Drop-out pulse test

Question

I have a switching MOSFET circuit.

Input to the switching MOSFET circuit is 11V at the drain. The gate is connected to a function generator. 5V peak-peak, frequency = 1kHz, 1ms (time period) & 90% duty cycle.

MOSFET Circuit (MOSFET IC Part Number : VN7140AJ) :

Probing at the gate & source (drain is at 11V,) I get the below waveform (only with the MOSFET) :

THE ABOVE WAVEFORM IS OBTAINED WITHOUT CONNECTING THE MOSFET SOURCE TO THE DOWNWARD CIRCUIT. ONLY 11V, FG INPUT is provided and the Output is taken at the source.

With the same MOSFET circuit, if I connect this circuit source terminal to the my new circuit (shown below,)

I get the below waveform:

Why doesn't the MOSFET source go to 0V after I connect to the new circuit? But it goes to 0V, if I don't connect to the new circuit.

It doesn't go to 0V, because of C0101 & C0102 capacitors, or what might be the reason?

Answer 1

So when the gate turns off and the source goes high impedance, you've got a 47uF capacitor and a 100mA current sink as the load. We can ignore the left half of the circuit because the capacitances are so small in comparison.

Over the 100us drop-out the load will draw out 100us*0.1A = 10uC of charge from the capacitor. Since capacitance is charge/voltage this will lower the voltage by 10uC/47uF = 0.21V. From the oscilloscope picture we see the voltage actually lowers by 0.72V. This could be explained by the capacitor having lower capacitance than it says, or more likely, the that the load draws more current than expected in these conditions, but we're in the right ballpark so I think this is the explanation.

Edit to address the clarifying comments:

If the source was actually a direct connection to GND when the gate is LOW, then yeah even with the capacitor circuit the voltage at the source would go LOW pretty immediately after the gate going LOW. But if the source becomes a high impedance connection to GND then what we actually see makes sense.

If the source behaves like a connection to GND through a large resistor then the charge on the capacitors can't disappear immediately and we'll see the slow decrease in voltage that we do see. In my calculation I assumed the source would behave as if it had infinite resistance, and we could disregard it completely so the only current flowing would be the 100mA load, but the voltage drops more than my calculations predicted. I speculated why this might be, but didn't consider that the source has a finite resistance, so some current can flow through it and explain the quicker than expected voltage decrease.

Answer 2

In the case when you connect the MOSFET to the load circuit try adding a 220 or 470 ohm resistor across C101/C102. In other words put the resistor from MOSFET source to GND. Then observe the difference in the switching waveform.

Answer 3

Your first circuit makes no sense. (sorry to be harsh) As (hand) drawn, the only thing that the Nfet can do is to short 11V to ground. The rest of the circuit won't matter, ground is ground. From your waveforms I expect that the fet is actually a P channel mosfet and is not connected to ground.
(going out on a limb here) Are you trying to charge the battery? If so, you should really either add an inductor to the charging path and a free-wheeling diode, or put the battery in a large bucket of sand so when it catches fire from the peak current loads the fire will be mitigated.

...and the series caps are near-useless. Again,out on a limb as this smells like a battery charger.

Answer 4

Putting aside for the moment your circuit, which should either use a P-MOSFET as a switch or a higher driver voltage for the N-MOSFET, based on your question and comments (and as requested), what you are really interested in, is to determine how long your pre-charged output capacitor can hold its voltage during a voltage interruption (e.g. 100us).

$$V_{CAP,final} = V_{CAP,initial} - \frac{I_{load}\cdot t}{C_{out}}$$

Where:

$$V_{CAP,initial} \approx V_{supply} - V_{Diode}$$

Plugging some values:

$$V_{CAP,final} = (11V - 0.5V) - \frac{100mA\cdot 100\mu s }{47\mu F}=10.287V$$

After this power shortage, the capacitor is then recharged to 10.5V again.

P.S. The formula in my comment assumed erroneously a load with constant power consumption, which is not your case.