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Let's talk about avalanche
This is my opinion, the way I see it, I would like to hear other opinions.
When we surpass the current limit of the cicuit, the Mosfet avalanches.
Mosfets have this inbuilt safety feature. The voltage limit, for example 1000V, is the working voltage of the Mosfet. A bit higher, somewhere between 10 to 25% the avalanche diode, like a Zener diode, shorts the mosfet to ground.
This protects the circuit and the Mosfet from overvoltage damage.
So, what does this mean for a PI TX circuit?
Insert picture of TX circuit:
The TX "charge" current comes from the battery, through the coil, to the ground. This current builds a magnetic field around the coil. This changing magnetic field induces eddy currents in the target. For better visualization, we consider these eddy currents to spin in the clockwise direction. (Eddy=Vortex)
Now we block the current with the Mosfet.
When the current stops flowing, the magnetic field collapses.
The current that is induced in the coil by the fast change of the collapsing magnetic field now has to find a different path, since the path to ground is blocked.
The path back through the coil has high impedance. The resistance of the path through the damping resisistor is relatively high. Therefore the Flyback voltage builds up.
When the Flyback voltage reaches about 1000V plus whatever the safety margin ( 10 to 25%) is, the avalanche diode discharges the surplus.
So now the Flyback has 3 different paths to discharge.
a) the coil with it's impedance
b) the damping resistor (including the RX path)
c) the avalanche diode
What exactly is the Flyback? I will leave my understanding of that for another day.
The only path that is useful for the PI TX, is through the coil. The Flyback current builds a new, fast changing magnetic field. The fast changing field induces new eddy currents in the target. These eddy currents now spin in the counter-clockweise direction. Initially these eddy currents have to cancel the eddy currents that are still running in the clockwise direction.
The decay of these eddy currrents is what gives us the target response.
The current going through the damping network bypasses the coil. It does not add to the magnetic field of the coil, but it helps making a higher rate of change of the magnetic field.
The current through the avalanche diode is a total waste, but it protects the circuit like a fuse.
To make it clearer, let's put some numbers (not calculated or meant to be exactly correct) to it:
Coil charge current: 3A
Avalanche current: 1A
Damping current: 1A
Useful coil discharge current: 1A This is the only useful current that induces a target response
Please correct me where wrong.
Let's talk about avalanche.pdf
This is my opinion, the way I see it, I would like to hear other opinions.
When we surpass the current limit of the cicuit, the Mosfet avalanches.
Mosfets have this inbuilt safety feature. The voltage limit, for example 1000V, is the working voltage of the Mosfet. A bit higher, somewhere between 10 to 25% the avalanche diode, like a Zener diode, shorts the mosfet to ground.
This protects the circuit and the Mosfet from overvoltage damage.
So, what does this mean for a PI TX circuit?
Insert picture of TX circuit:
The TX "charge" current comes from the battery, through the coil, to the ground. This current builds a magnetic field around the coil. This changing magnetic field induces eddy currents in the target. For better visualization, we consider these eddy currents to spin in the clockwise direction. (Eddy=Vortex)
Now we block the current with the Mosfet.
When the current stops flowing, the magnetic field collapses.
The current that is induced in the coil by the fast change of the collapsing magnetic field now has to find a different path, since the path to ground is blocked.
The path back through the coil has high impedance. The resistance of the path through the damping resisistor is relatively high. Therefore the Flyback voltage builds up.
When the Flyback voltage reaches about 1000V plus whatever the safety margin ( 10 to 25%) is, the avalanche diode discharges the surplus.
So now the Flyback has 3 different paths to discharge.
a) the coil with it's impedance
b) the damping resistor (including the RX path)
c) the avalanche diode
What exactly is the Flyback? I will leave my understanding of that for another day.
The only path that is useful for the PI TX, is through the coil. The Flyback current builds a new, fast changing magnetic field. The fast changing field induces new eddy currents in the target. These eddy currents now spin in the counter-clockweise direction. Initially these eddy currents have to cancel the eddy currents that are still running in the clockwise direction.
The decay of these eddy currrents is what gives us the target response.
The current going through the damping network bypasses the coil. It does not add to the magnetic field of the coil, but it helps making a higher rate of change of the magnetic field.
The current through the avalanche diode is a total waste, but it protects the circuit like a fuse.
To make it clearer, let's put some numbers (not calculated or meant to be exactly correct) to it:
Coil charge current: 3A
Avalanche current: 1A
Damping current: 1A
Useful coil discharge current: 1A This is the only useful current that induces a target response
Please correct me where wrong.
Let's talk about avalanche.pdf
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