IRF530 max input frequency

[awaiting]

Hi everyone,

I am using IRF530 in a half h bridge DC motor driver. The MOSFET is behaving erratically if the PWM frequency is increased above 20 KHz. I checked the datasheet thoroughly but could not find any thing related to maximum operating frequency. Is input capacitance related in any way to the max input frequency(It has input capacitance of 670pF).

 

[meBigGuy]

You need to be able to drive the 670pf input capacitance at the frequency you want to operate at. That requires a decent driver to achieve high slew rates. Look for power fet driver chips.

In order to achieve the sub 50ns switching times that the device can support according to the delay and transition times in the spec section ( and figure 10a and 10b), you need to slew the gate faster than 50ns.

Say for example you want to switch the gate from 0 to 10V in 50ns. That requires I = Cdv/dt = 670pf*10V/50ns = 134ma of drive current.

 

[awaiting]

I am using IR2104 as the driver. It has turn on propagation delay of 820 ns (max) and turn on rise time of 170ns (max). This I assume gives the fastest time in which the driver can respond. So, it means I cannot get a frequency higher than 1MHz. But I want to test the circuit in 20KHz to 1MHz range.
I also tried IRF520, it has input capacitance half of IRF530 and it had slightly better response. It was operating properly upto 22.5KHz.
I think you read 20MHz instead of 20KHz. I calculated the drive current for 20KHz and it was 0.16 mA.

 

[Baluncore]

Fast switching of high capacitance loads such as a mosfet gate requires that short bursts of charge flow as current to or from the gate. That charge comes from power supply bypass capacitors. Your driver will need plenty of fast supply capacitance to supply the charge for the transition. Without that the power rails will fall and upset the driver bias and logic.

Don't forget that the mosfet capacitance is largely miller capacitance between the gate and drain. Switching the mosfet requires that the gate–drain capacitance be charged through the entire drain voltage change.

 

[meBigGuy]

You don't calculate the drive current for the average frequency. You calculate the maximum drive current peaks in order to switch in the time specified. I calculated the current to switch 670pf 10 volts in 50ns (134ma). 0.16ma will slew 670pf by 10V in 41uS.

My choice of 50ns was arbitrary.

It doesn't matter if you switch 1 per minute, or 1 MHz, the current spikes during switching will be 134ma if you are ramping in 50ns.

Do you understand that the rate of change of voltage on a capacitor is proportional to the current flowing. The faster you want it to change, the more current you need to be able to supply. That means low resistance drivers, stout supply, and very good bypassing.

On a side note, realize also that the slower you switch, the more power the power fet will consume while switching since it has significant voltage across it while still conducting high current.

It looks like the IR2104 should give you decent response if your supply can support the currents and your grounds are good. Realize, however, that the currents specified are short circuit currents. You can calculate a series R from the short circuit current to get an idea of what your response will be.

130ma, 10V gives 76 ohms which give a time constant of Tau = 76*670pf = ~50ns . So, it should be pretty much there in, maybe, 3T or about 150ns.