There are two ways to solve the problem as what you said about the Ib and Ic, the one is to take the 1k Rbe away, and reducing the values of Rb from 4.7k to 4.3k, that is to match the official info of datasheet, when Ib=1mA then the Ic will be equal to 10mA and the Vce will get into the saturation region.

If you use a casc_o_ded MOSFET, there isn't a problem to be solved.

Simply connect the gate to Vcc and the source to the 5V pulse output.

Provide a drain load resistance to +28V and its done.

The only downside is no current gain from the grounded gate MOSFET - this is very simple to fix with an emitter follower driven by the MOSFET drain.

You can pretty much leave the calculator in the drawer.

 

Thanks. It seems that you chose Ic = 10mA and then design Rb and Rbe1 to make sure the transistor is in saturation. With Ic =10mA, so in order for Q1 is in saturation in all conditions, take Ib = 1mA and from that you figured out the values of Rb and Rbe1 as above.
Here is how I calculate R3, is it the right way?

R3 = (28V - Veb2 -Vce1)/10mA = (28V - 0.6 -0.2) = 27K.
How did you calculate R4 and R5?

 

There are two ways to solve the problem as what you said about the Ib and Ic, the one is to take the 1k Rbe away, and reducing the values of Rb from 4.7k to 4.3k, that is to match the official info of datasheet, when Ib=1mA then the Ic will be equal to 10mA and the Vce will get into the saturation region.

If you use a casc_o_ded MOSFET, there isn't a problem to be solved.

Simply connect the gate to Vcc and the source to the 5V pulse output.

Provide a drain load resistance to +28V and its done.

The only downside is no current gain from the grounded gate MOSFET - this is very simple to fix with an emitter follower driven by the MOSFET drain.

You can pretty much leave the calculator in the drawer.

Thanks.
There are many ways to solve the problems, I only provided a method that it can be done, it depends on the OP to decide it.

When we using the MOSFET that it doesn't like as you said, could you drawing a circuit make it more easy to discuss?

 

Thanks.
There are many ways to solve the problems, I only provided a method that it can be done, it depends on the OP to decide it.

When we using the MOSFET that it doesn't like as you said, could you drawing a circuit make it more easy to discuss?

The circuit is so simple, it doesn't need a schematic.

The only additional component the MOSFET needs is a load resistor to +28V

The emitter follower stage (if needed) is pretty much self explanatory.

Collector to +28V, base to drain - you can put a load resistor from emitter to GND if the load doesn't draw enough current to bias the B/E junction.

If you need a large output current - select a logic level MOSFET, most types upto about 1A have low enough VGSthr anyway.

Unless you need output current much more than about 5A, most of the math is counting on your fingers stuff.