The Pin 6 of photocoupler should be disconnect from Vcc.

The BJT is drive by Ib current, but the MOSFET is drive by Vgs, so they are complete different, where did you get the idea that it needs the 200 mA for frequency and how high of the frequency?

I think maybe you just need one PNP stage to drive the MOSFET and the Ig of MOSFET is <10 mA.

Thank you @ScottWang for the prompt reply.
Yes, my bad. the PIN 6 optocoupler should be connected. it is base.
I agree that mosfet is voltage drive instead of current. Having more current makes gate charge discharge faster.
I use application note attached to calculate power dissipation + switching loss and gate current to be minimum.
The frequency is 2KHz. The load is a rotating LED signage. I put the frequency a little bit higher since the LED is moving, hence it is flicker free.

I shall take your advice, having single PNP drive mosfet simplify the case. And I take note about Ig is <10mA and do a practical breadboarding the circuit. My calculation is just simply idealistic theory.
Thanks for your time and advices.

 

Having more current makes gate charge discharge faster.

Heavy current will makes gate charge more fast, but it will also makes gate discharge more slower.

The frequency is 2KHz. The load is a rotating LED signage. I put the frequency a little bit higher since the LED is moving, hence it is flicker free.

How are the smallest and largest duty cycle of pwm and how is the rated current of LED?

 

Heavy current will makes gate charge more fast, but it will also makes gate discharge more slower.


How are the smallest and largest duty cycle of pwm and how is the rated current of LED?

it is between 1 to 1000. Actually I don't need full range, since the human eyes is logarithmic in nature, the value becomes like 1,10,100,1000.

 

it is between 1 to 1000. Actually I don't need full range, since the human eyes is logarithmic in nature, the value becomes like 1,10,100,1000.

Are those duty cycles like 1:1, 1:10, 1:100, 1:1000?
And how are the Vf and rated current of LED?

 

Are those duty cycles like 1:1, 1:10, 1:100, 1:1000?
And how are the Vf and rated current of LED?

Correct, the duty cycles is a choice of 4, which is 1:1, 1:10, 1:100, and 1:1000.
It just a cut of a LED strip running at 12V. current is approximately 2A.

 

Since you were concerned about the discharge time of Vgs then you can try the new circuit, if you have an O'scope then you can measure the discharge time, you also can be decreases the values of R5 resistor which I labeled.

 

Since you were concerned about the discharge time of Vgs then you can try the new circuit, if you have an O'scope then you can measure the discharge time, you also can be decreases the values of R5 resistor which I labeled.

View attachment 140131

Thank you @ScottWang . That's very good. Thank you so much for the help. I am so grateful.

 

Thank you @ScottWang . That's very good. Thank you so much for the help. I am so grateful.

You are welcome.
I'm waiting your results and feedback ...

 

Thank you so much for your helping hand @ScottWang
Here is my circuit. Vcc is 12V and the gate current I just calculate using gate current calculator based on mosfet parameter and frequency applied, it is turned out to be 200mA.
Thank you in advance for the help.
View attachment 140102

Put your load North of the MOSFET (irf540). That is, between pin 2 and Vcc.

A MOSFET gat must be 10v above the voltage at Pin 3 (Source) to turn on completely. The way you have it wired, you'll never be able to do that.

Also, disconnect pin 6 in opto as Scott said.

Why is this sooo complicated?

 

Put your load North of the MOSFET (irf540). That is, between pin 2 and Vcc.

A MOSFET gat must be 10v above the voltage at Pin 3 (Source) to turn on completely. The way you have it wired, you'll never be able to do that.

Also, disconnect pin 6 in opto as Scott said.

Why is this sooo complicated?

I almost forgot this point, specially when the MSOFET working with the pwm signal and not the static status.

Why is this sooo complicated?

If you meant that why this issue spent so many posted, you just read it from the first then you will got the answer as other threads have the similar problem, the TS probably thought that just cut a part of circuit to show us then that is enough, until we asked and asked again.

 

Put your load North of the MOSFET (irf540). That is, between pin 2 and Vcc.

A MOSFET gat must be 10v above the voltage at Pin 3 (Source) to turn on completely. The way you have it wired, you'll never be able to do that.

Also, disconnect pin 6 in opto as Scott said.

Why is this sooo complicated?

I am so sorry, its my bad..
it should be:


I really missed while put the load on N channel instead thinking of P channel.
Thanks @GopherT

 

I almost forgot this point, specially when the MSOFET working with the pwm signal and not the static status.


If you meant that why this issue spent so many posted, you just read it from the first then you will got the answer as other threads have the similar problem, the TS probably thought that just cut a part of circuit to show us then that is enough, until we asked and asked again.

Got it @ScottWang . Next time I shall bring overall from start to the end. so get the overall picture.
Pardon me, Will never do that again..

 

Got it @ScottWang . Next time I shall bring overall from start to the end. so get the overall picture.
Pardon me, Will never do that again..

That's good.
Just do what you can do and if we need any info then the helpers will ask you.

All our helpers like to see the whole picture of the question, show all the circuit and questions then you will get the quick answer and bring more helpers to help you.