Experiment 3, Part A, Low Pass Filters
where s=jω
The shape of the transfer function depends on the low frequency gain |T(ω=0)| = K, the shape factor Q and the center frequency ωO. The figure below shows typical plots of |T(jω)|.
|T(jω)|
The analysis of the active low-pass filter given below (known as a Sallen and Key circuit) is taken from Analog Filter Design by M. E. Van Valkenburg (Oxford University Press, 1982).
The circuit shown in Figure 2 is one of a class of circuits that were described in 1955 by Sallen and Key, then at MIT’s Lincoln Laboratory. In the circuit the noninverting op-amp circuit provides a constant relationship between V2 and Va, which is
The controlled-source representation of the Sallen and Key circuit is given in Figure 3.
The circuit may be routinely analyzed using Kirchhoff’s current law. At node a the currents directed out of the node must sum to zero, or
Similarly, the sum of the currents out of node b is
We next rearrange these equations in a form for solution:
and
We now eliminate the voltage Vb and solve for the ratio V2/V1=T. The result is, after some algebraic simplification,
This transfer function is recognized as being of the general form
Experiment 3, Part B, Band Pass Filter
Updated November 6. 2001
The transfer function for a bandpass filter is given by
and B.W. denotes bandwidth.
We can draw a equivalent for the portion of the circuit located within the dotted lines
If we add all of the other components to the circuit, the circuit below results.
-If we can find VX, we will know VOUT (Vout = - (R2/Z1) Vx.
- We can find Vx by applying KCL at “x”.
This yields:
Hence,
and therefore,
Q=7Determines
f0 = 620 Hz = o/(2)DeterminesR1C (but not either alone)
After you determine the component values please be sure to calculate and plot the variation of |T(j)| versus . Your curve should look like the one below
In the lab you will measure this frequency response and compare it with the theoretical curve calculated from the transfer function.
Experiment 3, Part C, Oscillators
Sedra and Smith 4th edition (974-975, 980-981)
Or 5th edition (1166-1168, 1171-1172) or 6th edition (1336-1344)
We will analyze and build a Wien-bridge oscillator. An oscillator produces an output without any (signal) input. (Noise is responsible for driving the output of the oscillator.) We will examine the circuit shown below and at first assume that there is an input Vin. We will calculate the output and show that, under certain circumstances, there is an output even when the input is zero (or just noise).
Oscillation occurs, i.e. there is an output without any input if L(j)=1.
L(j) is a complex quantity, the statement L(j) = 1 implies |L| = 1 and that the phase of L(j) is a multiple of 2L(j) = 1 is known as the Barkhausen criterion. Does |L| or |A| have to be exactly = 1?
We can approach this problem in another way. We will follow noise at a particular frequency, n(f) around the circuit. The noise will experience gain = A then feedback (or voltage division). The gain will multiply the noise by A and the feedback will multiply its complex amplitude by . This means that after “one pass” through the system the output due to a noise component at a particular frequency will be given by its original value plus the value that results from the gain and feedback or
n(f) {1 + A);
after two passes the output will be given by
n(f){1 + AA
fter n passes the output will be given by
n(f) {1+ A + (A)2 + (A)3 + (A)4 + (A)5 + …+ (A)n}
Clearly each of the terms in the output of the above equation will add constructively to the proceeding terms i.f.f. the phase of Ais a multiple of 2. It is also clear, that if in addition |A| > 1, then the output will grow without bound. This means that there will be an output without any input (except for noise). In a real circuit what limits the size of the output?
If |A| < 1then each successive term is smaller than the preceding one and the sum is finite (and very small).
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When you do the experiment, you will observe that if the gain is too large, the output will not be sinusoidal; the peaks and valley will be clipped and for high gain the output will look like a square wave. Here is a way of stabilizing the output from the Wien-bridge oscillator. You might want to try this circuit in class. The required lamps will be available in the lab. The circuit below is drawn for a 28V, 24 mA lamp. If a different lamp is used the size of the feedback resistor (shown as 470 ohms in series with a 1000 ohm potentiometer) may have to be changed. (One common alternative lamp is a 12V, 60mA lamp.)
R and C should have the same values of those used in the original experiment.
Adjust Rvariable until the output sine wave stabilizes. The peak to peak output voltage is expected to have a value in the range of 10-16 Volts.
The WeinBridge oscillator has very low distortion because the action of the lamp prevents the op amp from saturating. The resistance of the lamp’s filament rises dramatically as the filament temperature rises. Even though the filament does not get nearly warm enough to produce a visible glow, the resistance increases as the oscillator’s output increases. The increase in resistance reduces the gain of the op amp circuit and at a specific output voltage a balance is reached where the oscillator has just enough gain to maintain oscillation but the output cannot increase further. Note that if the lamp were fully lit its resistance would be 28V/24 mA or 1.17 k and the gain of the op amp would be insufficient for oscillation. When the lamp is dimly lit its resistance is substantially less than 1 k and the op amp’s gain is >3.
Without a nonlinear resistance in the circuit the oscillator would either not run or would drive the op amp into clipping, causing increased distortion.
Prof. Steve Jacobs suggested the use of this stabilization technique in this experiment.
Prof. Jacobs has located a detailed discussion of Wien-bridge oscillators (including stabilization circuits) at
Updated July 17, 2006 and March 10, 2011