Hi, all
I want to drive eight IRFP064 with SG3525 directly (in each side). Please explain that is there any formula for calculating the number of MOSFET's?
Regards

 

When you add FETs you increase gate charge seen by the driver. Since the driver has a finite sink/source current, you will now require more time to charge / discharge the FET's gate. This results in increased time in the linear region of the MOSFET, thus, higher power dissipation.

There are other concerns, such as device mismatch which make it difficult to parallel MOSFETs.

The equation I used is : I = Q/dt
The current required by the driver to turn on the gate charge. In simpler terms,
I = Q/time.to.turn.on/off gates. You should be specifying about 20-40nS for an efficient setup.

Steve

 

It sounds like you are building a high current controller. What is your actual current need? Remember, the Id rating can be misleading. There are also package limitations to consider.

If you can accept a 60V rating, there are several options with lower gate charge and similar capacitance and current ratings, such as the IRF6648 and FDP10AN06A0.

Finally, remember to use a separate gate resistor for each, if you use gate resistors.

John

 

In many circuits (available on the net) I noted that, many transistors are used in parallel. Here is an example is attached. So it is requested to define a simple rule/formula for paralleling MOSFETS for SG3525 etc

 

Wasn't that question answered in post #2? The SD3525 can source/sink 500 mA. I use an LT1158, which also sources 500 mA, to drive five IRF1010E mosfets in parallel. It is not a continuous duty application, though.

John

 

One thing I didn't mention is the reasoning behind the 20-40nS suggestion. If you want to design it properly, you need to integrate the power loss during the switching, on-time, and off-time. This will reveal peak and average power losses. You need to ensure your thermal resistance of your setup is low enough to keep the temperature within operating range.

As far as simplicity goes, the equation I gave to you is about as simple as it gets.