img  szaxx
Re: Saturable reactors. « Reply #5 on: 2015-11-22, 22:18:26 »
Sr. Member
****

Posts: 276
An inductor core charged magnetically produces an opposing 'current' on supply disconnection as the magnetic charge is induced into the winding. The return 'current' depends on the value of the inductor and is proportional to the voltage across the load.
-L di/dt.
The above is basics we all know.
I think the correct term should be power or EMF, 'current' was the issue on the page I read on some random HV site in regards to I sqrd t with DOL switching inductive loads. The switchgear  contacts being where the bemf will cause damage if low velocity types are used as the arc current is higher with lower voltages. Suppressors or snubbers not mentioned.
It mentioned somewhere that this bemf acts like a constant current source under certain conditions.
Perhaps I misunderstood it somewhat at that time. The only constant I see, thinking about it, is large inductances can kick like a mule if you're too close lol.


The pulse sharpening aspect I've mentioned years ago. This is what I want to use with some external control. It's having the option of precision timing that can be adjusted with a bonus of the charge already in the inductor at switch on of the control coil I want to use.
You disconnect the power source and it kicks back so disconnecting the magnetic storage, it kicks forwards?

I should read up again on some of this, its been a few decades in certain areas of the subject. The internet works fine if you're on the correct page.

OUR is a very fine page in my book. :D
   
img  Smudge
Re: Saturable reactors. « Reply #6 on: 2015-11-24, 16:50:40 »

Group: Professor
Hero Member
*****

Posts: 1940
Quote from: szaxx on 2015-11-22, 22:18:26
An inductor core charged magnetically produces an opposing 'current' on supply disconnection as the magnetic charge is induced into the winding. The return 'current' depends on the value of the inductor and is proportional to the voltage across the load.
-L di/dt.
The above is basics we all know.
A lot of people get this wrong.  On supply disconnection the bemf from the falling current does not oppose the supply, it adds to it, so the current flow does not reverse. it keeps going in the same direction.  The open switch sees this enormous voltage spike that then ionises the air into conduction and the inductor continues to draw current in the same original charging direction.  But that current is reducing in value at a rate that sustains the arc voltage across the switch.

Smudge
   
img  szaxx
Re: Saturable reactors. « Reply #7 on: 2015-11-26, 12:55:34 »
Sr. Member
****

Posts: 276
The reverse diode in DC switched inductive circuits stops ringing and this may one of the reasons so many have misunderstood.

By saturating the core with a control pulse, the stored energy in the core gets added to the current flow in the wiring.
This would appear to occur as the cause of diminishing the rise time hence the sharpening of it.
Is the above correct ?

Assuming there will be two identical trigger circuits, if one has an applied delay pulse to the control winding the core charge will differ as will the inductors output voltage if each were measured before the pulse. Would this affect the rise time if the time difference was short between the pulses in two circuits?
I'm assuming the measured power will be identical in both circuits.
( Two off, Cap - sat reactor- inductive test circuit. )
   
Pages: [1]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-11-26, 22:21:05