What Factors Affect the Switching Frequency of a MOSFET?

  • Thread starter Thread starter JJ91
  • Start date Start date
  • Tags Tags
    Frequency Mosfet
Click For Summary
SUMMARY

The switching frequency of a MOSFET can be calculated using the total cycle time derived from the datasheet parameters: switching on time (ton), switching off time (toff), rise time (tr), and fall time (tf). For example, a total cycle time of 44.5 ns results in a switching frequency (Fsw) of 22 MHz. However, accurate calculations must consider the overlap of rise and fall times with turn-on and turn-off times, which can lead to a frequency of approximately 41.66 MHz. The driving conditions specified in the datasheet, particularly the input and Miller capacitance, significantly affect the achievable switching speed, with stronger drivers enabling faster transitions.

PREREQUISITES
  • MOSFET datasheet interpretation
  • Understanding of switching times: ton, toff, tr, tf
  • Knowledge of input and Miller capacitance
  • Experience with MOSFET driver circuits
NEXT STEPS
  • Research MOSFET driver design techniques for improved switching speeds
  • Learn about the impact of gate capacitance on switching performance
  • Explore advanced MOSFET applications in high-frequency circuits
  • Investigate the use of parallel driver configurations for enhanced performance
USEFUL FOR

Electrical engineers, circuit designers, and anyone involved in high-speed MOSFET applications will benefit from this discussion, particularly those focused on optimizing switching frequencies in power electronics.

JJ91
Messages
40
Reaction score
0
Hi,

I would like to ask hot o determine switching frequency for a MOSFET by looking at the datasheet.

Most of the datasheet don't provide such information but they provide the following times:
Switching on time [ton], raise time [tr], falling time [tf], switch off time [toff]

example:

- Toff (Switching Off time) [nS] 14
- Ton (Switching Oontime) [nS] 10
- Tr (Raise time) [nS] 13
- Tf (Fall time) [nS] 7.5

Total cycle time = 44.5nS therefore:
Fsw=1/T = 22Mhz.

Would this be right ?
 
Engineering news on Phys.org
I checked a MOSFET data sheet and it seemed like the rise and fall times were included in the turn-on and turn-off times. There was a diagram which showed the turn-on, turn-off times clearly overlapping the rise time and fall time.

So, you would not have to include these.
On your figures, I get about 41.66 MHz.

Doing this would result in seriously degraded pulses.

This calculation does not give you an ultimate upper frequency of the FET, though.
If the FET was used for small signals, the full rise or fall times would not be used like they are for pulse switching
 
Switching time of MOSFET are based on the drive condition SPECIFIED in the data sheet, nothing more. The major problem with driving MOSFET is the input capacitance and the Miller capacitance you have to overcome. If you have a stronger, lower impedance driver than the specification, you CAN make it faster. I did so many of the MOSFET pulsing circuit with sub nano second rise time and fall time and various delays. It is all about the drivers. I designed a very precise turn on time of a N-MOSFET by using a P-MOSFET to pull the gate up STRONG and I get such a fast rise time and short propagation delay it's incredible. It was way less than 1nS transition time using a medium size MOSFET that can pulse like an amp or more for 100V...Yes, no typo 100V. To do that, I used a strong MOS driver...Two in parallel to drive the P-MOSFET...to drive the N-MOSFET! All the speed concentrated on that one edge, the rise time.

It is all about the drive.

This is true inside a CMOS IC. That's the reason people get processor speed to 3 or 4 GHz speed. The FETs inside are very very tiny, gate capacitance is so low. They don't take much drive to switch the gate, that's the reason they achieve the fast switching time and short prop delay. But why are the external I/O pins so slow? Because in order to have any drive to drive the external circuits with parasitic capacitance, they need strong drivers at the I/O pins. Most end up doing multiple stages of what I did...a small FET drive a little bigger FET, then the little bigger FET drive the intermediate FET...so on and the last stage have big enough FET to drive the I/O.

This is the name of the game. At least this is absolute true at the time I was in the field.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
7
Views
15K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
8K
Replies
7
Views
3K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K