New PI technology proposals.

[UWLocator]

"Flyback" is a term derived from the line output transformer (LOPT) that is used in CRT displays, which is also commonly known as a flyback transformer. It acquired this name because the back-emf from the coil is used to generate the flyback pulse. Even in descriptions of how a flyback transformer works, the reverse voltage across the coil is referred to as the back-emf.
Unfortunately this is a common occurance in the English language, where words get misused over time, and then misunderstandings can arise.
At least we got it sorted out in the end.

The term "flyback" was used long before the invention of the cathode ray tube in 1897. But that really makes no difference.

As I said before, I understand what you are referring to when you use these terms. I also understand what counter-emf refers to.

I only got involved in this discussion because I do not agree that the emf created, on the rx coil, by a collapsing field is what causes a response from a target. Still don't agree with that claim as made in the original post. I also do not see a benefit to putting the tx pulse on only 1/2 of the coil. I believe that the negatives out weigh the positives with that approach.

On another note, while I really like Spice and several of the current simulator programs, I do not think they are ready to model complex inductor relationships. I hope that one day soon they will be. But, I don't think that day has come.

Also, think about that series resistance some. There is a benefit to sampling at a point where the polarity does not reverse.

HH,
Rip

[Qiaozhi]

Suppose a turbine input is connected to the mains and the output is connected to a water valve and the valve output is at atmospheric pressure and the turbine is coupled to an unloaded flywheel and the water valve, when opened, has some resistance (fet rds_on).

Upon closing the valve, the pressure at the turbine output drops to zero because the flywheel (inductance) resists change and is slow to react (a high impedance).

The unloaded turbine/flywheel slowly gathers momentum, and the pressure at turbine output rises towards mains pressure and eventually flat tops at a pressure below mains after the turbine attains maximum speed.

So, We then turn the output valve off but the energy stored in the flywheel (magnetic field) is dumped back into the system and causes the flywheel to keep spinning. This in turn results in an increase in pressure (emf) but a dramatically limited water flow (current).

This pressure (emf) initially rises until it equals mains pressure and if the process ceased at this point then the turbine would slowly grind to a stop in a time determined by it's own efficiency, but the flywheel continues to spin and the pressure continues to rise to well above mains pressure and the only way out is back thru the turbine to the much lower mains pressure. This causes water (current) to flow back thru the turbine, which in turn forces the turbine to come to an abrupt halt.
If we fit a one way valve (diode) to allow a low resistance path around the turbine back to the mains then the turbine would continue to rotate until the energy stored in the flywheel had dissipated to the point where it had lost efficiency. This valve though would only operate after the pressure noticeably exceeded the mains (= diode's 0.6v drop) so there would still be an abrupt initial rise in pressure (spike) which would exceed mains pressure until the valve opened (diode conducts) and the rest of the energy would be spread over time.

The peak high pressure "polarity" is the same as the mains, relative to atmospheric pressure, but higher. It doesn't swing negative unless it oscillates (rings).

We could also introduce air bubbles (capacitance and ringing) but if my analogy is correct then it would be a bit silly to say the tx coil back emf is a waste of energy and serves no purpose at all when it is the collapsing field and resultant emf which appears to drive the process to zero.

If it is not correct then I'm sure someone will tell me so, or at least I hope they will.

If I understand your water analogy correctly, I think you are attempting to prove that the back-emf is not a negative voltage. This is incorrect.

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.

Let's take an even simpler example than we did previously. i.e. a 10uH coil (with a 10 ohm DC resistance) connected in series with a switch (10 ohm ON, 100MEG OFF), and connected across a 10V supply - see SPICE netlist.

Plot 1: Switch closes. No surprises here - voltages across coil and switch both settle at half the supply voltage (5V), as we effectively have two 10 ohm resistors in series across a 10V supply.

Plot 2: Switch opens. There is a back-emf generated across the coil of -448V due to the collapsing magnetic field, and +458V appears across the switch. Note that this value is the sum of the back-emf + the power supply voltage (448 + 10). This has to be the case, in order to satisfy Kirchoff's laws.

Plot 3: Now this is the interesting one. This shows what happens to the current in the coil. When the switch was closed, a steady-state current was established in the coil of 500mA. i.e. 10V/20ohms. When the switch opens, the magnetic field collapses, and the current in the coil rapidly decays. At no time does this current become negative. Therefore, from Ohm's Law, if the voltage is negative and the current is positive, then the resistance must be negative. In this case, because we are dealing with a coil, it is actually the inductive reactance that goes negative. This simply means that the coil has now become a generator.

Personally I don't like water analogies when describing electrical circuits. It only seems to be relevant for simple passive circuits.

----------------------------------------------------------------
* Really simple coil test circuit

* 10uH coil with 10 ohm DC resistance
L1 VP10 1 10u
RL1 1 A 10

* Switch with 10 ohm ON resistance and 100MEG OFF resistance
YSW1 VSWITCH A 0 STIM 0 param: RON=10 ROFF=100MEG

* 10V supply
V1 VP10 0 DC 10

* Pulse that turns switch on and off
V2 STIM 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(VP10,A) V(A))
.PLOT I(RL1)
.END
----------------------------------------------------------------

[Qiaozhi]

Actually, I understand what you are talking about. And, I always have.

What you are calling back emf, I have always called flyback. I have always provided protection, in the form of reversed diodes, for relays and solenoids in my circuits. But, I have always called it flyback protection.

Just a difference in what I was taught I guess. I was always taught that back emf was an opposition to a changing current and flyback was the result of a collapsing field.

Rip

"Flyback" is a term derived from the line output transformer (LOPT) that is used in CRT displays, which is also commonly known as a flyback transformer. It acquired this name because the back-emf from the coil is used to generate the flyback pulse. Even in descriptions of how a flyback transformer works, the reverse voltage across the coil is referred to as the back-emf.
Unfortunately this is a common occurance in the English language, where words get misused over time, and then misunderstandings can arise.
At least we got it sorted out in the end.

[robby_h]

"Upon closing the valve" should be "Upon opening the valve".

[robby_h]

Suppose a turbine input is connected to the mains and the output is connected to a water valve and the valve output is at atmospheric pressure and the turbine is coupled to an unloaded flywheel and the water valve, when opened, has some resistance (fet rds_on).

Upon closing the valve, the pressure at the turbine output drops to zero because the flywheel (inductance) resists change and is slow to react (a high impedance).

The unloaded turbine/flywheel slowly gathers momentum, and the pressure at turbine output rises towards mains pressure and eventually flat tops at a pressure below mains after the turbine attains maximum speed.

So, We then turn the output valve off but the energy stored in the flywheel (magnetic field) is dumped back into the system and causes the flywheel to keep spinning. This in turn results in an increase in pressure (emf) but a dramatically limited water flow (current).

This pressure (emf) initially rises until it equals mains pressure and if the process ceased at this point then the turbine would slowly grind to a stop in a time determined by it's own efficiency, but the flywheel continues to spin and the pressure continues to rise to well above mains pressure and the only way out is back thru the turbine to the much lower mains pressure. This causes water (current) to flow back thru the turbine, which in turn forces the turbine to come to an abrupt halt.
If we fit a one way valve (diode) to allow a low resistance path around the turbine back to the mains then the turbine would continue to rotate until the energy stored in the flywheel had dissipated to the point where it had lost efficiency. This valve though would only operate after the pressure noticeably exceeded the mains (= diode's 0.6v drop) so there would still be an abrupt initial rise in pressure (spike) which would exceed mains pressure until the valve opened (diode conducts) and the rest of the energy would be spread over time.

The peak high pressure "polarity" is the same as the mains, relative to atmospheric pressure, but higher. It doesn't swing negative unless it oscillates (rings).

We could also introduce air bubbles (capacitance and ringing) but if my analogy is correct then it would be a bit silly to say the tx coil back emf is a waste of energy and serves no purpose at all when it is the collapsing field and resultant emf which appears to drive the process to zero.

If it is not correct then I'm sure someone will tell me so, or at least I hope they will.

[UWLocator]

Actually, I understand what you are talking about. And, I always have.

What you are calling back emf, I have always called flyback. I have always provided protection, in the form of reversed diodes, for relays and solenoids in my circuits. But, I have always called it flyback protection.

Just a difference in what I was taught I guess. I was always taught that back emf was an opposition to a changing current and flyback was the result of a collapsing field.

Rip

[Qiaozhi]

OK, think about it this way. Put a 100M resistor across the switch, and a 1K resistor in series with the coil. Now put a scope across the 1K resistor to view the DC signals. The polarity across the resistor will be the same with through out the cycle.

The voltage will go a lot higher for a short period when the switch is opened, but the polarity will not change. That is why we use a reversed diode across the coil of a relay. We are protecting the switching circuit from that high induced voltage.

Rip

BTW: Don't really do this, the battery will not like it.

Hi Guys,

You are right in both cases. The polarity does reverse and the current flows in the same direction as the original. What you need to consider is the battery is the source and the inductor the sink initially. When you open the circuit the magnetic field collapses and the the EMF is reversed but now the inductor is the generator and the battery is the sink. This is the confusing part,direction of current flow through the battery.

Regards,

Stefan

Hi Stefan,

You beat me to it.

The reason that UWLocator is not seeing the back-emf as a negative voltage is because he is measuring across the 1K resistor in his example.
Let us use this simple example, as follows:

Take a 12 supply, and connect the positive terminal to a 1mH coil with (let's say 10 ohms DC resistance), place a 1K resistor in series with the coil, and finally then take this via a switch (with an on-resistance of 10 ohms and an off-resistance of 100MEG) to the negative terminal of the battery. We will use the negative terminal as the 0V line. Close the switch and wait until the circuit reaches steady-state. There will be 117.6mV across the coil, 11.76V across the 1K resistor, and 117.6mV across the switch, totaling 12V. Then open the switch. A negative back-emf will be developed across the coil (from simulation) of -1141.74V. In all my analyses I have been consistent in stating that this back-emf is produced by the collapsing magnetic field across the coil. As we are discussing metal detectors and PI coils here, this is the correct place to do the measurement. Now this is where the misunderstandings can occur. The battery or power supply is holding each end of this circuit at a steady 12V, which means that there must be +1153.74V (1141.74 + 12) across the series combination of the 1K resistor and switch. In fact there is 4.31V across the 1K resistor, and 1149.43V across the switch. Now it is clear why arcing occurs across switches, when charged inductors are involved. The result is that these 3 voltage measurements add up to 12V. In order to measure the reverse voltage from the collapsing magnetic field, you must take measurements across the coil.

You can run this simple SPICE simulation to confirm this for yourself:

---------------------------------------------------------
* Coil test circuit

L1 VP12 1 1m
RL1 1 A 10
R1 A B 1K
YSW1 VSWITCH B 0 STIM 0 param: RON=10 ROFF=100MEG
V1 VP12 0 DC 12
V2 STIM 0 PULSE 0 1 100n 100n 100n 100u 50m

.TRAN 100u 100m
.PROBE ALL
.PLOT V(VP12,A) V(A,B) V(B)
.END
---------------------------------------------------------

[Stefan]

Hi Guys,

You are right in both cases. The polarity does reverse and the current flows in the same direction as the original. What you need to consider is the battery is the source and the inductor the sink initially. When you open the circuit the magnetic field collapses and the the EMF is reversed but now the inductor is the generator and the battery is the sink. This is the confusing part,direction of current flow through the battery.

Regards,

Stefan

[simonbaker]

y

[UWLocator]

OK, think about it this way. Put a 100M resistor across the switch, and a 1K resistor in series with the coil. Now put a scope across the 1K resistor to view the DC signals. The polarity across the resistor will be the same with through out the cycle.

The voltage will go a lot higher for a short period when the switch is opened, but the polarity will not change. That is why we use a reversed diode across the coil of a relay. We are protecting the switching circuit from that high induced voltage.

Rip

BTW: Don't really do this, the battery will not like it.

[Qiaozhi]

To me, it is in the same direction as original pressure source, so I don't think of it as "back emf", but different terminology maybe.

Regards,

-SB

I don't think this is just different terminology. The back-emf opposes the applied force, otherwise you would have positive feedback.

[simonbaker]

You are correct. It's the same thing.
Back-emf is the electric potential difference that opposes the current that induces it. It is this opposing electric potential difference that manifests itself when the switch is opened. During the time when the switch closes, a magnetic field is established around the coil, which opposes the applied current. You could imagine this being the same as winding up a large spring. It gets increasingly harder to wind the spring. The spring is now charged with potential energy. The same with the coil, although the energy is contained in the magnetic field. Once you let go, this energy is released. In the case of the coil, the collapsing magnetic field generates a current in the coil, and hence a voltage that is the reverse polarity of the initial applied voltage. This is also referred to as the back-emf.

Another analogy: water current going through a pipe with a "perfect" turbine (no leakage) attached to a huge flywheel. It takes a while to get the flywheel up to speed, the inertia is the "back emf", equal and opposite to the water pressure. If you block the water flow suddenly, the flywheel momentum drives the turbine, keeping it turning in same direction, creating a huge pressure to keep the water going through the blockage. To me, it is in the same direction as original pressure source, so I don't think of it as "back emf", but different terminology maybe.

Regards,

-SB

[Qiaozhi]

I don't think so. When the field collapses, the voltage produced should be the same polarity as the applied voltage.

The coil acts as a resistor to an increasing current and it acts as a battery when the current is decreasing.

Rip

Your statement is incorrect.
Rather than post the results of a SPICE simulation, I will point you to a PI article published on Geotech -> http://geotech.thunting.com/cgi-bin/...anks/index.dat

Note the coil waveform shown on page 1, and the section entitled "Theory", where it states:
----------------------------------------------------------------
When power is applied to the coil it generates a magnetic
field proportional to the number of turns of wire and the current
passing through them. When the power is removed the voltage
across the coil first drops to zero and then, as the magnetic field
decays, builds up in the reverse direction as a back e.m.f. is
induced into the coil.
----------------------------------------------------------------
If the back-emf was the same polarity as the applied voltage, then you would have positive feedback, and the voltage would continue to build in the coil, which is not the case. This is a consequence of Lenz's law.

[UWLocator]

You are correct. It's the same thing.
Back-emf is the electric potential difference that opposes the current that induces it. It is this opposing electric potential difference that manifests itself when the switch is opened. During the time when the switch closes, a magnetic field is established around the coil, which opposes the applied current. You could imagine this being the same as winding up a large spring. It gets increasingly harder to wind the spring. The spring is now charged with potential energy. The same with the coil, although the energy is contained in the magnetic field. Once you let go, this energy is released. In the case of the coil, the collapsing magnetic field generates a current in the coil, and hence a voltage that is the reverse polarity of the initial applied voltage. This is also referred to as the back-emf.

I don't think so. When the field collapses, the voltage produced should be the same polarity as the applied voltage.

The coil acts as a resistor to an increasing current and it acts as a battery when the current is decreasing.

Rip

[Qiaozhi]

Hi friends,

I made in the mean time lots of new experiments and found interesting things:

1. Csplit = C/N isn't definitely correct. The coils full capacitance is still there and is acting always. So we can not make the coil faster with cascaded critical damping on each coil section. And it is a fact, which makes the coil design easier now.

2. Splitting the coil (like CT coil) makes the coil a bit faster (~1µs) and has a bit more target response on the target. However, the max. response time point in the receive coil (MONO) is too early that cannot be processed by the front-end at this time. The target response decays then much more compared to a standard configuration (low target response but longer response time). So the possible benefits will be real drawbacks on MONO coils. Particularly, small time constant targets will suffer from this and I won't follow this anymore. Any benefits from CT coil configuration can only be used with a DD coil configuration.

I suggest, we should optimize first the standard PI configuration (standard MONO coil with clamping diodes). There are still more possibilities to squeeze out. I will develop new SPICE files, which answers some fundamental questions with respect to target response (that's what counts at the end). Finally, we know then where to get the real bonus (free lunch).

Aziz

So the bottom line is that you now agree there is no "free lunch"?
Or even free breakfast or dinner.