# Voltage of a toggled inductor

Discussion in 'General Electronics Chat' started by Hybird, Mar 22, 2011.

1. ### Hybird Thread Starter New Member

Jan 5, 2011
12
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I'm having a bit of trouble figuring out how to calculate the max voltage an inductor will get to when switching it with a N-channel MOSFET. My circuit is shown below. All it does is turn on the MOSFET for 5us as the inductor current rises to its max value.

RDSon(M1) = Drain source resistance of M1 = 0.012.
V1 = supply voltage of battery = 1v

So max current attainable would be V1/RDSon(M1) = 1v / 0.012Ω = 83A.

The gate of the MOSFET is driven by V2 which is on for 5us then off for the remainder of the time. So the current starts to build through the inductor and never quite reaches its max value of 83A. In fact the MOSFET is on for such a short time I have found out it can be approximated by a linear function:

Imax ≈ V1 * Ton / L

So for my example it would be:

Imax ≈ 1v * 5us / 500uH = 10mA

As you can see from the graph of the circuit below, this is precisely the max current. But I can't figure out how to get the max voltage of the inductor with this value. Any ideas?

To get my value for max current I simply took the inductor voltage equation and assumed the δI/δt could be replaced by ΔI/Δt, as below:

VL = V1 = L * δI/δt ≈ L * ΔI/Δt => ΔI = V1 * Ton / L.

If I could somehow find out how long it takes the max current to go down to zero as marked below, I could use the straight line approximation again and get a ball park value of what the peak voltage will be based on the max current but I can't see where to get that value from.

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Dec 26, 2010
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Don't do it like that!!! Switching off an inductor that way with no alternative current path will produce a voltage which is dependant on the FET switching speed, together with parasitic capacitances in the FET and the coil. Unless the FET turns off at a sufficiently slow rate, this can be hazardous.

In practice the voltage might well be limited by the breakdown voltage of the FET, or maybe the winding insulation if you have a miniature coil and a high voltage FET. The upshot of all this is that such a set-up could produce damaging voltages, most likely destroying the FET.

A well-designed system of this kind would usually provide another path for the current when the FET turns off, either a direct path connected to the coil or a virtual path by means of another winding linked to the same magnetic flux (a transformer secondary).

3. ### Jaguarjoe Active Member

Apr 7, 2010
770
90
You can use a high voltage (~350) MOSFET or IGBT along with a slightly lower voltage (~300) zener diode across the source and drain. The more robust zener will take the hit. IIRC, there are IGBTs with built in protection that are used as coil drivers in inductive ignition systems.
Placing diodes or snubbers, etc across the coil slows the discharge considerably.

4. ### Hybird Thread Starter New Member

Jan 5, 2011
12
0
Thats not the entire circuit, just a part of it. The other side is connected to a voltage ladder. But if this is highly related to the switching speed of the fet then at least I know where to start the calculation. I have thought about putting a zenar diode between the right side of the inductor and to ground, but I think it may kill the performance of my circuit. Ideally this would be my circuit:

Here the LT IC is used to toggle an internal FET at the SW terminal to generate the same effect. The voltage divider at FB simplely tells the IC to stop toggling the IC once the top of the ladder has reached 1000V.

I can't see any other way to make this work with DC supply. The SW pin of the IC is rated for about 75 volts max. So that's why I'm trying to figure out how toggling the inductor relates to inductor votlage. We have a circuit somewhat like this with a pulsing IC that toggles an external power mosfet rated for 500v, but I am trying to figure out how to use an IC with an internal mosfet as below.

I think my best bet might be to use a transformer instead of an inductor but the goal is to keep things really small.

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Dec 26, 2010
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You should not expect to limit the maximum FET voltage by controlling the turn-off speed. It is possible to do so, but this would result in poor efficiency. The key is to design the load circuit (your multiplier) so that the FET voltage remains within limits.

Unfortunately, with only a 75V peak rating on the SW pin you cannot safely get more than 75V per stage out of your multiplier, and losses will mean that you get a somewhat lower voltage. To get 1kV safely, you would therefore need to have 14 or 15 stages, maybe more. It might be more practical to begin with a more highly rated driver device.

With the circuit as shown, and just a few stages, the output voltage could be expected to build up until it was limited by the internal FET breakdown. Depending on the power levels and the robustness of the device, it might well fail catastrophically.

Note that your multiplier returns all the input capacitors to the driving point, and all the reservoir capacitors to ground. Although this has some advantages over the more usual laddered arrangement, remember that this will require some of the capacitors to be rated for the full output voltage.

I am not sure that you should be trying to tackle this as a practical problem. You seem to be at a fairly early stage of understanding these matters, and this sort of system is potentially hazardous.

6. ### Hybird Thread Starter New Member

Jan 5, 2011
12
0
This is the issue, most DC DC converter IC's don't have internal mosfet capable of tolerating anything more than approximately 80 volts max. Most of them drive at a frequency of 300+ kHz where our design drives at 45kHz. So thats another design consideration. My bose wishes to stay at that frequency cause he thinks 300kHz will be too fast, I'm not so sure I agree with him on that one. Surely a faster switching speed means less time for current to flow through the inductor in the on stage and less energy stored by the inductor leading to a lower voltage spike when mosfet is turned off.

I'm aware of this. Thats why I'm trying to limit the voltage. We currently have a circuit but instead of switching an internal mosfet in an IC as shown in my circuit, we are toggling an external mosfet rated at 500v. And the circuit is infact working.

The part I left out is that this is for a gieger circuit where the tube is set at 410 volts, a radiation event occurs and the tube is essentially a short circuit through a large resistive load (not shown in the circuit) and a signal is detected. This leads to very little current draw from the top of the ladder each time a radioactive particle is present.

Noted. We have all caps rated for 1000 volts. This is a paralell ladder as I have come to find out. I was told it was more efficient then the series ladder where the capacitors need only be rated for the max votlage output divided by the number of stages (plus some additional just in case).

Thanks for your concern, but I am well aware of the dangers of high voltage.

The way I see it is, if you want to get to an adjustible high voltage in the range of 500-1000v you have 2 options:

1.) Drive an external power mosfet rated for at least 500v. (The current circuit does this).

2.) Toggle the primary side of a transformer with an IC that has an internal mosfet and limit the votlage across that mosfet and hence primary by using a zenar diode. Then drive the other side of the ladder with the secondary which would make for a higher voltage at the bottom of the ladder.

Space is a concern so I am trying to make this with as few components as possible. Also, the boss wants this to be a variable ladder so it can be used for other similar circuits like Photomultiplies tubes, etc.

I'm leaning toward option 2. If I use the transformer, zenar diode, internal switching mosfet IC, then I think I only have the choice of going with a paralell ladder drawn in my circuit. My reason, is say I have a 1:12 transformer. Then I have a IC rated at say 50 volts. So I zenar the primary side of the transformer at 40 volts. So the ladder gets 40*12 = 480v at its first stage. In a series ladder that would mean each stage should be rated for at least 500v. Might as well just use the parallel which has to be rated that high anyway if its more efficient.

Am I completely full of @#\$! ?

7. ### #12 Expert

Nov 30, 2010
16,660
7,305
I recently studied a similar problem and this is the result: The enegy in the inductor is 1/2 L(I squared), where I is the current at the instant of turn-off. If you parallel the FET in your first drawing with a capacitor and use a fast diode to direct the inductor's energy into the capacitor, the energy from the inductor will go into the cap as 1/2 C(V squared), plus V1.

I actually don't think your second drawing will work, but hopefully, I have given you a tool you can use to quantify your work. The dire warnings you are getting here are about the fact that, without a capacitor, the voltage will tend to go to infinity.