Physics 127AL, Problem Set 2 Solutions
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Physics 127AL, Problem Set 2 Solutions
1 Bandpass filter
We can construct the circuit as follows:
C1 |
R2 Vout
Stage 2 |
|
Vin |
||
R1 |
||
Stage 1 |
|
To build Stage 1, we choose R1 such that the input impedance Zin , 1 is much larger than the 100Ω output impedance of the signal source. We can assume that the relevant input frequencies are larger than the 3dB point of this stage, since at this frequency
f3 dB , 1 = =⇒ C1 = = = iR1 ,
so that we can take Zin , 1 = C1 + R1 ≃ R1 . This allows us to set R1 = 1kΩ, so for f3 dB , 1 = 100Hz,
C1 = = 1.6 × 10 −6 F or 1.6µF.
For low input frequencies, Zin , 1 ≃ C1 , so we take XC1 = 1kΩ, and the same values for R1 and C1 follow. In either case, we should then have R1 = 1kΩ and C1 = 1.6µF.
For Stage 2, we will need Zin , 2 ≫ Zout , 1 to keep the impedance of this stage from loading the first stage. We can then take R2 = 10kΩ using the same arguments above, so for f3 dB , 2 = 10kHz,
C2 = = 1.6 × 10 −9 F or 1.6nF.
By construction, the output impedance is Zout , 1 + Zout , 2 ≃ Zout , 2 for the whole circuit. The maximum value of Zout , 2 is R2 , since Zout , 2 = R ∥C2 ,
so the worst-case output impedance is 10kΩ . The minimum recommended
load impedance must be greater than this value, so we can take it to be at least
100kΩ . |
2 Offset high pass
At low frequencies, the filter acts as a voltage divider with the input halved, and at high frequencies, as a high-pass filter. If we have a resistor R and ca- pacitor C connected in parallel, then its output impedance is
RC R(i/ωC) R
C
Vin
R
R
At high frequencies, the resistor parallel to the capacitor (as viewed from the input) can be ignored, so the capacitor forms a high-pass filter with the other resistor. In this case, the frequency response we have suggests that ω3 dB ≃ ω0 , so we must choose our values for R and C such that ω0 ≃ 1/RC . (This is not an exact prescription, since at ω = ω0 , the effective impedance of the RC-in-parallel
piece is
R 1 + i
1 − i 2
so we instead have
= | | = | 3 i | = 4 = 0.6325,
which is not the correct value at the 3dB point. It is possible to calculate ω3 dB exactly, so that we can set ω0 to this value.)
3 DC power supply
The output is a sine wave that is clamped at +0.7V and −0.7V by the two
V
+8.9 V
+0.7 V
-0.7 V
-8.9 V
4 Full-wave bridge
We can construct the circuit as in this diagram:
For us to deliver 10V DC with at most a 0.1V (pp) ripple, our rectifier output should be 10.1V. Since the rectifier adds two diode drops, this requires that the output amplitude for our transformer be 10.1V + 2(0.7V) = 11.5V, or an rms output of 11.5/^2V = 8.13V.
To choose the capacitor, we note that the ripple voltage ∆V is given by
Iload
∆V =
(See Horowitz and Hill, p. 33.) Then for the typical wall-power frequency of
60Hz and Iload = 10mA = 0.01A, we should have
C = = F = 8.33 × 10 −4 F = 830µF.
This is not a standard capacitor value, so we can adopt a higher capacitance, say C = 1000µF, since this only reduces the ripple voltage. For safety, we can:
• Put a bleeder resistor across the output of the supply to discharge the capacitor within a few seconds when the power is turned off. A 1kΩ resistor would discharge it in a couple of seconds.
• Put a fuse on the input to the transformer.
• Enclose the full circuit in a metal box that is connected to ground through a three-prong power cable.
• Cover any exposed wires on the transformer with a metal cover inside the main box with a warning label that the power cable should be disconnected before servicing.
2023-06-16