New PI technology proposals.

[Aziz]

Hi Qiaozhi,

Hi Aziz,

The problem is that I'm not using LTspice, so I don't have the inductor model. Examining the asc file with a text editor, tells me that L=300uH, C=100pF and R=2 ohms. I will use those values, unless you tell me otherwise.

L, C, R are correct values for the totals coil parameters. If you want to compare two systems, these values must be equal. If you split the coil, care must be taken. Following formulas can be used for the split coil:

Lsplit = L / (N*N),
Csplit = C / N,
Rsplit = R / N,
where N = number of splits.

The damping resistor is calculated:
Rd = sqrt(L / (4*C) ),
where sqrt = square root, L=coils inductance, C=coils capacitance.
If you split the coil, each section has its own damping resistor Rdsplit. In this case, use the Lsplit and Csplit values. If you have additional capacitances seen by the coil (like the drain diode, Mosfets Coss, coax cable, coil shielding) then these values must be taken into account for the damping resistance.

If you split the coil, the Mosfets and diodes on-resistance have to also taken into account, to achieve same exposed magnetic field energy (E = 0.5*L*I*I) for comparing two systems. In this case, a drain resistance is easy to adjust two systems switch-on currents.


L, C, R may differ from model to model i have published but not in comparing two systems.

Aziz

[Qiaozhi]

Hi Qiaozhi,



L, C, R are correct values for the totals coil parameters. If you want to compare two systems, these values must be equal. If you split the coil, care must be taken. Following formulas can be used for the split coil:

Lsplit = L / (N*N),
Csplit = C / N,
Rsplit = R / N,
where N = number of splits.

The damping resistor is calculated:
Rd = sqrt(L / (4*C) ),
where sqrt = square root, L=coils inductance, C=coils capacitance.
If you split the coil, each section has its own damping resistor Rdsplit. In this case, use the Lsplit and Csplit values. If you have additional capacitances seen by the coil (like the drain diode, Mosfets Coss, coax cable, coil shielding) then these values must be taken into account for the damping resistance.

If you split the coil, the Mosfets and diodes on-resistance have to also taken into account, to achieve same exposed magnetic field energy (E = 0.5*L*I*I) for comparing two systems. In this case, a drain resistance is easy to adjust two systems switch-on currents.


L, C, R may differ from model to model i have published but not in comparing two systems.

Aziz

Hi Aziz,

I've hacked together a SPICE netlist (without a schematic) as it was the quickest approach for this simple circuit. You will see the parameter statements are implementing the equations that you provided. However, I am not getting the same results you claim to be seeing. Please have a look at the netlist below, and see if you can spot any mistake. The split coil has a lower flyback voltage than the mono.

-------------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=2
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))

* Mono coil
L1 VP12 1 L
RL1 1 2 R
C1 VP12 2 C
RD1 VP12 2 RD
YSW1 VSWITCH 1 0 3 0 param: RON=20 ROFF=100MEG

* Split Coil
L2 4 5 Lsplit
RL2 5 VP12 Rsplit
C2 4 VP12 Csplit
RD2 4 VP12 RDsplit
L3 VP12 6 Lsplit
RL3 6 7 Rsplit
C3 VP12 7 Csplit
RD3 VP12 7 RDsplit
YSW2 VSWITCH 7 0 8 0 param: RON=20 ROFF=100MEG

* Coil coupling
K23 L2 L3 1

* Voltage sources
VP12 VP12 0 DC 12
V1 3 0 PULSE 0 1 100n 100n 100n 100u 50m
V2 8 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(2,VP12) V(7,4)
.END
-------------------------------------------------------

[Aziz]

Hi Qiaozhi,

both systems doesn't expose same magnetic energy due to different current flows of each coils.
System 1:
I1 = U/(Ron+R) = 12/(20+2) = 0.545 A

System 2:
I2 = U/(Ron+R/2) = 12/(20+1) = 0.571 A

As the system 2 drives only the half of the coil, it should drive with 2 time more current to the system, to be compareable.
So I2 should be 0.545*2 A = U/(Ron/2 + R/2) = 12/(10+1) = 1.091 A.
The switch on system 2 needs Ron resistance of 10 Ohms.

Aziz

[Aziz]

Hello friends,

can anyone check the self-capacitance on the half coil of the CT coil. I suspect, the humble simplification of Csplit=C/N isn't correct.
As the self-capacitance of the full coil doesn't disappear, it has a capacitive effect to the half driven coil. In this case, there is no free lunch.
Also the splitted coil critical damping wouldn't work in this case.

Aziz

[simonbaker]

Hi Aziz,

I've hacked together a SPICE netlist (without a schematic) as it was the quickest approach for this simple circuit. You will see the parameter statements are implementing the equations that you provided. However, I am not getting the same results you claim to be seeing. Please have a look at the netlist below, and see if you can spot any mistake. The split coil has a lower flyback voltage than the mono.

-------------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=2
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))

* Mono coil
L1 VP12 1 L
RL1 1 2 R
C1 VP12 2 C
RD1 VP12 2 RD
YSW1 VSWITCH 1 0 3 0 param: RON=20 ROFF=100MEG

* Split Coil
L2 4 5 Lsplit
RL2 5 VP12 Rsplit
C2 4 VP12 Csplit
RD2 4 VP12 RDsplit
L3 VP12 6 Lsplit
RL3 6 7 Rsplit
C3 VP12 7 Csplit
RD3 VP12 7 RDsplit
YSW2 VSWITCH 7 0 8 0 param: RON=20 ROFF=100MEG

* Coil coupling
K23 L2 L3 1

* Voltage sources
VP12 VP12 0 DC 12
V1 3 0 PULSE 0 1 100n 100n 100n 100u 50m
V2 8 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(2,VP12) V(7,4)
.END
-------------------------------------------------------

Hi Qiaozhi - I tried to draw your netlist for split coil in LTSpice.

(By the way, in mono coil, should switch be connected to node 2 instead of 1?)

In split coil, node 4 doesn't make circuit -- it floats at 12 volts. I don't get waveform for V(7,4) like yours, instead step up from -1.1V to 0.0 V.

I must be missing something from your netlist.

-SB

[Aziz]

Hello friends,

I found an error in my thoughts. Does anybody know, where the big mistake is? As Carl stated, there is no free lunch available! Except, you will work for it.

This is a good task for you all to find the error. Start thinking a little bit unorthodox. Later, I will show you where the error is and then start to squeeze out the available free lunch! (Only available after same work).

Hint:
Consider the time constant of the coil (T = L / R).


Aziz

[Carl-NC]

can anyone check the self-capacitance on the half coil of the CT coil. I suspect, the humble simplification of Csplit=C/N isn't correct.

Most likely if you test a coil segment in isolation, you will find that it has lower self-capacitance, maybe even approaching C/N. But when you put the segments together, you get inter-coil capacitance plus the mutual induction between segments, and the whole thing is too complicated to model.

- Carl

[UWLocator]

Most likely if you test a coil segment in isolation, you will find that it has lower self-capacitance, maybe even approaching C/N. But when you put the segments together, you get inter-coil capacitance plus the mutual induction between segments, and the whole thing is too complicated to model.

- Carl

I think you wind up with more negatives than positives with the transmit only on 1/2 of the coil.

You have the other 1/2 coil acting as a secondary load during transmit. And, being coupled fairly well to the primary, it has a major effect on the inductance of the primary.

I would not expect to see any benefit in the signal available from the target as a result of transmit on 1/2 of the coil. I would expect the current required to be more than double that of a mono coil.

Trading battery time for a higher voltage during field collapse does not seem like a good idea to me.

I do believe that the tapped coil with multiple damping resistors has some potential to allow faster target detection. But, I would still use the entire coil for both transmit and receive.

JMO,
Rip

[Aziz]

Hi Carl,

Most likely if you test a coil segment in isolation, you will find that it has lower self-capacitance, maybe even approaching C/N. But when you put the segments together, you get inter-coil capacitance plus the mutual induction between segments, and the whole thing is too complicated to model.

- Carl

that's correct. The mutual inductance isn't a problem here and is correct modelled. The capacitive coupling of the coil halves is still there and isn't modelled in the SPICE simulation. If I put this into the model, then there isn't really a free lunch. But improvements are still there. As you mentioned, the capacitive coupling of the coil halves must be reduced (which also reduces slightly the inductive coupling). This could be achieved with some spacer of each coil halves. There is an another big improvement when the MOSFET does not reach its avalanche breakdown voltage.

But I want to let people think about the mistakes I did before presenting a better solution. The good news is, that SPICE is working correct and can be realible used to simulate PI technology. Except the model is wrong.

Regards,
Aziz

[Qiaozhi]

Thanks Aziz. I was wondering whether that might be the problem. Now the results match.

Thanks Simonbaker. You are correct ... I accidentally used node 1 instead of 2, although it only makes a small difference to the results. While hacking together the netlist by hand, I typed in the value of RL1 which was next to the node number (2). Doh!
I suspect you're not getting the correct output, because your coils are in antiphase. The coupling coefficient of unity means they are cancelling each other exactly.
And finally ... the updated netlist is shown below, with 3 sets of results. I am not happy with the coupling coefficient set to unity, and have re-run the simulation using 0.5 and 0.1. As you can see, the waveform settles even faster, but with a reduced peak amplitude.
Comments?

---------------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=2
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))

* Single coil
L1 VP12 1 L
RL1 1 2 R
C1 VP12 2 C
RD1 VP12 2 RD
YSW1 VSWITCH 2 0 3 0 param: RON=20 ROFF=100MEG

* Split Coil
L2 4 5 Lsplit
RL2 5 VP12 Rsplit
C2 4 VP12 Csplit
RD2 4 VP12 RDsplit
L3 VP12 6 Lsplit
RL3 6 7 Rsplit
C3 VP12 7 Csplit
RD3 VP12 7 RDsplit
YSW2 VSWITCH 7 0 8 0 param: RON=10 ROFF=100MEG

* Coil coupling
K23 L2 L3 1

* Voltage sources
VP12 VP12 0 DC 12
V1 3 0 PULSE 0 1 100n 100n 100n 100u 50m
V2 8 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(2,VP12) V(7,4)
.END
---------------------------------------------------------

[simonbaker]

Thanks Aziz. I was wondering whether that might be the problem. Now the results match.

Thanks Simonbaker. You are correct ... I accidentally used node 1 instead of 2, although it only makes a small difference to the results. While hacking together the netlist by hand, I typed in the value of RL1 which was next to the node number (2). Doh!
I suspect you're not getting the correct output, because your coils are in antiphase. The coupling coefficient of unity means they are cancelling each other exactly.
And finally ... the updated netlist is shown below, with 3 sets of results. I am not happy with the coupling coefficient set to unity, and have re-run the simulation using 0.5 and 0.1. As you can see, the waveform settles even faster, but with a reduced peak amplitude.
Comments?

---------------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=2
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))

* Single coil
L1 VP12 1 L
RL1 1 2 R
C1 VP12 2 C
RD1 VP12 2 RD
YSW1 VSWITCH 2 0 3 0 param: RON=20 ROFF=100MEG

* Split Coil
L2 4 5 Lsplit
RL2 5 VP12 Rsplit
C2 4 VP12 Csplit
RD2 4 VP12 RDsplit
L3 VP12 6 Lsplit
RL3 6 7 Rsplit
C3 VP12 7 Csplit
RD3 VP12 7 RDsplit
YSW2 VSWITCH 7 0 8 0 param: RON=10 ROFF=100MEG

* Coil coupling
K23 L2 L3 1

* Voltage sources
VP12 VP12 0 DC 12
V1 3 0 PULSE 0 1 100n 100n 100n 100u 50m
V2 8 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(2,VP12) V(7,4)
.END
---------------------------------------------------------

I forgot completely about coil phasing, thanks.

I still don't understand "floating" coil that is attached to V+ at one end and nothing at other (except damping resistor across it).

I think you could completely disconnect second coil section and make it a transformer instead of center-tapped half and get same result (without DC bias).

It seems there should be another damping resistor that spans both coils to help out somehow.

* Very revealing Aziz simulations and how tricky to model real coils - interesting stuff.

Regards

-SB

[Qiaozhi]

I forgot completely about coil phasing, thanks.

I still don't understand "floating" coil that is attached to V+ at one end and nothing at other (except damping resistor across it).

I think you could completely disconnect second coil section and make it a transformer instead of center-tapped half and get same result (without DC bias).

It seems there should be another damping resistor that spans both coils to help out somehow.

* Very revealing Aziz simulations and how tricky to model real coils - interesting stuff.

Regards

-SB

Let me see if I can explain what's going on ...

With reference to the two graphs below, and the updated netlist.
1st Graph - the switch closes
2nd Graph - the switch opens

These graphs show both the current and voltage waveforms for the split coil configuration. You will note in the 1st graph that the voltage in the lower coil V(TX2,VP12) jumps to nearly -12V and then decays to a value of around -600mV, which represents the voltage drop across the resistance of the switch (10 ohms). The voltage in the upper coil also jumps initially to nearly -12V, and decays to 0V. The reason the top coil jumps to nearly -12V, when it appears to be floating, is because L2 and L3 are closely coupled. As time progresses a current path is established through L3 (the lower coil) and RL3 from VP12 to TX2, which decays to 600mV, but no similar path is available for current to flow through L2 (the upper coil). In the latter case there is a very small current that is dissipated through L2 and RL2, in parallel with RD2. The result is that a magnetic field is established around the lower coil (L3), but the upper coil (L2) does not contribute to this field.

Now ... the switch opens (2nd graph) and the magnetic field collapses, causing the current flow through the lower coil (L3) to drop quickly to zero. The collapsing magnetic field also generates a very large back emf across both coils. Again, remember that L2 and L3 are closely coupled, and the coupling coefficient (K) has a large effect on the behavior. The result is that both coils will experience this large voltage from the back emf, and since they are in series, there will be double the voltage across the whole coil assembly. So what's happening to the current? The current in the lower coil starts to decay faster than with the mono coil because the upper coil (L2) is providing a charge path in addition to the usual damping resistor. So the current in L3 is dropping, but the current in L2 is increasing. Once the current flow is equalized, the decay proceeds at its normal expected rate, and both coils will discharge through their respective damping resistors. This is why the damping is faster with this configuration, when compared to the mono coil, as the upper coil provides an additional current path.
If you look at the netlist, you will see a third coil configuration with a single damping resistor, instead of two separate ones. The simulation results for both configurations were the same. i.e. no improvement in decay time between the two types.

Updated netlist:

-----------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=1
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))
.param RON_mono=20
.param RON_split=10
.param ROFF=100MEG

* Single coil
L1 VP12 1 L
RL1 1 TX1 R
C1 VP12 TX1 C
RD1 VP12 TX1 RD
YSW1 VSWITCH TX1 0 STIM 0 param: RON=RON_mono ROFF=ROFF

* Split coil with separate RD
L2 A 2 Lsplit
RL2 2 VP12 Rsplit
C2 A VP12 Csplit
RD2 A VP12 RDsplit
L3 VP12 3 Lsplit
RL3 3 TX2 Rsplit
C3 VP12 TX2 Csplit
RD3 VP12 TX2 RDsplit
YSW2 VSWITCH TX2 0 STIM 0 param: RON=RON_split ROFF=ROFF

* Split coil with single RD
L4 B 4 Lsplit
RL4 4 VP12 Rsplit
C4 B VP12 Csplit
L5 VP12 5 Lsplit
RL5 5 TX3 Rsplit
C5 VP12 TX3 Csplit
RD4 B TX3 {RDsplit*2}
YSW3 VSWITCH TX3 0 STIM 0 param: RON=RON_split ROFF=ROFF

* Coil coupling
K23 L2 L3 1
K45 L4 L5 1

* Voltage sources
VP12 VP12 0 DC 12
V1 STIM 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(TX1,VP12) V(TX2,A) V(TX3,B)
.PLOT V(TX2,VP12) V(VP12,A)
.END
-----------------------------------------------------

[UWLocator]

Let me see if I can explain what's going on ...

With reference to the two graphs below, and the updated netlist.
1st Graph - the switch closes
2nd Graph - the switch opens

These graphs show both the current and voltage waveforms for the split coil configuration. You will note in the 1st graph that the voltage in the lower coil V(TX2,VP12) jumps to nearly -12V and then decays to a value of around -600mV, which represents the voltage drop across the resistance of the switch (10 ohms). The voltage in the upper coil also jumps initially to nearly -12V, and decays to 0V. The reason the top coil jumps to nearly -12V, when it appears to be floating, is because L2 and L3 are closely coupled. As time progresses a current path is established through L3 (the lower coil) and RL3 from VP12 to TX2, which decays to 600mV, but no similar path is available for current to flow through L2 (the upper coil). In the latter case there is a very small current that is dissipated through L2 and RL2, in parallel with RD2. The result is that a magnetic field is established around the lower coil (L3), but the upper coil (L2) does not contribute to this field.

Now ... the switch opens (2nd graph) and the magnetic field collapses, causing the current flow through the lower coil (L3) to drop quickly to zero. The collapsing magnetic field also generates a very large back emf across both coils. Again, remember that L2 and L3 are closely coupled, and the coupling coefficient (K) has a large effect on the behavior. The result is that both coils will experience this large voltage from the back emf, and since they are in series, there will be double the voltage across the whole coil assembly. So what's happening to the current? The current in the lower coil starts to decay faster than with the mono coil because the upper coil (L2) is providing a charge path in addition to the usual damping resistor. So the current in L3 is dropping, but the current in L2 is increasing. Once the current flow is equalized, the decay proceeds at its normal expected rate, and both coils will discharge through their respective damping resistors. This is why the damping is faster with this configuration, when compared to the mono coil, as the upper coil provides an additional current path.
If you look at the netlist, you will see a third coil configuration with a single damping resistor, instead of two separate ones. The simulation results for both configurations were the same. i.e. no improvement in decay time between the two types.

Updated netlist:

-----------------------------------------------------
* PI Coil Simulation

* Coil parameters
.param L=300u
.param C=100p
.param R=1
.param RD=SQRT(L/(4*C))
.param N=2
.param Lsplit=L/(N*N)
.param Csplit=C/N
.param Rsplit=R/N
.param RDsplit=SQRT(Lsplit/(4*Csplit))
.param RON_mono=20
.param RON_split=10
.param ROFF=100MEG

* Single coil
L1 VP12 1 L
RL1 1 TX1 R
C1 VP12 TX1 C
RD1 VP12 TX1 RD
YSW1 VSWITCH TX1 0 STIM 0 param: RON=RON_mono ROFF=ROFF

* Split coil with separate RD
L2 A 2 Lsplit
RL2 2 VP12 Rsplit
C2 A VP12 Csplit
RD2 A VP12 RDsplit
L3 VP12 3 Lsplit
RL3 3 TX2 Rsplit
C3 VP12 TX2 Csplit
RD3 VP12 TX2 RDsplit
YSW2 VSWITCH TX2 0 STIM 0 param: RON=RON_split ROFF=ROFF

* Split coil with single RD
L4 B 4 Lsplit
RL4 4 VP12 Rsplit
C4 B VP12 Csplit
L5 VP12 5 Lsplit
RL5 5 TX3 Rsplit
C5 VP12 TX3 Csplit
RD4 B TX3 {RDsplit*2}
YSW3 VSWITCH TX3 0 STIM 0 param: RON=RON_split ROFF=ROFF

* Coil coupling
K23 L2 L3 1
K45 L4 L5 1

* Voltage sources
VP12 VP12 0 DC 12
V1 STIM 0 PULSE 0 1 100n 100n 100n 100u 50m

* Simulator commands
.TRAN 100u 100m
.PROBE ALL
.PLOT V(TX1,VP12) V(TX2,A) V(TX3,B)
.PLOT V(TX2,VP12) V(VP12,A)
.END
-----------------------------------------------------

I guess I need to go back to school. I thought back emf was the opposition to an increase in current as a voltage was applied to a coil.

HH,
Rip

[simonbaker]

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

[Qiaozhi]

I guess I need to go back to school. I thought back emf was the opposition to an increase in current as a voltage was applied to a coil.

HH,
Rip

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.