Hi Gyula,
I agree with your first comment. I was implying that the gate current was zero for a DC input. Sorry for the omission of that detail. Certainly you need current to charge and discharge the input capacitance. Whether or not the input capacitance charging/discharging is going to affect what you are trying to do depends on your excitation frequency and what kind of ON/OFF slew rate you need. You are probably aware that there are MOSFET driver chips that take care of the annoying capacitor charging and discharging for you if that is critical and your signal source can't handle it.
The other thing to keep in mind about the charging and discharging of the gate input capacitance is that it is propagating AC energy into the circuit. A spike of current to charge the input capacitance will generate a spike of current on the source and/or drain pins. That energy will continue going into the circuit somewhere.
Going back to the capacitive effects of the gate input itself, it can be part of the fun of the investigation into a given setup. You want to be aware of the capacitive effects of the gate and decide if they are going to affect what you are trying to accomplish or not. For example, if you want to use the MOSFET in a switching application at 60 Hz and you can tolerate the slew rate limitation from the gate input capacitance, then you can ignore it altogether. On the other hand, if you want to do a switching application at 20 KHz, then it may come into play and you may need the MOSFET driver chip or do it yourself with transistors, etc.
For your second comment I just have a few things to say without getting too deeply into all of the design issues. For the gate input let's say that you can define an "OFF" region, a "linear" region, and an "ON" region. This is a function of the source-drain current and source voltage.
So for example if the source-drain voltage is 10 volts and the current is 100 milliamperes, then "OFF" might be 0-2 volts, "linear" might be 2-6 volts, and "ON" might be 6-10 volts. Therefore if you excite the gate input with a sine wave, it all depends on the voltage sweep of the sine wave to determine how the MOSFET is going to react to this stimulus.
Just as a caveat, I am being pretty general and probably oversimplifying things here. Certainly if I was testing a similar circuit I would be referencing the datasheet for the MOSFET and investigating all of these aspects.
I don't know how the gate input LC resonator is acting in Luc's setup, and just putting a scope probe on it will probably change it. It would still be nice to see anyways to get a sense of the voltage sweep on the gate input. I am making the assumption that the MOSFET is operating partially in switching mode and partially in linear mode by observing the waveform.
Anyway, with respect to Luc's observation of negative supply current to this setup, that could be an interesting investigation. I am assuming that the multimeter is averaging the AC component superimposed on the DC current and returning a value that is slightly incorrect. We are talking microamps here. A suggestion for Luc would be to replace the power supply with a large capacitor and see what happens. I am assuming a self-resonant setup here because you absolutely must remove the signal generator from the equation. I saw some references by Luc about the a capacitor as the power source charging while the setup was running, and I will assume that he was also using a signal generator in those cases. You cannot forget that the separate signal generator is injecting AC power into the circuit. Going back to the case where the capacitor powers everything and the current consumption is only microamps, then you may be able to get away with using a smallish capacitor to power the whole thing so that the observed voltage drop takes place in a reasonable amount of time.
It make sense that the current consumption is very small because an RL (R very small) resonant circuit is constantly storing and returning energy to the source for a net current consumption of zero. The very small R is dissipating energy and hence that energy ultimately has to come from the power supply (or capacitor acting as the power supply).
And just to be complete an LC resonant circuit is constantly transferring energy back and forth between the L and the C, you don't need a power source at all. Luc may be interested in taking a charged capacitor and connecting it to one of his coils to observe the oscillation on his virtual scope. He can capture the waveform in "one shot" mode and then measure the frequency. That can then be compared to his measured L and C to see if the expected resonant frequency is in accord with the measured resonant frequency.
MileHigh
« Last Edit: 2010-04-04, 21:44:06 by MileHigh »
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