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EE302Lecture15

# EE302Lecture15 - DC Transistor Circuits NPN PNP What...

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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

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2 silicon I E = I B + I C I C = β I B (note the direction of I C ) β is the current gain
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 .

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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
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

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