This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

Op Amps Part 2

Learning Objectives: 1. Examine additional operational amplifier applications. 2. Examine filter transfer functions.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics, this Dr Robinson.

Â In this lesson we're going to look at a filtering demonstration where we make

Â measurements using test equipment on an actual circuit.

Â In your previous lesson you were introduced to second order filter

Â circuits.

Â And our objectives for this lesson are to examine frequency spectre of signals and

Â to demonstrate filtering by a second order filter circuit.

Â A frequency spectrum is a representation of a time domain waveform

Â on voltage versus frequency access.

Â Here I'm showing both a one kilohertz,

Â one volt amplitude sine wave in the time domain.

Â And it's representation in the frequency domain or its frequency spectrum.

Â This is voltage versus time graph.

Â This is a voltage versus frequency graph.

Â This sine wave with amplitude 1 and

Â frequency 1 kilohertz is represented here on this spectrum.

Â By a single spike at a frequency of 1 kilohertz having amplitude of 1 volt.

Â Any single spike on a frequency spectrum represents a sine wave in the time domain.

Â Now let's look at a somewhat more complicated waveform.

Â Here, I'm adding two sine waves together.

Â A 1 kilohertz sine wave, shown in blue,

Â and a 2 kilohertz sine wave shown in green.

Â The red curve is the summation of the blue curve and the green curve.

Â So at any time, we can obtain the red curve by adding

Â the 1 kilohertz sine wave amplitude and the 2 kilohertz sine wave amplitude.

Â Here I'm showing the spectrum of the sum of the two sine waves, or

Â the spectrum of this red curve.

Â And you can see, as you might expect, it's composed of two sinusoidal waveforms.

Â One at 1 kilohertz with amplitude 1, and one at 2 kilohertz with amplitude 1.

Â The spectrum is indicating that this

Â somewhat complicated waveform is obtained by adding two sine waves together.

Â One at 1 kilohertz, one at 2 kilohertz.

Â Here I'm showing both a square wave in the time domain and

Â a partial representation of its frequency spectrum.

Â The react should be an infinite number of these sinusoidal components.

Â And must be combined together to give us this square wave in the time domain.

Â Now this component here at 1 kilohertz is known

Â as the fundamental at n=1, this is the third harmonic.

Â This is the fifth harmonic, this is the seventh harmonic.

Â This sine wave occurs at a currency of 3 kilohertz or

Â three times the frequency of the fundamental at 1 kilohertz hence, n=3.

Â This is occurring at 5 kilohertz, this sound wave at 7 kilohertz etc.

Â Now if we were to say have five function generators,

Â one function generator generating a 1 kilohertz sine wave with amplitude 1 volt.

Â Another generating a 3 kilohertz sine wave with this amplitude.

Â Another at 5 kilohertz, 7 kilohertz, 9 kilohertz etc.

Â And we took the output from each one of those function generators and

Â added them with, say, an opamp summing circuit.

Â Then the output would approximate this square wave.

Â We would need many,

Â many more components to make the approximation a better approximation.

Â Here I'm showing how increasing the number of sine waves to generate

Â the approximation affects the shape of the square wave in the time domain.

Â So, for example,

Â this red curve here, I'm adding this sine wave plus this sine wave to generate it.

Â And you can see that just two sine waves added together

Â are already a reasonable approximation to our square wave.

Â The green curve, I'm adding the first four spectral components,

Â the fundamental plus the third, fifth and seventh harmonic.

Â And then the blue curve, a better approximation,

Â I'm adding the first six spectral components.

Â But you can see that a square wave can be thought of

Â as a sum of individual sinusoidal components.

Â This square wave is composed of these sinusoids added together.

Â We are now shown the circuit schematic of an opamp circuit known as a relaxation

Â oscillator.

Â This circuit produces an output with no signal input.

Â The only input to this circuit is a DC voltage used to

Â supply power to this opamp.

Â When the power supply is turned on at the output, a square wave will be produced.

Â That oscillates between the V plus rail In the V minus rail of the opamp.

Â The frequency of the square wave is related to the time constant set by this

Â R and this C.

Â It's a simple circuit to build and

Â actually you could experimentally determine values here to produce

Â the square wave of the frequency that you desire.

Â And the square wave could be used say as the input to a MOSFET switch,

Â turning something on off on off.

Â But say instead of a square wave you wanted to generate a sinusoidal

Â output voltage.

Â Remember the square wave, we can think of it in terms of its spectrum,

Â where the square wave is composed of individual sine wave components.

Â Then say that we apply the square wave to the input of a band pass

Â filter where we filter out using an ideal band pass filter this component.

Â And eliminate these additional spectral components.

Â Well then at the output of the band pass filter,

Â because the spectrum now consist of a single spike on the frequency spectrum.

Â We know that it would be a sinusoidal wave form.

Â So, by band pass filtering a square wave, we can generate a sine wave.

Â If we band pass filtered out this component, we would generate a sine wave

Â at 3 kilohertz, rather than at the fundamental of 1 kilohertz.

Â If we filtered this one, we could generate a 5 kilohertz sine wave.

Â Now let's look at the measurements we're going to make.

Â I've built a relaxation oscillator that generates a 1 kilohertz square wave.

Â We're going to apply that square wave to the input of a Sallen-Key band pass

Â filter.

Â Having a center frequency of 1 kilohertz and

Â a Q equals 5 design using techniques we talked about in our previous lesson.

Â We'll look at the frequency spectrum and the signal and time domain here.

Â And we'll also look at the spectrum and

Â time domain waveform here to see how the filter affects the input square wave.

Â We're then going to cascade this filter with another identical filter to improve

Â the filtering and then examine the output waveform.

Â We'll then cascade a third Sallen-Key band pass filter.

Â And look at the output wave form after its been filtered through three band pass

Â filters and compare it to the input square wave.

Â Now as we make this measurements, we're going to look at a figure of merit known

Â as the Total Harmonic Distortion or THD.

Â THD is a numerical indicator of the amount of distortion present in a sine wave.

Â A pure sine wave, having a frequency spectrum that looks like this,

Â a single spike, would have a THD equal to zero.

Â In the formula here, v2 is the voltage of the second harmonic,

Â v3 is the voltage of the third harmonic, fourth harmonic, etc.

Â A pure sine wave at n = 1 here would have no upper harmonics.

Â The numerator would be zero, and its THD would be zero.

Â Indicating a that it's a pure sine wave with no distortion.

Â Now in this application we're talking about here,

Â the conversion of a square wave to a sine wave.

Â We can consider that square wave to be a highly distorted 1 kilohertz sine wave.

Â So here we have an n = 3 component, an n = 5 component an n = 7 component, etc.

Â So our THD from this formula would be non-zero.

Â Now as we filter it, filter this component through one band pass filter

Â through a second band pass filter, and through a third band pass filter.

Â We would expect these components to go to 0.

Â And the THD of our waveform would approach 0%.

Â Ideally approaching 0% if we were able to eliminate all of these

Â upper harmonics from the square wave.

Â Now let's look at some measurements.

Â So here's the constructed circuit.

Â This portion here is the relaxation oscillator its output is

Â applied to a potentiometer so that I can adjust the level of the square wave.

Â I have a unity gain buffer here that isolates the output of the relaxation

Â oscillator from this cascade of three Sallen-Key band pass filters.

Â Let's first use the NI Elvis Bode analyzer to generate a Bode magnitude plot for

Â one of the Sallen-Key band pass filters

Â As the analyzer runs, it applies sine waves to the input of the filter.

Â Measures the output sine wave and

Â then calculates the gain at each of those frequencies.

Â You can see that the Bode magnitude plot

Â is a band pass filter centered at 1 kilohertz.

Â And if we measured the quality factor of this band pass filter,

Â it would be close to a q equal to five.

Â And let's examine the output of the relaxation oscillator

Â both on the oscilloscope and

Â using the dynamic signal analyzer to display the frequency spectrum.

Â Click run, and there's the output of the relaxation oscillator.

Â It's approximately a 1 kilohertz square wave with a peak to peak amplitude of 20.9

Â volts or so.

Â Now on the dynamic signal analyzer.

Â There is its frequency spectrum.

Â So there's the fundamental component at 1 kilohertz, like we'd expect.

Â And here are the upper level harmonics at 3 kilohertz, 5 kilohertz, 7 kilohertz,

Â 9 kilohertz.

Â I've limited the spectrum to just DC to 10 kilohertz.

Â Now I'm going to take that relaxation oscillator output and

Â apply it to the input of one of our band pass filters.

Â And then we are going to examine the output of the band pass filter.

Â Now let's monitor the output of the relaxation oscillator on

Â channel 0 on the oscilloscope.

Â And look at the output of the band pass filter on channel 1.

Â So in green is the input to the band pass filter and the output is in blue.

Â And you can see that just by filtering through this single band pass filter

Â the second order band pass filter centered at 1 kilohertz.

Â The output is already approaching that of a sine wave.

Â If we look at the dynamic signal analyzer to look at the spectrum

Â of the output signal, let me change the source channel

Â to 1, And let it finish averaging.

Â You can see that on this scale already the fundamental component is much,

Â much larger than the distortion components.

Â And the distortion components are essentially 0 on this scale.

Â If I change the.

Â Scale, so that we have.

Â There we go, that's a reasonable scale.

Â From 0 to 1, you can see that there's still the 3 kilohertz component is visible

Â on the scale and the 5 kilohertz, there's a small blip there.

Â Now let's take this output, the output of this band pass filter and

Â apply it to another bandpass filter to see how the waveform is affected.

Â So, there's a cascade of two band pass filters.

Â You can see the output is essentially a sine wave with the square wave input.

Â So, we've cascaded two band pass filters second order at 1 kilohertz.

Â And applied a square wave to the input to the cascade,

Â the output looks like this sine wave.

Â Let me increase the gain a little bit.

Â And adjust the time base a little bit, there we go.

Â So the input square wave, output sine wave, and on the frequency spectrum,

Â You can see that essentially you have only the fundamental component at 1 kilohertz.

Â The THD after this filter is 0.89%, which indicates a pretty clean sine wave.

Â If we go back and look at the output after the first filter.

Â With these additional components here, let's wait for the averaging to stop.

Â And you can see that we have a THD at 6.5% which is still a reasonably clean

Â sine wave as you can see in the time domain waveform.

Â So after one filter we have a THD of 6% after two filters,

Â we have a THD of approximately 0.88%.

Â And let's add in the third filter to see what it's effect is.

Â Go back to the oscilloscope, hit run.

Â And change the scale, there we go.

Â So input square wave, this is our output after three band pass filters.

Â And on the dynamic signal analyzer.

Â We have a THD of 0.12 or essentially a very clean,

Â pure sine wave after filtering three times.

Â Now we've looked at various wave forms both before and after filtering.

Â In both the time domain and the frequency domain.

Â Now I wanted to do one more test for you.

Â I wanted to play this 1 kilohertz tone for you.

Â Both of the output of the relaxation oscillator,

Â a square wave with all the harmonic content.

Â And I was going to play for you the 1 kilohertz tone after filtering.

Â To see if you can hear the difference that the harmonic content makes.

Â So first let's listen to the output of the relaxation oscillator, the square wave.

Â [SOUND] Do you hear the harshness of that?

Â Compared with, here's the output of the third band pass filter.

Â [SOUND] Essentially a pure sine wave.

Â So there's our pure 1 kilohertz tone.

Â And we can compare that with [SOUND] our 1 kilohertz

Â square wave tone that contains all of the harmonic content.

Â So one more time.

Â [SOUND] A pure sine wave.

Â [SOUND] Harmonious, melodic,

Â you can listen to it all day and

Â a 1 kilohertz square wave,

Â a harsh, grating tone.

Â So in summary, during this lesson we introduced frequency spectra.

Â And we looked at how a real circuit could be used to extract a sinusoidal component

Â from a square wave input.

Â So thank you and until next time.

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