From page 9 of the go along with the video document (see link in the description) "Putting in these known values, we get 1.5 mA = 5 ∗ (−1/2.141K) + (some constant)" We have the 1.5 mA on the left. We add 5*(1/2.141k) to both sides and we get 3.836 mA = (some constant). So, now we have Ic = Vce ∗ (-1/2.141K) + 3.836 mA Does this help? 🙂
Man it was 49 years ago I leaned this in school. Had totally forgot all of it but it is comming back looking at your video. Thanks for great content.
Yeah, it was a LONG time ago for me, too. I'm glad that you enjoyed the video! 🙂
I followed your explanation up until the value for (some constant). Where did 3.836ma come from?
From page 9 of the go along with the video document (see link in the description)
"Putting in these known values, we get
1.5 mA = 5 ∗ (−1/2.141K) + (some constant)"
We have the 1.5 mA on the left. We add 5*(1/2.141k) to both sides and we get
3.836 mA = (some constant).
So, now we have
Ic = Vce ∗ (-1/2.141K) + 3.836 mA
Does this help? 🙂
Got it I wasn't adding 5 * (1/2141) to both sides, I'm a little slow on the math sometimes. Thanks
@@U812-k7j I've done that! You are welcome! 🙂