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A Vgs = 9 volts will saturate drive an Ids > 110 amps as per Fig_3.....
In honesty, I am not sure of the feasibility of using a huge lot of batteries and attempting to use a small TO-220 MOSFET to drive the coils.
Reason 1: If your gate driving circuit fails, you will end up with a MOSFET stays turned on too long and burns out, then causing a short circuit condition for the lithium batteries.
Reason 2: Large inductive spikes from the coils will likely kill the MOSFET without appropriate dampening. Look up the functioning of a buck converter and pay close attention to the diode and how it is used. You will need a large rectifier to use to prevent damage to the MOSFET (which again can lead to an overload condition and a catastrophic failure.)
Reason 3: With voltage lower than 50V, are you sure the coils will draw that current when pulsed? Remember, the golden rule with inductors, the current flowing through then CANNOT change instantly. That means even if they measure 0 ohms with a meter, when you initially power them up, they will act like an OPEN CIRCUIT, and current flowing through them will rise quickly something gives out from too much current. The speed at which the current draw rises over time is dependent on the voltage across the coil, and the inductance (in other words, smaller coil and more voltage = faster current rise.) The same goes for when you turn OFF the MOSFET. What happens is because current cannot change instantly, and now the impedance of the MOSFET is almost infinity, the voltage will climb up and up and up and up and UP... until something gives out and lets the current continue flowing from one side of the inductor to the other side. Usually the MOSFET gives out and it is killed.
Recommended: You should use a smaller, cheaper battery to power a small inverter to charge a large capacitors bank up to several hundred volts, and then discharge that bank with a chicken stick into a coil of like 5 to 10 turns of wire, because the initial voltage on the capacitors is SO high, and the inductance is pretty low, that will discharge almost instantly, and the pulse of current will be huge. You will not need to worry about the lithium battery safety, since the capacitors can only hold sooo many joules of energy anyway, and it is all discharged safely. However you will be dealing with lethal voltages and capacitors. Make sure you understand the safety procedures and it should be safe. (Use the buddy system, always have one hand behind your back when working on the HV side, have a AED on standby for your buddy, and keep in mind the capabilities of HV and just use commen sense.)
1.this is why i need a fuse
2. voltage supression
Fuses that limit such high current will be very expensive, I suppose you can get a 100A circuit breaker, and that will be less sensitive to large surge of current since they rely on heating of a bimetallic strip. (the strip will have thermal inertia, and takes time to heat up, that means that under a short fixed duration of time, it will not trip even past the rated current, it would take a lot more to cause it to trip. The exact amount will be a function of how hot it already is, the thermal resistance between it and the ambient air, and the voltage drop developed across it times the current though it.)
What is your planned pulse duration? and the voltage you are pulsing the coils with? and I would recommend if you were to go with this setup, using a small IGBT brick, are at a bare minimum, A few parrelelled TO-247 package IGBTs. a single TO-220 packaged device just does not have the thermal capacity to support that abuse. Also if we are talking about a very short pulse, there may not be enough time to fully saturate the coil to maximize the magnetic field with low voltage, meaning that although it does not explode, it will not work very well, either.
Fuses DON'T "limit" current, they prevent it flowing for very long....
Subtle distinction, but an important one. What kills typical fuses in a short circuit can be a current of 25,000 Amps. Its not limited at ALL.
Fuses do in fact "limit" current [to a degree] because they proportionately increase the voltage drop across themselves with increasing current, and the lower voltage automatically can only mean that a lower current must thence obtain. Punky current [just like Man] "has gotta know its limitations" before it becomes con'fused'.
Good point, I didn't word things correctly to make that clear, but true. It takes time for the filament inside a fuse to burn out, and during that time, a huge lot of current can flow. That is why it would be very difficult to predict the instantaneous current flow and if the fuse will burn out or not.
We'd need the inductance and the resistance to know, and, depending on the exact configuration, I'd like to know how the slug movement affects the impedance too. Presumably energy lost from the field has to be converted to energy in the slug's kinetic energy, so the spike may not happen.
200 micro henries and about100 m Ohms.
and what configuration..?
So you have a 2mSec time constant, and 55V, your current will rise to 200A in how long ?
i'l have to use high gauge wire as a fuse
Rather than using a fuse it's much better to use a small relay that's holding itself just-closed with a lower voltage than that which is specified for it. For example, if you're using a 12 volt relay then have it hold itself in with just 8-9 volts. Use your variable voltage power supply to see just how low a voltage you can use before the relay snaps its contacts open again. Using the minimal amount of voltage will allow the relay to open very quickly indeed and you won't be worried about whether a fuse will blow quickly enough, and using a relay means you won't need to diagnose and/or replace blown fuse$ all the time. Use an opamp to drive the transistor that drives the relay. Even better and simpler, if you're able to obtain a 12 volt relay that has its own manual button/lever then you can engage the relay by hand and then let the 8-9 volts do the easy bit in keeping the relay's contacts just-closed and no more.
AED? Erm... This AED:
First consider the MOSFET may more likely EXPLODE and fail by blowing itself to an open junction !
Inductive spikes are a function of how fast you turn the MOSFET Off. A 555 just does not have the speed ability to discharge the 3247 pF Gate capacitance fast enough to raise a serious HV transient but running a coil I would strongly recommend a 350 Vds MOSFET...
In my time I was driving Motorola fets fast enough that the inductive kick back of a straight 6" length of litz wire would generate 100 volts.
I am guessing you are looking at the same data sheet I am. I found mine here:
This document was written by International Rectifier (r), and it has a "PD-91279E" written on the first page. I mention this just to make sure we're looking at the same graphs and stuff.
From graphs 1,2,3,4 on page 3, you can see the largest Vgs they were willing to try was +15 volts. Also from these graphs, a Vgs =10 V gives something that looks not too different from Vgs = 15V.
I am guessing any Vgs between 10 and 15 volts will turn it pretty much fully on. Also guessing any Vgs within a few volts of zero will turn it fully off.
By the way, the gate kind of looks like a capacitor, electrically, so turning it on or off quickly may require driving it with something with small output impedance.
I'd say you'll end up with one fried MOSFET!
What load is it driving? if it is a coil, the coil will be inductive, and by putting a square wave into the gate from a 555 in a monostable configuration, you will cause really nasty voltage transients (big voltage spikes) that will kill your transistor, possibly in a closed circuit condition, causing an overload on the lithium battery pack you mentioned in the previous question, if I recall correctly.
To answer your question exactly, you can turn a MOSFET on by applying a voltage to the gate of the MOSFET relative to the source.12V relative to the source should be plenty, that will drive the MOSFET as far possible into saturation. The gate of a MOSFET is very sensitive to ESD and overvoltage, so you do NOT want to exceed the voltage that is specified.
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