Objectives:

1. To understand how to design and build a basic active low-pass filter circuit using an op-amp as the active element.

2. To determine the effects of adding a capacitor in parallel with the feedback resistor in an inverting amplifier configuration.

3. To observe the effects of phase shift that the low-pass filter exhibits on the output of the circuit at various frequencies.

Parts and Equipment:

Resistors: 1 K, 3.9 K
Capacitor: 3 - 0.1 µF
Operational Amplifier: LM348 or equivalent
    Pin Connections
Signal Generator
Bi-Polar D.C. Power Supply (±15 VDC)
Oscilloscope
Digital Multi meter
Capacitance on Multimeter or LCR meter
Breadboard and wire

Textbook Reference:

Hambley, Allan R.,  "Electronics", Prentice Hall, 2nd Ed. 2000, pp. 61-120, 727-734.
Serda and Smith, "Microelectronic Circuits", Oxford University Press, 4th Ed. 1998, pp. 58-108, 884-917.

Material Review:

The circuit used in this experiment will pass low frequency signals on to its output with a gain A_v, but will attenuate signals of higher frequencies. The frequency at which the gain starts to decrease is controlled by the product of C and R_f. This type of circuit is usually configured as an inverting amplifier and is referred to as an active low pass filter. The circuit used in this lab is the simplest form of an active low pass filter.

Where

Letting V_x go to 0 according to the ideal op-amp assumption gives

Rearranging gives

Solving for V_out

Then Solving for the voltage gain A_V gives

As w or f goes to 0 this gives the DC gain as

When w = w_H of f = f_H  the magnitude of the voltage gain drops to one over the square root of two of the DC or low frequency value.  This is the high cutoff frequency or half power frequency.  It depends only on the values of the feedback resistor and the capacitor giving

The phase angle of (1 + jwRC) = tan^-1(wRC).

The phase angle of A_V = ±180º - tan^-1(wRC).  Use +180º to match the way Micro-Cap calculates the phase angles.

The magnitude of the gain stays near the DC value until the radian frequency becomes appreciable compared to 1/(R_fC) and then it begins to drop off as the frequency increases approaching zero as omega approaches infinity. The phase angle of the output starts at +180^o near zero frequency and decreases to +90^o as the frequency approaches infinity. This phase range could also be given as -180^o to -270^o.  The frequency at which the magnitude of the gain drops to 1 over the square root of 2 of its D.C. value is called the cutoff frequency or the half power frequency. This frequency may be found by setting

or 2 = 1 + (wR_fC)²  which gives (wR_fC)²  = 1   or wR_fC = 1.  Solving for w gives w_c = 1/R_fC for the radian cutoff frequency.  Since w = 2pf , the cutoff frequency in Hz is f_c = 1/(2pR_fC)

Note that the phase angle at this frequency will be:

Procedure 1: Inverting Amplifier configuration

1. Using power supply voltages of ±15 VDC for the op-amp, construct an inverting amplifier circuit with a gain of -3.9 using an input resistor of 1 K Ohms. Install the 0.1 µF capacitor in parallel with the feedback resistor as seen in Fig. 1. Calculate the cutoff frequency (f_c) for the circuit using the measured values of the components.  It should be around 400 Hz.  Add two extra 0.1 µF capacitors to the circuit. One should be connected between the + DC supply(pin 4) and ground the other should be connected between the - DC supply(pin 11) and ground. These capacitors are to help prevent oscillation in the amplifier circuit due to interaction between the circuit and the power supply. They should be placed as close to the Op-Amp itself as physically possible.  Make sure that the circuit is correctly connected before turning on the power supply voltages.  Failure to do so may cause the op-amp to saturate and in some cases cause permanent damage to the op-amp.

2. Set the signal input, v_in, to zero.  That is replace the signal source, v_in, with a short circuit to ground. Carefully measure the DC output voltage.  Make sure you record the proper sign,  It should be between +50 mV and -50 mV, usually very small.  This output with no input is called the output offset voltage.  It is an error in the output of the circuit.  It can be treated as an equivalent input offset voltage applied to the non-inverting input of the op-amp.  The equivalent input offset is calculated by dividing the measured output offset by the gain of the amplifier from the non-inverting input, A_v = (1 + R_f /R_in).  This offset has no effect on the ac operation of the circuit, but can cause errors in dc measurements of small voltages.

3. Use a signal input voltage, v_in, of 0.1 VDC and connect it to the amplifier signal input as V_in.  Using a digital Multi-meter, measure and record both V_in and V_out as accurately as possible.  Calculate the DC voltage gain both with and without correcting the output voltage by subtracting the output offset voltage measured in step 2 from the measured output voltage.  Be sure to use the correct sign on the offset voltage.  Repeat this measurement and calculation with V_in = 1.0 VDC.  Which gain calculation would you expect to be more accurate?  Why?

4. With an oscilloscope connected to both the signal input and the output, apply an A.C. signal such that the output voltage has an amplitude between 5V and 10V.  Then measure V_in, V_out, T, and Dt(for phase measurement) at your calculated cutoff frequency and at each of the following frequencies: 20Hz, 50Hz, 100Hz, 200Hz, 500Hz,  1kHz, 2kHz, 5kHz and 10kHz, 20kHz, 50kHz.  Print a copy of the waveforms at 20 Hz and 10 kHz, and at your calculated cutoff frequency.  As the gain starts to drop increase the input voltage trying to keep the output voltage amplitude between 5V and 10V until you reach the maximum output of the signal generator.  Also take these measurements at your calculated cutoff frequency.   If the output waveform starts to look like a triangular wave instead of a sine wave your amplifier has reached the slew rate limit and you will have to reduce the input voltage until this effect is eliminated to get accurate gain measurements.  These voltages can be printed from Wavestar or read from the measure mode of the oscilloscope.  Use the cycle AC rms values.   For the voltage measurements remember to keep from 2 to 5 cycles of the waveform on screen with the vertical scale adjusted for maximum on scale waveform for each frequency to maximize the accuracy of the voltage measurements.  When measure the time Dt between the rising edge zero crossing of v_in and the rising edge zero crossing of v_out for use along with the period, T,  from Wavestar in calculating the phase shift.  Remember that to maximize the accuracy in measuring Dt you must expand the time scale so the distance between the zero crossings is at least half the width of the screen.  Also you must check to see that both channels are set for zero divisions vertical displacement,  Use ac Coupling on both channels to eliminate any effects of DC voltages that might be present.  Increase the vertical gain on both channels so the slope at the zero crossings will be steep enough to easily determine the exact position of crossing zero.

5.  Calculate the AC voltage gain and phase shift of the circuit at each frequency.

6.  Set your signal generator to square wave output at 100 Hz with an amplitude of 1 V.  With this input observe and record the output waveform.  Repeat at a frequency of 500 Hz.

CALCULATIONS / GRAPHS:

1. Using a spreadsheet plot a graph of the gain (dB) vs. frequency response. Set the x-axis scale to log instead of linear for the frequency. This can be done in Excell by selecting the x-axis and then right clicking on it.  Then click on Format Axis in the menu that appears.  Next click on the scale tab in the Format Axis menu.  Finally check the logarithmic scale box.  Set the lower limit on the x-axis to 10Hz or 0.01kHz.  The minor grid lines can be turned on by Right clicking inside the chart area, then clicking on Chart Options, then select the Gridlines tab and check the gridlines to be turned on.  The gain in dB is calculated: Gain (dB) = 20 log(V_o/V_in ) = 20 logA_V. Determine the slope in dB per decade (of frequency) in the capacitive "roll-off" region (the straight line portion above the cutoff frequency).  Do this by drawing a straight line tangent to the curve at the highest frequency point and extend the line to cover one full decade (5kHz to 50kHZ) of frequency.  Then determine the difference in gain across the frequency decade for this line. 

2. Plot a graph of phase vs. frequency response with the frequency on a log scale.

3. From both graphs, determine the half-power frequency.  On the gain graph use a straight edge and draw a horizontal line 3 dB below the level of the low frequency gain.  The point where this line intersects the gain curve is the upper 3 dB or half power frequency.   On the phase graph draw the horizontal line at 135°.  Do the two values agree?   Show  by drawing lines and labeling them on your graph how you determined the cutoff frequency.

4. Run a Micro-Cap simulation of this circuit using the actual measured values of all components.  Run an AC Analysis using a frequency range from 100kHz to 10Hz.  Set the Frequency step to log and the number of points to 201.  Include a copy of your Micro-Cap circuit and the printout of the ac analysis.  Save the numeric output to a disk file, import it to your spreadsheet, and plot the gain on the same graph as your experimental gain data for comparison do the same with the phase. Plot the Micro-Cap data as a line with no symbols. Plot the experimental data as symbols with no line, choosing a symbol like an x or an o.

5.  Run a Micro-Cap Transient analysis showing the input and output voltages.  Use an input frequency of 1 kHz with an amplitude of 1 V.  Plot the first 2 ms of the response using a maximum time step of 0.01 ms.  Repeat with an amplitude of 10 V.  What is different about the results of the two runs?

6.  Can you explain the results of  step 6 in the procedure, given that a square wave can be represented by the following Fourier series?

or

Exercises/Summary:

1. For a similar circuit with the capacitor replaced by an ideal inductor of 500 mH, derive the gain versus frequency expression. Use the same procedure as in the material review section of this handout.

What is the cutoff frequency for this circuit?

What type of filter will this be?

What might be a useful application?

2.  What changes in operation would be produced if the original circuit were changed to a non inverting amplifier circuit.  That is if the input voltage source was moved to the non-inverting input of the op-amp  and the ground connection moved to R_in.  How would this effect the gain and frequency response?  Try a Micro-Cap simulation of this circuit.  Derive the gain expression from the circuit.

3. Describe why frequency response curves (Bode Plots) are commonly plotted as Gain (dB) vs. log (f); that is give advantages as to why these curves are plotted this way. Look up Bode plot in the index of your text book.

4.  What is different about the first cycle of the output waveform on the transient analysis?  Can you think of a reason for this difference?

Conclusions:

Drawn from the results of this experiment.

This page last updated 18 September 2007