{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE302Lecture15 - DC Transistor Circuits NPN PNP What...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
DC Transistor Circuits What happens in a transistor is a small base (B) current results in a large collector (C) current. The collector current is dependent on the base current. The transistor base current is amplified in the collector current. Because there are diode junctions, there is a voltage drop associated with the base to emitter junction. This suggests that the DC transistor can be represented with the model shown below: 1 + NPN PNP
Image of page 1

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

View Full Document Right Arrow Icon
2 silicon I E = I B + I C I C = β I B (note the direction of I C ) β is the current gain
Image of page 2
100k+200k -200k I 1 2 = -200k 200k I 2 -0.7 I 1 = 13 μ A I 2 = 9.5 μ A I 3 = - β I 2 = -(150)(9.5 μ A) = -1.425 mA KVL Loop 3: - v o + 1000 I 3 + 16 = 0 or v o = -1.425 + 16 = 14.575 V 3 200k( I 2 I 1 ) + V BE = 0 Fig 3.44 .
Image of page 3

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

View Full Document Right Arrow Icon
Nodal Eqns: Working with Fig. 3.44(b) KCL Node 1: (0.7 2)/100k + 0.7/200k + I B = 0 I B = 9.5 μ A KCL Node 2: 150 I B + ( v o 16)/1000 = 0 v o = 16 1000 (150)(9.5 μ A) = 14.575 V Which was easier? The nodal method. Next we will discuss these topics: Superposition Thevenin Equivalent Norton Equivalent 4 Fig 3.44
Image of page 4
Superposition: Given the circuit: V=IR V is always proportional to I provided R is constant. Now consider this circuit: V=I 1 R + I 2 R V is proportional to I 1 and I 2 independently . V responds linearly to I 1 and I 2 sources. The R could be a more complex circuit with many elements so we could think of the R as a matrix, i.e. [V] = [R][I] If the [R] circuit had had 2 nodes then this would have been true: V 1 I 1 I 1 0 V 2 = [R] I 2 = [R] 0 + [R] I 2 From this example we see that we could solve for a part of the V due to only I 1 when I 2 is removed and get a value for V due to the I 1 . Likewise we could solve for a part of the V due to only I 2 when I 1 is removed and get a value for
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern