ece0132_hw1_soln_fall2007

# ece0132_hw1_soln_fall2007 - ECE 0132 Homework Assignment#1...

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ECE 0132 Homework Assignment #1 – Solutions 1. DATA BOX. (a) When the DATA BOX is in state 01010101 0 1 2 3 4 5 = Q Q Q Q Q Q , the 1.0 k , 2.2 k and 4.7 k resistors are connected to +5V, and the 1.5 k , 3.3 k and 6.8 k resistors are connected to ground. Therefore, the circuit model for this state is 5V 1.0 k 2.2 k 4.7 k 1.5 k 3.3 k 6.8 k V out . By combining resistors in parallel, this circuit can be simplified as follows. 5V R 1 R 2 V out The parallel resistance in the upper half of the resistor network (between the output node and +5V) is . k 5997 . 0 mS 6673 . 1 7 . 4 1 2 . 2 1 0 . 1 1 1 7 . 4 // 2 . 2 // 0 . 1 1 1 1 = = + + = = R R R

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The parallel resistance in the lower half of the resistor network (between the output node and ground) is . k 8955 . 0 mS 1168 . 1 8 . 6 1 3 . 3 1 5 . 1 1 1 8 . 6 // 3 . 3 // 5 . 1 2 2 2 = = + + = = R R R Next, we solve the simplified circuit to obtain the output voltage out V . The current flowing through 1 R and 2 R is 2 1 5 R R I + = . The output voltage out V is the potential difference between the output node and ground, which is the same as the voltage across 2 R . Therefore, using Ohm’s Law, we have + = + = 2 1 2 2 2 1 5 5 R R R R R R V out . This is a general and very useful result known as the voltage divider . Whenever two resistances are connected in series, and we wish to know the voltage across one of them, the voltage divider equation expresses the desired voltage as a fraction of the total source voltage, which in this case is 5V. Using the parallel resistances obtained previously, we have 994 . 2 8955 . 0 5997 . 0 8955 . 0 5 = + = out V V. (b) Repeat part (a) for states 011011 and 011001. When the DATA BOX is in state 011011 0 1 2 3 4 5 = Q Q Q Q Q Q , the 1.0 k , 1.5 k , 3.3 k and 4.7 k resistors are connected to +5V, and the 2.2 k and 6.8 k resistors are connected to ground. Therefore, the parallel resistance in the upper
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