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  • Originally posted by simonbaker View Post
    I think in your first split coil arrangement (floating end), you can disconnect the two coils electrically at the center tap with no difference to the behavior, as the coupling is purely magnetic - do you see what I mean?

    Center tap is only meaningful if you have some return current path in the non-driven coil, otherwise it might as well be pure transformer (electrically separate coils).

    Is that not so?

    Regards,

    -SB
    Actually, this is not my split-coil arrangement. It is a SPICE simulation of the circuit first suggested by Aziz.

    Like you, I was a little confused about the floating section, although I could see that there was a connection between the two coils because of the close magnetic coupling. That's why I decided to investigate further.

    I've had some more thoughts on this ... but will post later.
    I'm still not convinced about the advantages (or otherwise) of this approach.

    Comment


    • Originally posted by simonbaker View Post
      I think in your first split coil arrangement (floating end), you can disconnect the two coils electrically at the center tap with no difference to the behavior, as the coupling is purely magnetic - do you see what I mean?

      Center tap is only meaningful if you have some return current path in the non-driven coil, otherwise it might as well be pure transformer (electrically separate coils).

      Is that not so?

      Regards,

      -SB
      OK - I can see what you're suggesting. I've isolated the upper coil electrically, but kept the inductive coupling. The voltages and currents remain essentially the same, but there is (of course) no large voltage available across the series combination of the coils, as they are now electrically disconnected. I seem to remember that Aziz said that this was necessary for improved target response. Also, it is now not possible to use a balanced receiver. To be honest, it has become rather confusing, with the totally floating upper coil, and any potential improvement can only be confirmed or denied by building a test coil. Personally I think this is a leading up a blind alley.

      Comment


      • The result of some further experiments:

        Since I remain to be convinced on the potential benefits of this split coil approach, I decided to run a simple simulation to compare the decay times for the standard mono coil (300uH), the split coil (which is essentially the mono coil, but with a center-tap), and a single 75uH coil (i.e. one half of the split coil).

        The first plot below shows what happens to the current in the various coils when the switch opens. It is interesting to note that the single 75uH coil has the fastest decay.

        The second plot shows what happens to the voltage across the various coils when the switch opens. Again (as you might expect) the 75uH coil has the fastest decay.

        The conclusion is that there is no advantage whatsoever to using the split coil arrangement over a single coil. The faster decay time is attributable simply to the reduction in coil inductance and coil capacitance. The disadvantage is the increased power consumption required to drive this coil.

        As Carl stated earlier ... "there is no free lunch".
        Attached Files

        Comment


        • 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

          Comment


          • Originally posted by Aziz View Post
            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.

            Comment


            • Originally posted by Qiaozhi View Post
              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

              Comment


              • Originally posted by UWLocator View Post
                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.

                Comment


                • Originally posted by Qiaozhi View Post
                  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

                  Comment


                  • Originally posted by simonbaker View Post
                    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.

                    Comment


                    • 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.
                      Last edited by UWLocator; 12-12-2008, 06:16 PM. Reason: Add the BTW

                      Comment


                      • y
                        Last edited by simonbaker; 12-12-2008, 07:06 PM. Reason: too hasty

                        Comment


                        • 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

                          Comment


                          • Originally posted by UWLocator View Post
                            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.
                            Originally posted by Stefan View Post
                            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
                            ---------------------------------------------------------

                            Comment


                            • 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

                              Comment


                              • 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.

                                Comment

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