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**Unformatted text preview: **Physics 3330 Prelab 4 solutions 1. All but two of the parameters can be found in the tables labeled DC Electrical Characteristics or AC Electrical Characteristics on pages 2–3. The two parameters which require more digging are the maximum output current I O max and the output impedance R O . The maximum output currents can be found in graphs on page 5. At a temperature of 25 ◦ the maximum current (positive or negative) is about 25 mA at fairly low output voltage (less than 5 V). The output impedance for an amplifier circuit is given in the top-left plot on page 7 for the LF156. If we pick a closed loop gain of A V = 10 and a frequency of 10 kHz, we see that the output impedance is about 0.8 Ω. To find the output impedance of the op-amp we need the equation R O = R O / (1+ AB ) which gives the amplifier circuit output impedance R O in terms of the op-amp output impedance R O . Solving for R O gives R O = R O (1 + AB ). Now B ≈ 1 /G = 1 /A V = 0 . 1. We can get A from A = A / (1 + jf/f ) (Eq. 2). Note also that f T = A f so f = f T /A = 5 MHz / 2 × 10 5 = 25 Hz. Keeping in terms of the symbols, we find A = A / (1 + jfA /f T ). Note that the denominator has a real and imaginary component. What we are really interested in is just the magnitude. Technically, we get this by multiplying by the complex conjugate and taking the square root. However, in this case we can notice that the imaginary part of the denominator is much larger than the real part as long as the frequency we are concerned with f is much larger than...

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