Coil field collapse peak current

[A Gumby]

I wanted to get an idea of what current was flowing in a coil at the end of a PI detector's excitation pulse taking into consideration the time constant of the coil, so I created an Excel spreadsheet to take a look based upon:

I = V/R*(1-1/e^P/(L/R))

I = coil current at end of pulse
V = supply voltage
R = total resistance of coil and switching device
e = Euler's Number (approx. 2.7183)
P = pulse width in seconds
L = inductance of coil

I also wanted to get an idea of what kind of repetitive avalanche stress was being forced upon the MOSFET switching device, so I created another Excel spreadsheet to calculate that based upon:

Es = 0.5*L*I^2

Es = stored energy in coil at end of pulse (Joules)

Substituting for I:

Es = 0.5*L*(V/R*(1-1/e^P/(L/R)))^2

It appears that at the operating point of interest, the avalanche energy would be just over 7mJ which is within the 13mJ repetitive avalanche energy ratings of both the IRF740 and IRF9640. The 11A repetitive avalanche current ratings of both devices is also junction temperature limited. Since the duty factor would be very low with a 6us damped voltage spike from a field collapse at 100Hz, heating may not be as much of a factor as the peak current.

Does anyone know the formula for calculating the peak current created by the collapsing field?

[bbsailor]

I wanted to get an idea of what current was flowing in a coil at the end of a PI detector's excitation pulse taking into consideration the time constant of the coil, so I created an Excel spreadsheet to take a look based upon:

I = V/R*(1-1/e^P/(L/R))

I = coil current at end of pulse
V = supply voltage
R = total resistance of coil and switching device
e = Euler's Number (approx. 2.7183)
P = pulse width in seconds
L = inductance of coil

I also wanted to get an idea of what kind of repetitive avalanche stress was being forced upon the MOSFET switching device, so I created another Excel spreadsheet to calculate that based upon:

Es = 0.5*L*I^2

Es = stored energy in coil at end of pulse (Joules)

Substituting for I:

Es = 0.5*L*(V/R*(1-1/e^P/(L/R)))^2

It appears that at the operating point of interest, the avalanche energy would be just over 7mJ which is within the 13mJ repetitive avalanche energy ratings of both the IRF740 and IRF9640. The 11A repetitive avalanche current ratings of both devices is also junction temperature limited. Since the duty factor would be very low with a 6us damped voltage spike from a field collapse at 100Hz, heating may not be as much of a factor as the peak current.

Does anyone know the formula for calculating the peak current created by the collapsing field?

A Gumby,

Use the program MiscEl at this web link. http://www.miscel.dk/MiscEl/miscel.html

In particular, use the inductor charge feature, enter the inductance, voltage and coil TX circuit resistance. Move your mouse over the graph and see the coil current at any point on the coil charge curve.

For the inductor discharge feature of this program enter the coil inductance, current at TX turn off and the value of the damping resistor to see the discharge time constant as well as the peak flyback voltage.

I hope this helps.

bbsailor

[Bill512]

Does anyone know the formula for calculating the peak current created by the collapsing field?

The answer, in my opinion, is very simple... Imax=V/R (at any time:t>>L/R).
The collapsing field does not "create" any bigger current , just starts from the Imax value and drops exponentially to zero. The voltage across the coil is higher one or two orders of magnitude ,trying to overcame the open circuit. The coil now acts as a voltage source and the voltage polarity is reversed (not the current).
Bill

[A Gumby]

The answer, in my opinion, is very simple... Imax=V/R (at any time:t>>L/R).
The collapsing field does not "create" any bigger current , just starts from the Imax value and drops exponentially to zero. The voltage across the coil is higher one or two orders of magnitude ,trying to overcame the open circuit. The coil now acts as a voltage source and the voltage polarity is reversed (not the current).
Bill

A slap my forehead moment. You are correct, of course. There is a voltage spike, but no current spike. The maximum current on discharge will be same as the maximum current flowing at the end of the excitation pulse.

[Monolith]

The answer, in my opinion, is very simple... Imax=V/R (at any time:t>>L/R).
The collapsing field does not "create" any bigger current , just starts from the Imax value and drops exponentially to zero. The voltage across the coil is higher one or two orders of magnitude ,trying to overcame the open circuit. The coil now acts as a voltage source and the voltage polarity is reversed (not the current).
Bill

This would be correct, with an ideal inductor. However, have you analyzed the situation with a real PI coil? Have you considered the capacitance, inter wire, Mosfet, coil to shield, coax and stray?
How does the equation look like then?

Monolith

[Bill512]

This would be correct, with an ideal inductor. However, have you analyzed the situation with a real PI coil? Have you considered the capacitance, inter wire, Mosfet, coil to shield, coax and stray?
How does the equation look like then?

Monolith

This is correct in all inductors ,real or not real (the non real components is a approximation to make our life more easy until a certain point).
These capacitances that you mentioned, simply effects the rate of change (dI/dt, positive or negative) not the absolute maximum current value.
There is no reason to get a higher current than this which established at the end of the on time. Further more , there is a serious reason not to get any bigger current.
Have you ever considered the Lenz law, which is a consequence of the law of conservation of energy???
The Lenz law simply "wants" to sustain the magnetic field and for sure not to increase it !
Now,the magnetic field is directly related to the current which flows in the coil. Any bigger current will lead to a bigger value of magnetic field ,a result that is opposed to the Lenz Law ...

[Monolith]

This is correct in all inductors ,real or not real (the non real components is a approximation to make our life more easy until a certain point).
These capacitances that you mentioned, simply effects the rate of change (dI/dt, positive or negative) not the absolute maximum current value.
There is no reason to get a higher current than this which established at the end of the on time. Further more , there is a serious reason not to get any bigger current.
Have you ever considered the Lenz law, which is a consequence of the law of conservation of energy???
The Lenz law simply "wants" to sustain the magnetic field and for sure not to increase it !
Now,the magnetic field is directly related to the current which flows in the coil. Any bigger current will lead to a bigger value of magnetic field ,a result that is opposed to the Lenz Law ...

Agreed on the conservation of energy. There will not be more Joules come out than went in.
Now for the current. When the Flyback goes up to 500V, it charges the capacitances to 500V. Therefore the total amount of current coming out of the conductor is less during the time of capacitor charge. Then the capacitors discharge again as the inductor voltage decays at an exponential rate. Therefore the current now increases? This does not seem the case, since the time of decay or discharge increases because of the capacitors.

Is this reasoning right?

Monolith

[Bill512]

Agreed on the conservation of energy. There will not be more Joules come out than went in.
Now for the current. When the Flyback goes up to 500V, it charges the capacitances to 500V. Therefore the total amount of current coming out of the conductor is less during the time of capacitor charge. Then the capacitors discharge again as the inductor voltage decays at an exponential rate. Therefore the current now increases? This does not seem the case, since the time of decay or discharge increases because of the capacitors.

Is this reasoning right?

Monolith

In generally, yes. This depends from several factors as coil inductance ,capacitance and the chosen damping resistor . The typical PI coil capacitance is small, but as you said it charges up to several hundred volts and the stored energy is proportional to the square of this voltage. Initially , the energy was stored in the form of magnetic field (inductance) and then a small percentage is stored in the form of electrical field (capacitance) . The missing part of energy is converted to heat on the resistor.
As you wrote , the inductor starts to discharge but at a lower rate because of the capacitance (charge) . Then, a little later , the current starts to increase as the capacitor discharge (at the same time charges back the inductor). The second current peak is definitely smaller in compare to the initial Imax because of the big energy dissipation on the resistor. Maybe all this above description is well known ? definitely yes, is a RLC circuit in parallel and the current oscillates. In general,is a damped oscillation.
In order to properly damp the circuit, we must change the resistor -the current curve must be a genuinely descending function but not overdamped.
Finally, this parasitic capacity will force the designer to sample at longer times, thus this coil is slow...

[DOOLEY]

underdamped coil benifits

hello bill , just looking at your pic of coil damping.

have been doing simple experiments just with a pi front end and a scope.

test piece of metal is a 3"long x 3/4" square hollow box aluminium.

when i damp sufficiantly to get a good curve above the + rail , as it's supossed to be like that by everyones rekoning ,i get a 0.3V deviation when placing test piece on coil.

however if i damp the coil to suppress oscillations , but still underdamped , i get the waveform you have in your pic at the bottom.which explains to me what was happening.

anyway , when underdamped i am getting 1.2V deviation on the bottom of the "dip" with same test piece, which is more sensitive , yes ??

and , as an added bonus as the bit i'm sampling / looking at is bellow top rail i can feed it straight to the op-amp with the same single supply and not have to use a split supply, which simplifies circuit.

so a simple / novice mistake of underdamping has resulted in a more sensitive front end.

incedentaly , if realy underdamped , but still not allowing full oscillations , the first flyback "dip" can nearly touch the bottom rail , and same test piece shows a 10V deviation , tempting to go down that route for sensitivity ,BUT the more underdamped the coil is then the "thinner" the scope trace is , so i suspect the signal is realy weak , and poss be to weak to use.

on a 12V drive supply , i can underdamp the "dip" down to 8V with no apparant weakening of scope trace. deviation at this damping level with same test piece is aprox 5V but output from amp is a little noisey.

underdamping so "dip" is about 10V seams about right , amp output clean.

what are your thoughts ????

[DOOLEY]

poss discrimination

hello , think iv'e killed this thread.

on the purple line above on the graph , the underdamped coil , interesting thing , the first dip below the top rail goes up with non ferrous and the second "wobble" above the top rail remains unchanged.

but with a ferrous target , they both go up , looks like and easy way do discrimination to me if both are sampled.

only thing is there deviation is differnt , the below line wobble is aprox twice the deviation of the above line wobble suggesting less sensitivity , poss due to energy disapating with time.

[Carl-NC]

on the purple line above on the graph , the underdamped coil , interesting thing , the first dip below the top rail goes up with non ferrous and the second "wobble" above the top rail remains unchanged.

but with a ferrous target , they both go up , looks like and easy way do discrimination to me if both are sampled.

Same thing happens regardless of damping. TDI uses this effect for ground balancing.

[Bill512]

hello , think iv'e killed this thread.

on the purple line above on the graph , the underdamped coil , interesting thing , the first dip below the top rail goes up with non ferrous and the second "wobble" above the top rail remains unchanged.

but with a ferrous target , they both go up , looks like and easy way do discrimination to me if both are sampled.

only thing is there deviation is differnt , the below line wobble is aprox twice the deviation of the above line wobble suggesting less sensitivity , poss due to energy disapating with time.

DOOLEY, sorry for the extra long "delay" on reply (aprx: 2.16E12usec) ,there are many interesting things on your experiments!
I was very busy and spent some free hours working on Chance PI and minelabs Ground Balance method... ...

[DOOLEY]

HELLO BILL , sorry caps lock on , shouting there for a while,
humour takes many forms , nice play on the "power of",

seems like i'm constantly discovering what people allready know.

new into this field of metal detecting millarky (spelt wrong , is there a right way ?)

i keep trying to think of differant ways of doing things.