I'm just to figure this stuff out by working backwards from the solution and I can't figure out why he decides to select Ic = 1mA.

Attached is his solution and the datasheet the students were provided with.

Can someone clarify? I'm just trying to understand the thought process behind his solution.

Thanks again!

 

I'm just to figure this stuff out by working backwards from the solution and I can't figure out why he decides to select Ic = 1mA.

Attached is his solution and the datasheet the students were provided with.

Can someone clarify? I'm just trying to understand the thought process behind his solution.

Thanks again!

It could be from a simple 'math' viewpoint, in order to keep the math simple and allow the underlying design issues to become apparent. It is also the smallest DC gain example listed in the datasheet.

 

P total must be 10mW or less.
If the collector current is 1mA then the collector power dissipated is 9V x 1mA= 9mW plus some power dissipated in the base bias resistors which is close to the max allowed power dissipation.

 

P total must be 10mW or less.
If the collector current is 1mA then the collector power dissipated is 9V x 1mA= 9mW plus some power dissipated in the base bias resistors which is close to the max allowed power dissipation.

Ah this must be what he was thinking when he selected Ic!

Thank you!