Construct the circuit below.  Use matched, 1%, precision resistors. To match resistors, start with ten to twenty of each value.  Measure the resistance of each resistor, and select the two that have the closest values.  

Note that the offset adder needs to be explicitly grounded.  One way to do this is illustrated in the photo at right.  The offset adder input is not otherwise used in this problem.

Begin investigating this circuit by connecting the meters to point A. Vary the offset adder voltage, and prove that the $10\,\mathrm{k}\Omega$  resistor does indeed behave appropriately. (This purpose of this step is to check your meter polarities.)

Next connect the meters to point B. Vary the offset adder voltage and measure the current. Prove that the NIC appears to be a $-10\,\mathrm{k}\Omega$ resistor. 

A gyrator is a circuit that converts an impedance $Z$ to $R^2/Z$ . Gyrators can be synthesized from two NICs as show in the block diagram at right.  The components directly below each individual NIC form the impedance $Z$ of that particular NIC.

When the explicit Z in the gyrator comes from a capacitor,  to $Z=1/j\omega C$, and the gyrator will look like it has an impedance $j\omega CR^2$; i.e. it will look like an inductor of inductance $CR^2$. As high quality inductors are difficult to obtain, bulky, and nearly impossible to fabricate on integrated circuits, inductors are often replaced with gyrators, particularly in filters and resonant circuits. 

Construct the resonant (fake $L$)$C$ circuit below. Use matched 1% precision resistors: all three $330\,\Omega$ resistors must match, but the pair of $2.21\,\mathrm{k}\Omega$ resistors in the top NIC need not match the pair of $2.21\,\mathrm{k}\Omega$ resistors in the bottom NIC. This is one of the most touchy circuits that you will build this semester; the better matched the resistors, the better the circuit will work. Note that 47k resistor is used to isolate the wavefunction generator from the rest of the circuit. 

Drive the circuit with a sine wave in the neighborhood of $5\,\mathrm{kHz}$. What is the circuit’s resonant frequency?  What is its calculated resonant frequency? Adjust the variable resistor to maximize the $Q$.  

Change the $0.1\,\mu\mathrm{F}$ capacitor C1 in the RLC circuit to a $1\,\mu\mathrm{F}$ capacitor. Does the resonant frequency change appropriately? 

Build the relaxation oscillator at right.  Save images of the signal at both the op amp output and across the capacitor. How does the circuit work?  (Hint‑Think about the hysteretic comparator circuit.) What is the oscillator’s output frequency? How does it compare to theory?

An easy way to decouple the stages in a multistage filter is to use a follower. Build the circuit at right.  It consists of an initial $RC$ filter, a follower, and an identical $RC$filter.  The circuit has two outputs: one, which gives the signal after only the first $RC$filter, and the second which includes both $RC$ filters. Feed the outputs into channels 2 and 3 on  your scope.  (Send the input signal into channel 1.)  Set up auxiliary triggering.  Casually explore the behavior of this circuit near its $3\,\mathrm{dB}$ frequency of about $160\,\mathrm{Hz}$.  You should observe that above the passband, the 2-pole output is more strongly attenuated than the 1-pole output. (Multistage filters are more commonly called multi pole filters.  This terminology comes from the complex analysis of the filter response; every stage will contribute a pole in the complex plane.)

Note that the $10\,\Omega$ resistor reduces the effects of the changing input impedance of the filter, but does not otherwise affect the performance of the filter.

Do not disassemble your circuit, it will be used again in this exercise.

With a more sophisticated topology, it is possible to construct a filter with a sharper cutoff than the two stage $RC$ filter above.  Construct the 2-pole Chebyshev Sallen-Key filter at right.  (Sallen-Key refers to the topology of the components. Chebyshev refers to the the particular optimization used to select the values of the components; there are several common optimizations, all having somewhat different properties.)  

Feed the output of this circuit to channel 4 on your scope. Drive the input of this filter with the same input that you are using to drive the RC filter; you should be able to see all four output traces on the screen at the same time.

Casually explore the behavior of this circuit near its $3\,\mathrm{dB}$ frequency, which should also be near $160\,\mathrm{Hz}$. 

Filters, as well as many other circuits, are frequency dependent, and it can be painfully slow to explore their behavior.  The signal generator has a frequency sweep mode that can greatly speed up such measurements.  Normally, the signal generator outputs a single frequency.  In sweep mode, the signal generator scans through a preset range of frequencies.  Watch the video at right to understand how to set up a frequency sweep.

• At very low frequencies, all the the filters pass the input signal without attenuation.
• At high frequencies, the 1-pole $RC$ signal is substantially greater than the 2-pole $RC$ signal; the stop band attenuation is not as large.
• Also at high frequencies, the 2-pole $RC$ signal has approximately the same amplitude as the 2-pole Chebyshev filter.  In general, far from their $3\,\mathrm{dB}$ frequencies, all filters of the same order (same number of poles) behave approximately the same.
• The transition band, also called the skirt, is the band of frequencies near the$3\,\mathrm{dB}$ frequency. The Chebyshev filter has the steepest skirt, followed by the 2-pole $RC$ filter.  The 1-pole  filter has a broad skirt.

From the frequency response curves, you might conclude that the Chebyshev filter, having the steepest skirt, is always the best filter.  However, the Chebyshev filter has some imperfections.  For instance, as shown in the spice simulation at right, the Chebyshev filter has a ripple in its passband; just before it begins to attenuate, the signal actually increases by about 5%.  The $RC$ filters do not have a ripple.

Measure the approximate size of the Chebyshev ripple.  Since the ripple is small, this is difficult to do precisely.  It may be easiest to accomplish by changing your signal generator to output a single frequency, and setup an amplitude measurement on the Chebyshev channel.  Manually scan the signal generator frequency until you find the peak near $100\,\mathrm{Hz}$.  Then use averaging to measure this peak.  Also measure the signal at $10\,\mathrm{Hz}$ for comparison.  Note that the size of the ripple depends on the actual values of the resistors and capacitors in your circuit, and may not agree precisely with computer simulations.

Sometimes, these sorts of passband ripples are acceptable, and a Chebyshev filter (which will always have ripples) is acceptable.  Other times, ripples are not acceptable, and you should use an $RC$ filter (which will never have a ripple), or another type of rippleless filter, like a Butterworth filter.

So far, we have only considered the frequency response of the filters.  We should also consider the transient response of the filters, i.e. the response in time to an abrupt change to the input signal.

Set your signal generator to output a $12\,\mathrm{Hz}$ square wave.  Display the square wave on channel 1 of your scope. You should see traces like those at right.

Notice that while the $RC$ filters asymptote to their final values smoothly, the Chebyshev filter overshoots.  Measure the size of the overshoot.

Sometimes, one wants to optimize a filter to most faithfully reproduce an input transient.  Yet another type of common filter, a Bessel filter, does this best. All these filters [Butterworth, Chebyshev, Bessel (there are others)], as well as low pass, high pass, bandpass, and bandreject filters, can be built using the Sallen-Key typology by adjusting the component values and swapping the placement of the resistors and capacitors.

The first module in the RMS converter takes the absolute value of the signal. Construct the circuit at right. 

Drive the circuit with a variety of signals, amplitudes, and offsets from the waveform generator. Save some illustrative images.  How well does the circuit work?  

Disassemble the second op amp and its $1\,\mathrm{k}\Omega$ feedback resistor in this module, leaving the first op amp and its associated feedback network in place for the RMS converter.

Next, we need to square the signal after taking its absolute value. This operation is performed by taking the log of the signal, multiplying it by two, and then exponentiating.  

Build the circuit at right, which takes advantage of the exponential dependence of diode current on voltage,

$\displaystyle i(V)\approx i_\mathrm{sat}\exp\left( \frac{eV}{nkT}\right)$

to take the logarithm of its input signal.

As with all the remaining circuits, a good initial debugging signal is an offset triangular wave. Note that this circuit expects a negative signal; your signal generator can be easily set to make an entirely negative signal with the High Level and Low Level options on the Amplitude menu.

Configure the signal generator to output a DC voltage by putting it on the More->More->DC setting.  Drive the circuit with a series of DC input voltages, and record the input and output voltages at each point. Plot your data; how close is it to a true natural logarithm?  Do you have to scale and offset the data to make it match a logarithm?

Build the circuit at right, which multiplies the logarithm of the initial signal by negative two.  The circuit also offsets the signal by a constant $V_\mathrm{offset} $.

In Exercise 7.6, you should have found that the signal needs to be scaled and offset.  Adding in $V_\mathrm{offset} $ accomplishes these shifts such that the output of the squaring section of the RMS converter is equal, not merely proportional, to its input.

Rather than trying to estimate the proper shift, we can use a circuit to calculate the proper value. It turns out that the proper value for the offset is whatever the circuit in 7.6 would output for a $-1\,\mathrm{V}$ input.

Build the circuit at right to calculate the offset.  The $330\,\Omega$ and $200\,\Omega$ resistors are the only the resistors in the RMS converter that should not be 1% resistors.

The voltage across the Zener diode is nominally $6.2\,\mathrm{V}$, however there will be some variation between Zeners, and this voltage will vary. 

The divider network which immediately follows should reduce the Zener voltage such that the output of the first op amp is $-1\,\mathrm{V}$; if it is not $-1\,\mathrm{V}$, adjust the value of the $200\,\Omega$ resistor to make it $-1\,\mathrm{V}$. You may have to change the location of this resistor to the bottom of the divider.

You should find that the output of this circuit is approximately $+0.6\,\mathrm{V}$.

Drive the multiplier with the log output, and check that it doubles and offsets its input. You do not need to record anything.

Note that the Zener diode and the 1N4448 diodes are almost identical in appearance.  Make sure that you keep track of which is which; you will become hopelessly confused if you swap them. 

To exponentiate the signal, build the circuit below.

Use an offset triangle wave to debug your circuit. (The circuit requires a negative input.) Then drive the circuit with a series of DC input voltages, and record the input and output voltage at each point.  Plot your data; how close is it to a true exponential?

Finish the squaring module by hooking all the circuits together, as shown below.

Drive the circuit with a series of DC input voltages, and record the input and output voltages at each point. Plot your data; how well does the module square its input signal? It should be accurate to better than $\pm10\%$.  What is the largest input signal accepted by the module before it saturates?

Remove the input resistor R1, but keep the rest of the circuit.

Build the integrating averager at right.

Drive the circuit with a variety of waveforms, and show that its output is the inverse average of its input. Over what frequency range does the circuit work? What sets this range?

Finally, we need to take the square root of the averager signal. The required circuit is very similar to the squaring circuit.

Build the circuit at right. Debug each section independently. Use the offset voltage from 7.7.

Drive the circuit with a series of DC input voltages, and record the input and output voltage at each point. Plot your data; how well does the module take the square root of its input signal?

The behavior of the absolute value circuit is easiest to understand by considering negative and positive input signals separately.  Begin by recalling that the Golden Rules insist that the $V_-$ inputs to both op amps are held at virtual grounds; this is true regardless of the sign of the input signal.  

With a negative input signal, we would expect the output of the first op amp to be positive.  Then, the grayed out diode shown in the absolute value circuit at right is reversed biased and will carry no current; we will ignore it henceforth.  How much current will be be carried by the $500\,\Omega$ resistor?  What is the total current entering the $V_-$ node of the second op amp?  What output voltage on the second op amp  is required to carry this current away?