PI Monocoil - sampling under 4us almost impossible?

[Tinkerer]

Hi Simon,
I tried that both in practice and as simulation and I get exactly the same values for the resistor, no matter what the peak voltage is.

Regards,
Nicolae

Hi Nicolae,

When you increase the pulse length you store more energy in the coil's magnetic field. You have more Amps at switch OFF. So you are comparing apples with melons.
For a specific amount of energy stored in the Coil's field, a higher resistor value will drive the Flyback higher and the decay will be faster.
However, if the Flyback voltage goes above the Mosfets reverse diode, this will slow down the decay.
Take Misc.El and open the "Inductor discharge" page and play a bit with various A and R values. It is very interesting.

Tinkerer

[Ray-NM]

Just a question or 2.
If you increase the pulse width and decrease the coil inductance could you then sample earlier. The other question is what is the minimum inductance you can get away with and still have a good coil?
RayNM

[Aziz]

Hi Nicolae,

A few people already mentioned that getting a higher voltage on the coil will shorten the decay time. Can anybody explain to me why would this happen? To me it's not obvious.

Some Kirchhoff's Law for remember:
U = R*I
P = U*I = R*I² = U²/R

where,
U=voltage (i.e. flyback voltage)
I=current (i.e. coil current)
R=resistor (i.e. dampening resistor Rd, we can omit the other resistors like mosfet's Rdson, coils series resistor, ... in the circuit and they are not dominating here)
P=power (the pulse energy, which will be converted into heat in the dampening resistor, we also omit other losses here)

Generally, we want a critically damped search coil configuration. The damping resistor Rd depends only on the coils inductivity L and the total capacitance C, which includes coils capacitance, shielding and cables, mosfet's parasitic capacitances and other parasitic capacitances in the circuit.

The damping resistor is calculated as follows:
Rd = sqrt(L/(4*C))
We omit the coils series resistance in the chain (not dominating).

The flyback voltage is dependent on the coil current I at switch-off, the switch-off slope dI/dt (how fast) and damping resistor Rd (also dependend on L and C therefore). The higher the damping resistor Rd (for critical damping), the higher the flyback voltage. The lower the capacitance, the higher the flyback voltage. The higher the coil current, the higher the flyback voltage. There are limitations to the switching device: The avalanche breakdown voltage of the mosfet, which limits the flyback voltage.

Generally, the switch-off behaviour and search coil parameters L, C, Rd can be seen as constant for a given configuration. The exposed magnetic field energy E = 0.5*L*I² must be damped on the resistive elements (mainly in the Rd).

Back to Kirchhoff:
P = U*I = R*I² = U²/R

Consider, what happens to power P, if we double the flyback voltage. It grows quadratic, thus four times more. We can damp more. Power P could also be increased, if we reduce R. But the flyback voltage won't grow and the coil won't be critically damped anymore.

Note, that the flyback voltage U is not constant. It grows to maximum level and then decreases in a not linear manner. Only during avalanche breakdown of the mosfet it is nearly constant.

So, making the flyback voltage higher, makes the magnetic field energy E conversion into heat more efficient and hence faster.

I hope, this helps you for understanding.
Aziz

[nick_f]

Hi Nicolae,

I hope, this helps you for understanding.
Aziz

Hi Aziz,
Thank you very much for your detailed explanation. I will check it again, but at the moment unfortunately, it doesn't convince me the pulse will be faster. You could convince me a lot more if you show me some Spice simulation, where you measure the signal of the antiparallel diodes. If you can show me that for a higher voltage you can increase the resistance without causing underdamping, I will believe you. So far I can't simulate such an effect, for more than double in the flyback voltage, I can't increase the damping resistor not even 5% (in fact any amount), because I get underdamping. My experiments matched well the simulation, once I found the optimal damping resistor, I could vary the flyback voltage up and down and I couldn't optimise the resistor any better.
Your formula, Rd = sqrt(L/(4*C)) clearly shows the damping resistor doesn't depend on the voltage, only on the value of L and C. As long as changing the flyback pulse won't change the value of L and C, I believe you are not right. Have you done a practical test and got to the conclusion that if you increase the flyback pulse you can increase the damping resistor?
Even worse, by increasing the flyback voltage, I get an increase of the decay time, which I consider can be normally expected. It is more energy to dissipate on the damping resistor, and it takes a bit longer.
I see it similar with discharging a capacitor through a resistor. The higher the voltage on the capacitor, the longer will take to discharge. I don't think it matters if the charge is hold in a LC circuit, in a coil or in a capacitor (but I could be wrong here).
When I was playing with my board and chasing minimal pulse with, I was reducing the flyback voltage, while maintaining constant the damping resistor.

I insist that for the real test, you have to use the antiparallel diodes, and measure on them, otherwise you won't be able to pick the underdamping caused by a slight increase of the damping resistance.

Best regards,
Nicolae

[nick_f]

Hi Nicolae,

Take Misc.El and open the "Inductor discharge" page and play a bit with various A and R values. It is very interesting.

Tinkerer

Hi Tinkerer,
I checked the MiscEl and I changed the maximum current to different values (obtaining different flyback voltages). It agrees with what I am saying. The higher the initial current, the long it will take to discharge the voltage to a specific value (let's say 10V).
I think that by increasing the flyback voltage to obtain faster damping is a myth and I am waiting for the first person who will realise/admit I am right
Which is, if you want a shorter pulse, use less peak voltage (= less initial current through the coil).
Even if we take into consideration the capacitor in parallel with the inductor, both of them store energy and the more energy is in the circuit, the more time takes to discharge it through a resistor.

Regards,
Nicolae

[nick_f]

Hi Nicolae,

Consider, what happens to power P, if we double the flyback voltage. It grows quadratic, thus four times more. We can damp more. Power P could also be increased, if we reduce R. But the flyback voltage won't grow and the coil won't be critically damped anymore.

Note, that the flyback voltage U is not constant. It grows to maximum level and then decreases in a not linear manner. Only during avalanche breakdown of the mosfet it is nearly constant.

So, making the flyback voltage higher, makes the magnetic field energy E conversion into heat more efficient and hence faster.

Aziz

Hi Aziz,

I checked closer your demonstration. I don't have any objections on the formulas you mentioned there. My objection is that there is no relation between the formulas you presented that would demonstrate the relation between flyback voltage and decay time.
You should have arrived to a formula like t = function (1/V), and you didn't.

Regards,
Nicolae

[Aziz]

Hi Nicolae,

please, don't confuse. For the evidence of the physical fact, you have to expose same magnetic field energy E=0.5*L*I². Transmitting longer will cause more coil current and therefore more magnetic field energy E. This coil current will saturate for much longer transmit pulses (resistances will limit the DC current then). The higher magnetic field energy, the longer the damping process time. So you cannot compare two systems with different transmit pulse energy.

For a constant magnetic field energy E of two systems, the flyback voltage might be different depending on the factors I already mentioned. Try to find two equal mosfets (same Rdson, same parasitic capacitances, same switching behaviour) but with different breakdown voltage and simulate this. You will see the evidence.

Aziz

[nick_f]

Hi Aziz,
We should never reach the breakdown voltage of the transistor. Once we reach that, it makes no sense to pump more current through the coil, because it will only be wasted battery without any benefit to the detector.
If the breakdown voltage is reached for 1us, that 1us will only add to the decay time, without any advantage.
If we assume an equal amount of energy and the same inductor, and same parasitic capacitance, how can we generate a higher flyback voltage? We can't increase damping resistor, because the circuit will get underdamped. I would say that equal energy and higher flyback voltage for the same circuit is not possible (definition of madness: keep doing the same thing and expect a different result ) - sorry, I was just joking here).
In theory, we could obtain a higher flyback voltage for the same amount of energy by increasing the quality factor of the coil (decreasing the series resistance in circuit).

Regards,
Nicolae

[nick_f]

Nicolae,

Yep i'm an Aussie, sure we can arrange a prospecting trip if you don't mind walking a lot of Km's & digging a lot of holes. It is an adventure as you put it but it is hard hot work so there will be lots of blisters on the feet & hands.

I usually go away for long periods at a time, last time was nearly a year but we can organise something, i am home for a while at the moment & will be until i get this modification sorted & understood.

I'll let you know when i am about to make a move & we can sort something out, you will get hooked on it so be prepared, it will be a while until i make another trip but not to long hopefully.

I'll look forward to the trip.

PS,

I hope you have a sense of humour because mine sucks hahaha!!

Hi B^C,
Thank you very much for the offer, I will consider this. Unfortunately I am permanently very busy at work and the most I could come is like 2 days, over a weekend. I don't mind the walking and I am not too scared of blisters.
I also don't mind if we can find again in two days what you found last time in the one year trip When you plan something, please send me an email or PM, because I only discovered your message today.

I don't have much of a good sense of humour, but I will try to borrow some before the trip.

Best regards,
Nicolae

[bbsailor]

Critical Damping Versus Slightly Overdamping

Hi Aziz,
We should never reach the breakdown voltage of the transistor. Once we reach that, it makes no sense to pump more current through the coil, because it will only be wasted battery without any benefit to the detector.
If the breakdown voltage is reached for 1us, that 1us will only add to the decay time, without any advantage.
If we assume an equal amount of energy and the same inductor, and same parasitic capacitance, how can we generate a higher flyback voltage? We can't increase damping resistor, because the circuit will get underdamped. I would say that equal energy and higher flyback voltage for the same circuit is not possible (definition of madness: keep doing the same thing and expect a different result ) - sorry, I was just joking here).
In theory, we could obtain a higher flyback voltage for the same amount of energy by increasing the quality factor of the coil (decreasing the series resistance in circuit).

Regards,
Nicolae

Nicolae and Aziz,

Typically, PI circuits are slightly overdamped to allow sampling at the earliest time. One experiment that would be most helpful is to use a variable resistor to find the optimal PI damping resistance, then add 10pf, 50pf, 100pf, 200pf, and 300pf fixed capacitor to the TX circuit and re- adjust the optimal damping for each capacitor value and then graph the result for the calculated value of Rd versus the optimumly adjusted value of Rd.

Look at all the things that add capacitance to the TX circuit.
Coil winding capacitance
Shield-to-coil capacitance
Coax Capacitance
MOSFET COSS
Clamping diode capacitance
Use a series diode between MOSFET drain and coil to minimize MOSFET COSS

Aziz and Nicolae, can you verify something that I have very crudly measured? For each 100pf I reduce in the TX circuit capacitance, I can sample about 1us faster. This is not linear and is mostly approximate above 10us delay. Below 10us delay less than 100pf capacitance reduction seems to improve the delay by about 1us.

The best way to experiment with this is to observe an actual coil resonance without the damping resistor by measuring the period between pulsed resonance oscillations which represents all the things above that add capacitance. Make sure you remove the first opamp stage to not blow it when doing this experiment. You may want to use the probe 10X setting and/or insert a 1 to 3 pf capacitor in series with the probe to minimize probe loading for a more accurate measurement. Just convert the pulsed resonance peak period to an equivalent self resonance and calculate the total TX circuit capacitance with a known coil inductance.

This experiment will be most helpful to see the theory and the reality of the values of Rd. Also note that an optimum Rd value (for earliest sampling) will only be valid for a narrow range of TX pulse widths. Try measuring Rd with pulse widths of 25us, 50us, 100us, and 150us and see how this affects the adjusted optimum value of Rd for each pulse width and flyback peak. Doing this experiment will get to the heart of the matter. A higher value of optimumly adjusted Rd will make for a more sensitive coil.

bbsailor

[nick_f]

Nicolae and Aziz,

Typically, PI circuits are slightly overdamped to allow sampling at the earliest time. One experiment that would be most helpful is to use a variable resistor to find the optimal PI damping resistance, then add 10pf, 50pf, 100pf, 200pf, and 300pf fixed capacitor to the TX circuit and re- adjust the optimal damping for each capacitor value and then graph the result for the calculated value of Rd versus the optimumly adjusted value of Rd.

Look at all the things that add capacitance to the TX circuit.
Coil winding capacitance
Shield-to-coil capacitance
Coax Capacitance
MOSFET COSS
Clamping diode capacitance
Use a series diode between MOSFET drain and coil to minimize MOSFET COSS

Aziz and Nicolae, can you verify something that I have very crudly measured? For each 100pf I reduce in the TX circuit capacitance, I can sample about 1us faster. This is not linear and is mostly approximate above 10us delay. Below 10us delay less than 100pf capacitance reduction seems to improve the delay by about 1us.

The best way to experiment with this is to observe an actual coil resonance without the damping resistor by measuring the period between pulsed resonance oscillations which represents all the things above that add capacitance. Make sure you remove the first opamp stage to not blow it when doing this experiment. You may want to use the probe 10X setting and/or insert a 1 to 3 pf capacitor in series with the probe to minimize probe loading for a more accurate measurement. Just convert the pulsed resonance peak period to an equivalent self resonance and calculate the total TX circuit capacitance with a known coil inductance.

This experiment will be most helpful to see the theory and the reality of the values of Rd. Also note that an optimum Rd value (for earliest sampling) will only be valid for a narrow range of TX pulse widths. Try measuring Rd with pulse widths of 25us, 50us, 100us, and 150us and see how this affects the adjusted optimum value of Rd for each pulse width and flyback peak. Doing this experiment will get to the heart of the matter. A higher value of optimumly adjusted Rd will make for a more sensitive coil.

bbsailor

Hi bbsailor,

I think you are refering to the slightly underdamped circuit (where the damping resistance is larger than critical and the voltage goes through zero faster, but then it goes on the other side).
All the experiments you described are indeed very interesting and we should do them. At the moment I am involved in programming a microcontroller PIC16F628 and besides my board is already fitted in the metal detector. I am unhappy about the saturation recovery time of the NE5534 with a gain of 1000 and I want to play with the two stages opamp (as in Tinkerers Preamp). I will do all these tests, but I can't promise when. I am yet to order my batch of components (I already have on the list high voltage 100pF capacitors and MOSFET transistors).
What you are saying that the damping resistor is valid only for a narrow range of currents is something unexpected to me, I have to check that. I thought it is independent on the current/peak voltage and related only to L, C, R in circuit.

Best regards,
Nicolae

[simonbaker]

Hi Tinkerer,
I checked the MiscEl and I changed the maximum current to different values (obtaining different flyback voltages). It agrees with what I am saying. The higher the initial current, the long it will take to discharge the voltage to a specific value (let's say 10V).
I think that by increasing the flyback voltage to obtain faster damping is a myth and I am waiting for the first person who will realise/admit I am right
Which is, if you want a shorter pulse, use less peak voltage (= less initial current through the coil).
Even if we take into consideration the capacitor in parallel with the inductor, both of them store energy and the more energy is in the circuit, the more time takes to discharge it through a resistor.

Regards,
Nicolae

I think part of the idea of increasing the Rd is that you have to get the capacitance down too, which is why people strain to reduce the capacitance when making coils. So my guess is that you may be right with a given coil -- not just as easy as changing Rd -- may require making different coil too with low capacitance (or some other ingenious solution beyond me...).

Cheers,

-SB

[B^C]

Gday Guy's,

I have been following this with interest, the inductor calculation
V= L dI/dt may solve the issue. Someone with a bit more mathematical skill may be able to write out a few examples.
Belt me over the head at any time if it makes no sense!
A capacitor discharges current & the voltage ramps to 0 & an inductor discharges voltage & the current ramps to 0. So we need to discharge voltage for the inductor.
An inductor holds it current at 0 terminal voltage & the capacitor holds it voltage at 0 current, so the higher the voltage spike of the inductor the quicker the current will fall.

The flyback is determined by the duty cycle & the amperage not so much total amps but how fast the amperage changes direction, take note of this fact.

Nicolae might not be seeing much of a change due to the amperage changing at the same rate each test?.
Or there may be to much capacitance, or the flyback is not being "forced" enough to see an actual difference, remember it would take an infinite voltage for the current to drop in an instant.

Force the flyback hard & make it discharge when you want it to not when it wants to & you will see a difference & don't cut it short, let the sucker fly to drop the current.

Now i think Nicolae may have been seeing the current drop quicker with little flyback because this can be the case if the flyback is not high enough.

But!
We require a good "inductive kick" & if we just let the flyback just go "zing---fizzz----plop" like a chinese missile attempt well we can expect no difference or if the flyback just rolls off the edge & falls over we are not going to get a good inductive kick.
Kick it in the backside & let it fly, it's called flyback not zing-plop, if you can relate to what i'm saying without the inductive kick we may detect a car at 12".

A good storage of current is required to kick this in the butt when we "need" it.

If all the neccessary ingredients we need, like the duty cycle, the amperage, the amperage change of direction & then the final "kick" are tuned right you will see good flyback & the current drop fast as well.
If all things remain the same & she's a fizzer then you just will not see it!

I have attached a screen shot of forced flyback for want of a better word.

The Light Green--Lime is the switch off, the Red line is the Inductor Amperage, Dark Blue is the Inductor Voltage & the Light Blue is what i get after the parrallel diodes with the circuit i have at the moment.

It seems to me that the more capacitance you have the higher the flyback has to be to compensate.

[simonbaker]

Hi Aziz,
We should never reach the breakdown voltage of the transistor. Once we reach that, it makes no sense to pump more current through the coil, because it will only be wasted battery without any benefit to the detector.
If the breakdown voltage is reached for 1us, that 1us will only add to the decay time, without any advantage.
If we assume an equal amount of energy and the same inductor, and same parasitic capacitance, how can we generate a higher flyback voltage? We can't increase damping resistor, because the circuit will get underdamped. I would say that equal energy and higher flyback voltage for the same circuit is not possible (definition of madness: keep doing the same thing and expect a different result ) - sorry, I was just joking here).
In theory, we could obtain a higher flyback voltage for the same amount of energy by increasing the quality factor of the coil (decreasing the series resistance in circuit).

Regards,
Nicolae

I think I agree -- flyback voltage is proportional to how fast we can stop the current. The limiting factor must be the capacitor, because in ideal case with no capacitance, we could increase resistor and make voltage spike go towards infinity.

So I think you just have to get capacitance down, otherwise fooling with Rd to no avail.

Aziz seems to be talking about different case, where MOSFET breakdown is limiting the flyback voltage, so naturally a higher breakdown would help.

I think I agree also that going into breakdown is non-productive because di/dt seems to be zero -- wouldn't that make a double "pulse" as seen by the target, one pulse before MOSFET breakdown and one after? But maybe two quick pulses give a useful kick to typical targets for detection purposes.

I have another question about differential design, which I'll ask in another message...

Cheers,

-SB

[hobbes_lives]

Using ideal components, the coil and damping resistor can be analyzed as a parallel RLC circuit. Critical damping occurs when Q = 1/2. If we take the equation for Q in a parallel RLC circuit:



insert a value of 1/2 for Q and then solve for R, here's what we get:



This formula demonstrates the relationship of C to R. An increase in the value of C requires a reduction in the resistance of R for critical damping.

Here are a few examples:

If we solve for L=300uH and C=50pF, then R=1224.74 ohms for critical damping.

If we solve for L=300uH and C=100pF, then R=866.03 ohms for critical damping.

If we solve for L=300uH and C=250pF, then R=547.72 ohms for critical damping.