Inductive Switching Problem 2

Suggestion: It appears that the diode fast switching creates noise, oscillations, and harmonics.

I think I'm just about anybody and can give a shot in the dark. I can only guess at the problem of the overheating of the mosfet at one amp. My inclination is that the Vgs is not high enough to turn the rated .1 of an ohm RDS on. Therfore it seems to me that the field effect channel is not yielding the lowest RDS it is somewhere above this and is dissipating heat I^2RDS. RDS greater than a .1 of an ohm. Good Luck.

Hi all,
Thanks for a quick reply to my question, I feel I may have the answer. The circuit uses a sense resistor on the source pin of the mosfet. The voltage on the pin will rise with increasing current (2V for 4Amps) therefore is it possible that my 5V logic level input to the logic level mosfet is now going from 5V down to 3V. I am basing this on my understanding that the Input is with referance to the Source.

Do you think this is correct.
Mark Lewry

I am currently using a pre-driver & FET circuit to drive some inductive loads. It sounds like you and Xray have the right idea. If you are not fully "turning on" the FET (by decreasing Vgs & increasing Vs while keeping Vg constant), the resistance of the MOSFET will not be the 100 mOhm value.

Can you move the sense resistor to the Drain of the FET? This should still allow for over current detect (which is why I am assuming you are using the sense resistor), and will eliminate the souce voltage modification.

Also, you might want to make sure that the clamping mechanism on your pre-driver works while the IC is not powered or you must use self-clamping FET's. If your circuit is ever unpowered and needs to dissapate the energy from dropping out the low side of an inductive load (creating a very large positive voltage spike), then your FET should disapate the energy. Some pre-driver's can detect the over-voltage condition and turn on the FET to disapate the energy. To be safe, it would probably be a good idea to use self-clamping FET's. This will prevent damage to the pre-driver, FET or inductive load if your pre-driver does not operate while the IC is unpowered.

Hope this helps!

I must state that in my previous post, I assumed that you were controlling the low side of the load. If you are supplying the high side, any fluctuations in the load voltage will cause the Rdson to vary. This can also create the heated parts that you were seeing.

Also, the previous post stated that you should use self-clamping FET's. However, if you use the recirculation diode, there should not be any volatge spikes higher than the diode drop. Therefore, the previous post is more of an FYI.

Thanks for your reply Melone,
I am using the mosfet to control the low side as you assumed. The mosfet is an SGS-Thompson Omnifet VNP10N07 and is a fully auto protected logic level FET.

I like your idea about putting the sense resistor on the drain side but I fear that the voltage will rise when the mosfet is turned of and will damage the L297 controller chip. I had also tried using an external 10V supply to pull up the logic level of the 74LS08 AND gate (driven by the L297's Inhibit and Phase lines) that drives the mosfet. When no supply is present to the coil the gate input rises to the 10V but when the supply is present the gate input gets squashed down to about 4V ??? I cant understand this.

Anyway tonight I will try some other things, please keep your suggestions coming I do appreciate them.

Mark

Mark,

Could you email me a schematic of what you are doing? I must be missing something because I just can't picture what your circuit looks like.

Thanks,

Mark_Melone@yahoo.com

Suggestion: I agree with Mark Melone request for schematics. Generically, when there is a high voltage (10V) at no power supply and noticeable voltage dip with the power supply, the circuit must have a component that causes the voltage drop of the power supply to that point. E.g. if you have no load transformer voltage high and loaded transformer voltage low, the transformer has a fairly high internal impedance/resistance. Perhaps, the Thevenin's or Norton's equivalent circuit could be applied to simplify the circuitry.