HomeworkAssignment02Solution

To turn off a current source we remove it from the

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Unformatted text preview: we remove it from the circuit. The circuit to current . Then the right indicates the setup to determine . The output voltage is yields , no current flows into the inverting input and applying Ohm’s law ( Since ⁄ , it follows that ⁄( . 6 55:041 Electronic Circuits. The University of Iowa. Fall 2011. Question 6 Show that the output resistance seen by the load in the voltage-to-current converter circuit below is (1 5 points) Solution Determining the output resistance is equivalent to determining the Thevenin equivalent resistance, which we do as follows: turn off independent sources ( in this instance), add a test voltage and determine ⁄ the resulting current . Then The circuit to the right indicates the setup. Note that turning off a voltage source means shorting it. Furth, and KCL at the inverting input node gives ( ) ( If one recognizes that the op-amp and form a noninverting amplifier, one could also write this expression down by inspection. KCL at the noninverting input gives ( Combining equations (1) and (2) yields ( ( Finally, ) ) ( ⁄ , so that 7 ) 55:041 Electronic Circuits. The University of Iowa. Fall 2011. Question 7 The circuit shown is a representation of the common-mode and differential-input signals to a difference amplifier. One can write the output voltage as , where is the differential-mode gain and is the common-mode gain. Assuming an ideal op-amp, and setting show that the common-mode gain is ( ) ( ⁄ Hint: Consider replacing the source with two identical sources a nd as a nd respectively. Then use superposition to solve for the output voltage. connected (1 5 points) Solution Using the supplied hint, we add two sources and then turn on on one at a time and add the results (superposition). (a) (b) (c) With input , the circuit is a inverting amplifier and the output voltage is (⁄ ( With input , the circuit is a noninverting amplifier and the output voltage is ( ⁄ ( ⁄ Using superposition the, the output voltage is ( ⁄ ( ( ⁄ ) ( : ( ( ( ⁄ ⁄ ( ( ( ) ⁄ ⁄ ( ⁄ 8 ⁄ ( ⁄ ( ) ⁄ 55:041 Electronic Circuits. The University of Iowa. Fall 20...
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This note was uploaded on 10/26/2013 for the course ECE 55:041 taught by Professor Kruger during the Fall '11 term at University of Iowa.

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