HW9 - UC Berkeley EECS Department EECS 40 B E Boser HW9 RLC...

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UC Berkeley, EECS Department B. E. Boser EECS 40 HW9: RLC Circuits UID: You will get a full score (100) for at least 80 % of all problems solved correctly. 1. Voltage comparators signal when the input exceeds a threshold. Sketch v o versus v in for v re f = 2.5 V for the comparator circuit shown in Figure 1. Note that there is no feedback and hence the voltage difference at the input of the opamp is not zero! Calculate the output voltage for different values of v in . Assume that the operational amplifier achieves true “rail-to-rail operation” with supplies V ss =0 V and V dd =5 V. v o ( v in = 0.7 V ) = 1 pt. 0 v o ( v in = 2.4 V ) = 1 pt. 1 v o ( v in = 2.6 V ) = 1 pt. 2 Figure 1 A voltage comparator produces rises its output high ( v o = V dd ) if v in > v re f , and low ( v o = 0 V) otherwise. A normal operational amplifier with high gain and the output voltage limited by the supply can be used for this purpose. However, opamps are slow in this configuration, most taking several milliseconds for the output to change. Comparator circuits are much faster but cannot be used as amplifiers. The symbol in the diagram indicates that a comparator rather opamp is used. 2. Capacitors are used as transducers for touch sensors, fluid level sensors, electronic levels, acceleration and pressure sensors, and many other important applications. But how do we turn a change in capacitance into a corresponding electrical signal for further processing e.g. with a microcontroller? The diagram shown below shows a solution. The unknown capacitor C x (e.g. touch sensor, electronic level, etc.) is connected to a known reference capac- itor C re f . Circuit operation is as follows: a) Close S 1 and S 3 , open S 2 to discharge C x and charge C re f to v 2 = V re f . b) Open S 1 and S 3 and close S 2 . Some of the charge on C re f gets transferred to C x and v 2 drops as a consequence. c) Open S 2 and close S 1 ( S 3 remains open) to dump the charge C x received from C re f . d) Repeat steps (b) and (c) until v 2 < v 3 and the comparator output v o goes high. The number of steps is a measure of the value of C x . Plot v 1 , v 2 and v o for C x = pF, V re f = 3 V and R 1 = R 2 . Find an expression n C re f , C x for the number of cycles after which the comparator output v o goes high and solve for C re f = pF.

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