V2 Is an AC Voltage Source with a Phase of 0, and Amplitude of 2.5 Volts

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V2 Is an AC Voltage Source with a Phase of 0, and Amplitude of 2.5 Volts

Kyle Schlansker ENGR 311 A Lab 1 September 12, 2002

RLC circuit 1

V1 is a sinusoidal voltage source with amplitude of 2.5 volts. It has a 0 dc voltage offset, and its frequency is set to 1000 Hz.

To analyze the circuit, a transient analysis was performed. As shown in the window below, the time scale was chosen to be 0 seconds to 2 milliseconds. This scale was chosen in order to accurately view the 1000 Hz frequency of the sinusoidal voltage source. 4 . 0

2 . 0

0

- 2 . 0

- 4 . 0 0 s 0 . 2ms 0 . 4 ms 0 . 6 ms 0 . 8 ms 1 . 0 ms 1 . 2 ms 1 . 4ms 1 . 6 ms 1 . 8 ms 2. 0 ms I ( R1 ) V( V 1: + ) Ti me RLC circuit 2

V2 is an AC voltage source with a phase of 0, and amplitude of 2.5 volts. R1 was chosen to be smaller than the resistor from RLC circuit 1, in order to achieve a more well-defined resonance peak. Likewise, the inductance of L1 was increased to yield the same results.

Because an AC source was used in this circuit, an AC sweep analysis was required. The resonant frequency was calculated by hand, and then the start and ending frequencies were chosen in order to have the resonance peak centered for the plot. 500 points per decade were chosen so that an accurate graph of the peak could be visible and smooth. 6. 0mA

4. 0mA

2. 0mA

0 A 1 . 0KHz 3. 0KHz 10KHz 30 KHz 10 0KHz 300KHz 1. 0MHz I ( R1) Fr e qu e nc y

In the probe cursor box, A1 denotes the approximate frequency that resonance occurs. It shows it as 22,594 Hz. The hand calculation below re-assures the computer calculation by yielding the same ballpark resonance of 22,507 Hz.

L = 50 * 10^-3 H C = 1 * 10^-9 F w = 1 / Sqrt(L * C) f = w / (2 * PI) w = 1 / Sqrt((50 * 10^-3) * (1 * 10^-9)) = 141,421 radians/sec f = w / (2 * PI) = 22,507 Hz Inverting Operational Amplifier with a 3 To 1 Output to Input voltage ratio

The +-Vrails were tied to 12 volt dc sources to limit the output voltage to a maximum of +- 12 volts. For a simple inverting op-amp, the + terminal was set to ground, while the negative terminal was connected both the source line, and the output voltage line. A sinusoidal voltage source was chosen as in RLC circuit 1 with a 1000 Hz frequency and a 2.5 volt amplitude. Rf was chosen to be 3000 ohms, and Ri was chosen to be 1000 ohms. These resistor values were chosen such that the ratio of Rf/Ri would be 3/1, yielding an output/input voltage ratio of 3/1. As in RLC circuit 1, a transient analysis was chosen with the values shown in the window below.

In order to accurately view the plot, the final time was set to 2 milliseconds. Again, this was chosen by analyzing the 1000 Hz frequency of the sinusoidal voltage source. The following two graphs and their corresponding probe cursors show the output voltage and input voltage. The output voltage is in red, and the input voltage is in green. As expected, the output voltage is 3 times the amount of the input voltage which agrees with the circuit design of a 3/1 ratio for Rf/Ri. Also note that since this is an inverting op- amp, the output voltage is the inverse of the input voltage (plus increased by a factor of 3).

8. 0V

4. 0V

0 V

-4 . 0V

-8 . 0V 0s 0. 2ms 0. 4ms 0. 6ms 0. 8ms 1. 0ms 1. 2ms 1. 4ms 1. 6ms 1. 8ms 2. 0ms V( V5: +) V( U1A: OUT) Ti me

8. 0V

4. 0V

0 V

-4 . 0V

-8 . 0V 0s 0. 2ms 0. 4ms 0. 6ms 0. 8ms 1. 0ms 1. 2ms 1. 4ms 1. 6ms 1. 8ms 2. 0ms V( V5: +) V( U1A: OUT) Ti me

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