HW6%20solutions_EG - MAE 3780 MECHATRONICS FALL 2009 H O M...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MAE 3780 MECHATRONICS FALL 2009 1 HOMEWORK 6 SOULUTIONS 1. First, assume that the transistor is in the active region. Then we have: KVL around the left loop gives 0 V 7 . 5 = γ V R I B B Plugging in the appropriate values and solving for I B gives μ A 102 = B I Therefore, mA 4 . 20 μ A 102 200 = = = B C I I β Now, we solve for V A to see if the assumption that the transistor is in the active region is valid: mA 4 . 20 10 = = C A C R V V I , which yields V 36 . 10 = A V This is a contradiction, since V V A < . Now, assume that the transistor is in the saturated region. Then we have:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
MAE 3780 MECHATRONICS FALL 2009 2 Once again, we have μ A 102 = B I KVL around the right loop gives 0 0V 1 = CEsat C C V R I , which yields mA 8 . 9 = C I . To verify that the assumption that the transistor is saturated is correct, we find the ratio 1 . 96 μ A 102 mA 8 . 9 = = B C I I , which is significantly less than β . Therefore, the assumption is correct. 2. a. (15 points)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

HW6%20solutions_EG - MAE 3780 MECHATRONICS FALL 2009 H O M...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online