New PI technology proposals.

[Aziz]

Hi Carl,

"Why is able to change so fast when turn on for the same voltage change is so slow?"

Different time constants (taus). Instead of charging the coil with a voltage source, charge it with a current source. Now you have the same time constants, with a very fast turn-on accompanied by another flyback voltage.

- Carl

I am not totally agreeing with you. Even the current source can not start instantly.

The time constant of the coil is never changing (unclamped diodes period not taken into account). For both transmit on and transmit off, we have allways same time constants of the coil arrangement.
The answer is in the coils self inductance. Any change of the current in the coil will cause self induction. This is at the beginning of the transmit pulse high and we can not instantly increase the coil current therefore.

On the switch-off period, the current change on the coil off is much higher. So we have a higher self induced flyback voltage. Now the damping resistor takes as much as current from the coil, not causing much self induction effects. What would have an effect to a short circuited superconducting coil? Magnetic energy will not be converted to heat and a steady self induction appers (coil ringing). (we are not taking any loss effects into account).

The effect of magnetic field energy conversion into heat in the flyback period is much more efficient due to Kirchhoff's law (P=U*U/Rd) compared to the charge of the coil during on-periods. But there is still self induction during flyback damping period which prevents us to kill the stored magnetic field energy instantly. The critical coil damping says us, you can damp the magnetic field energy with fastest possible speed without ringing the coil.

underdamped coil: ringing ->not convenient for processing, ringing takes too much time, the magnetic field will collaps many many times and will cause many many times of self induction and the magnetic field energy conversion into heat is low.

overdamped coil: two different time constants for discharging magnetic field energy -> difficult for processing due to higher self induction effects

critical coil damping: easiest way of processing

So we need more time for charging the coil to some level of magnetic field energy and we can - thanks to Kirchhoff - damp this stored magnetic field energy with shorter periods even the time constant of the coil is the same.

Aziz

[gwzd]

Of course, the ultimate reality check is: build it!

- Carl

I couldn't agree more!
Just see how the nanovolt-lock-in-amplifier-laptop-metal-detector turned out!
regards,

[Aziz]

Hi Robby,

..
This might be more obvious by the attached picture but I'm sure it will lead to more confusion. You can see though that what occurs during the on period is very important and does play a large part in the outcome.

that's correct. This gives a good opportunity to determine the targets time constant (discrimination) with different pulse width timings.
So the first (short) pulse should be quite low compared to the second (long) pulse. The second pulse should reach a constant current flow through the transmit coil. The first pulse should be at very low TC (0.5..1 TC) to have a higher dI/dt in the transmit coil. Very large TC targets will have considerable effects for processing. But normally, we have low TC targets and omitting the effects during transmit on phase.

I think ML is using such techniques.

Aziz

[Carl-NC]

I am not totally agreeing with you. Even the current source can not start instantly.

I know, it's a mathematical exercise. IF you have an ideal current source and IF you switch it on to a coil with a very fast switch, you will get a di/dt and a flyback voltage that looks just like our turn-off.

The time constant of the coil is never changing (unclamped diodes period not taken into account). For both transmit on and transmit off, we have allways same time constants of the coil arrangement.

Oh, no, not at all!

The turn-on time constant is L/Rs where Rs is the series resistance of the coil (and the switch and the battery and whatever else is there, but let's just say the coil for simplicity). Assumes Rd is much much larger.

The turn-off time constant is L/Rd where Rd is the damping resistor. Assumes Rs is much much smaller.

This is why the coil is slow to turn on, and fast to turn off.

Example: L=300uH, Rs=3, Rd=600
turn-on tau = 100us
turn-off tau = 0.5us

This, of course, neglects the effects of C. This is a good exercise for an over-simplified Spice simulation!

- Carl

[Aziz]

I know, it's a mathematical exercise. IF you have an ideal current source and IF you switch it on to a coil with a very fast switch, you will get a di/dt and a flyback voltage that looks just like our turn-off.

Oh, no, not at all!

The turn-on time constant is L/Rs where Rs is the series resistance of the coil (and the switch and the battery and whatever else is there, but let's just say the coil for simplicity). Assumes Rd is much much larger.

The turn-off time constant is L/Rd where Rd is the damping resistor. Assumes Rs is much much smaller.

This is why the coil is slow to turn on, and fast to turn off.

Example: L=300uH, Rs=3, Rd=600
turn-on tau = 100us
turn-off tau = 0.5us

This, of course, neglects the effects of C. This is a good exercise for an over-simplified Spice simulation!

- Carl

You were right! It is confusing some times..
The coils Rs and Rd determines the turn-off time constant. And this is differing much more. Thanks for remembering.

Aziz

[Aziz]

Hi Carl,

the change (decrease) of time constant on the turn-off period, is still saying, you can get the energy faster out of the coil and burn into heat. Doesn't it?

Aziz

[Aziz]

I think, we have to consider the following (to cause many confusions now):

On the turn-off period, we have a voltage source now (L,Rs) and a load resistance (damping resistor Rd). From the point of coil as an energy storing element, the time constant is truely NOT changing! But the energy conversion is more efficiently now.

Any comments?

Aziz

[Qiaozhi]

Your reply: "I think you are attempting to prove that the back-emf is not a negative voltage. This is incorrect".
But it is correct if I'm using a conventional circuit with the top end of a coil tied to the positive rail and the bottom end switched to the negative rail. In this case a negative spike would just switch the coil on harder for a period at switch off.

"If you monitor the voltage across the coil you will see that the amplitude is positive when the switch is closed. But when the switch is opened the collapsing magnetic field generates a large negative voltage (back-emf) in the coil".
Yes but I assume you must be switching the positive end of the coil. It is a large positive voltage (back-emf) in my case.

So I suppose I'm agreeing but also disagreeing.

It would be more correct to say - the voltage generated across the coil (when the switch is opened) has the opposite polarity to the applied voltage (when the switch was closed).
The simple SPICE example was not a simulation of a conventional PI circuit, and was only intended to demonstrate what happens when you have a switch in series with a coil. Of course, the polarity of the back-emf (flyback) will depend on which way round the power supply is connected, and also where you have placed the measurement probes. But that should be evident from the SPICE deck.
I think you are correct, that we are in agreement. It was simply that I could not understand the point you were trying to make in your previous post.

[Qiaozhi]

I couldn't agree more!
Just see how the nanovolt-lock-in-amplifier-laptop-metal-detector turned out!
regards,

I was trying to be nice, by not mentioning that.
Still ... now you've done it ... I do think this is yet another "rainbow chasing" episode.

Aziz - when you have these theories, you need to test them in the real world.
However (having said that) I don't want to dent your enthusiasm. These sorts of discussions are very interesting, if only to clear up some misconceptions.

[Aziz]

Hi Qiaozhi,

..However (having said that) I don't want to dent your enthusiasm. These sorts of discussions are very interesting, if only to clear up some misconceptions.

As I said, I will not be worry about of my wrong thoughts (I am also still learning). The discussion shows, there is still much understanding lag of PI theory. I think, regardless of the truth of some ideas, the outcome of this discussion has an invalueable worth.

By the way, lock-in ampfier is really working. It is working pretty much, that I had coped with really interesting effects (ground, mechanical stability, etc.). To handle these disturbing effects would lead to a yet another VLF detector, which does not make sense to follow further at the moment. As I know, you all are PI related, I have decided to contribute to this area for a while.

So we should continue with the debate without loosing the respect to each other (avoiding discredit). Sometimes there is a change of view required. Different way of thinking and of course by doing many mistakes. This is a normal engineering process. Isn't it?


Aziz

[Ferric Toes]

Hi All,

Regarding the avalanche breakdown and what happens during that time, if you put a small resistor, say 0.1 ohms in series with the ground end of a coil, you can observe on a scope what the coil current is doing. The magnetic field from the coil is behaving in exactly the same way. You see the current growing exponentially due to the turn on TC's and at the end of the pulse you see it die away in the form of a linear ramp. The ramp down is linear because of the clamping from the mosfet avalanche. dI/dt is forced to be constant. High voltage mosfets have a greater slope, while low voltage ones have a lesser slope. This means that the target is energised all the way from the start of the TX switch off, till it reaches zero. Of course, once the mosfet comes out of the avalanche situation, the residual energy is mopped up by the damping resistor and other circuit losses. At that point the slope will change from a linear to an exponential decay. One basic point from PI theory is that provided the rate of field switch off is faster than 1/10 of the object time constant, then it is as if the field had been removed instantaneously. In other words you reach a point for a given target where switching off any faster is a waste of time.

Eric.

[Unregistered]

Hi Eric !!!

Hi All,

Regarding the avalanche breakdown and what happens during that time, if you put a small resistor, say 0.1 ohms in series with the ground end of a coil, you can observe on a scope what the coil current is doing. The magnetic field from the coil is behaving in exactly the same way.

Eric.

Hi Eric,

Glad to see you still posting, hope you are doing well, Merry Christmas.

Thank you very much for the information, you stated it all very well.

Better than I could have, and I was wanting to get into what the current

in the coil is doing after turn off. The 0.1 ohm is an excellent suggestion.

And the magnetic field is defined by the current in the coil as you said,

this is important for all to understand.

My experience has been that using a high enough avalanche MosFet that

it does not avalanche yields the fastest ramp down. (when possible)

The lower the voltage MosFet the longer it stays in avalanche

and the longer it takes the current to reach zero.

[JC1]

Hi Carl !!!

I know, it's a mathematical exercise. IF you have an ideal current source and IF you switch it on to a coil with a very fast switch, you will get a di/dt and a flyback voltage that looks just like our turn-off.


- Carl

Hi Carl,

Yes the current source would need to start out with a

very high voltage just like the turn off, say 2 kV.

and of course the very high flyback voltage at turn off is what

is causing the current to change rapidly.

So the fastest turn off (of the current in the coil) would be to

turn off without a damping resistor or avalanche MosFet and let

the current fall as in an undamped state. Then as the voltage

falls to say 10 volts switch in the damping resistor to make it a

critically damped coil and will slide into zero.

Does this work?????? I never tried it.

What does everyone think?





The above unregistered post is Mine, I forgot to put JC1.

Any other unregistered posts are not Mine.

[JC1]

tiny gold nuggets

Oh and I am looking for tiny gold nuggets,

so I want a fast turn off. Now in and undamped state

the flyback voltage is really going to jump up, especially in

your ideal spice simulations, so say we put a 1 MegOhm damping

resistor and then switch in a 600 Ohm damping resistor in parallel.

[Tinkerer]

2KV current source

Could we use the Flyback from an inductor as the 2KV current source?

Tinkerer