# How to make falling time shorter for mosfet switching for high voltage and high frequency ultrasonic pulser

I have a case that I need to build a pulser switching for my ultrasonic transducer which operate in 180 V and 1 MHz. I use mosfet IRF250 for the switching, and below is the circuit:

This is pretty good with response like this:

The green is input signal and the blue is output signal in the load, the problem is R1 resistor just 100 ohm, and the power needed will be 324 Watt, I want to reduce the power since the transducer only uses the voltage not the current, so I raise the resistance of R1 to 10K so the power will be 3.24 Watt, but the response is like this:

The falling time is increased drastically, what causes this ? Is it the capacitor between drain and gate inside of mosfet? How can I change the R1 to 10K but with same response as 100 ohm?

The large fall-time is most likely caused by the drain-source capacitance of the mosfet (700pF in datasheet). When turning off you have essentially a 10k resistor in series with 700pF of capacitance which takes time to charge up.

If you only need round about 20mA I would choose a smaller mosfet with less capacitance.

• This, and the capacitance of the transducer itself (not modeled). Commented Apr 14, 2018 at 6:44
• The capacitance of the transducer is actually going to dominate the 700pF in the simulation, so this rise/fall time issue will be much worse in reality. A push pull drive would work so much better, I think. Commented Apr 14, 2018 at 7:31
• @hatsunearu is the push pull drive for the mosfet gate driver? Commented Apr 14, 2018 at 7:37

You could just have a push pull drive / half bridge drive. Connect one side of the transducer to ground. Have one switch from 180V to the other side of the transducer, and another switch from the other side of the transducer to ground.

simulate this circuit – Schematic created using CircuitLab

I just chose the random PFET/NFET in the circuit diagram builder in the website, but you should pick whatever one you want.

This arrangement allows the load node to be pulled up or pulled down directly via the FET, so you won't run into those issues.

• is the push pull drive for the mosfet gate driver? Commented Apr 14, 2018 at 7:38
• No, M1 and M2 form a push pull drive into R1, which models the transducer load you have there. You need separate circuitry to drive the gates of M1 and M2. Commented Apr 14, 2018 at 7:41
• can i use only simple function generator with series resistor to drive the gate of M1 and M2? Commented Apr 14, 2018 at 7:46
• You can for M2, but for M1 you need a gate drive that goes from 180V to 180V - (Vx) where Vx is your gate drive amplitude. Making a circuit that does that is not trivial. Commented Apr 14, 2018 at 7:49
• A real ghetto way is to have another high voltage NFET with a resistor on the drain connected to 180V: 180V |--- resistor --- NFET --- gnd and use the node between the resistor and NFET as a crude gate drive for M2... but you see this arrangement is literally the same as the one you proposed in the original question, so do this only for testing. I suggest you get a dedicated gate drive IC for this. Commented Apr 14, 2018 at 7:54