Session 9 - Friday, February 27

An AC microphone amplifier

Microphone + amplifier + speaker = fun! :D
In this lab we used all the knowledge we've gained thus far in the semester to build a circuit that takes in a signal from a microphone, amplifies that signal, and then feeds it to a speaker.

Setting the Gain

To get us started we were given the diagram above. We were then tasked with designing CKT #1 so that there would be a reasonable voltage gain and reasonable was defined as having the output voltage be in the range 0V to 3.5V. 

Given that the input voltage coming from the microphone will be approximately 15mV we need to create CKT #1 so that the op amp will have a gain of roughly 100 in order to put our output voltage at 1.5V which is within the reasonable range described in the previous paragraph. To accomplish this we chose to make CKT #1 a resistor with a value of 10kΩ. With a 10kΩ resistor the ratio of CKT #1 to the 1MΩ resistor in the formula Vout = (1+ R1/R2)*Vin gives us a gain of 101 which is approximately 100.

We then built this circuit with a 10kΩ resistor in the place of CKT #1 and drove it with a 1kHz sine wave at Vpp = 20mV and no DC offset in order to test that it works as expected. As expected from the lab write-up the output is clipped on the low-voltage (negative voltage) side since the input is symmetric about zero due to being a sine wave and the output cannot go below the value of ground which is 0V. Also we found that the ration of input to output amplitudes in the unclipped regions was approximately 100 as we expected as can be seen in the photo below:

 The scale of the yellow is 500mV while the scale of the blue is 5mV showing that there is in fact a 100x increase.

Next we increased the input amplitude gradually in steps of 5mV and watched what happened to the output. When we reached input values above 4V the output was clipped at roughly 4V. This makes sense because the maximum output voltage is approximately 1.5V below the positive supply which is +5V. You can see this clipping in the image below:

Volume Control

The next step in our mission to build an awesome microphone to speaker circuit is to have a volume control knob. First, we needed some sound output so we connected a buzzer as our speaker. The buzzer had one pin grounded and the other connected to the amplifier output. The presence of the buzzer in the circuit changed the output signal as can be seen in the photo below: 

Now that we had a volume output, we modified our CKT #1 to give us control over the volume. We did this by adding a potentiometer in series with the resistor that was previously all of CKT#1 with the potentiometer's other side being connected to ground. Since the potentiometer has a resistance of its own ranging from 0Ω to 10kΩ, we swapped the 10kΩ resistor that was previously in CKT#1 with a 5kΩ resistor. This will give our amplifier a gain between 200x when potentiometer is set to 0Ω and 67x when potentiometer is at 10kΩ. Gains were calculated from the ratio of 1MΩ and the resistance of CKT #1.

Avoiding Clipping
Now we must change the circuit so that the signal doesn't get clipped below 0V in order to "have room" for the voltage swings of the audio signal. To do this we will make the signal sit on a DC offset. The lab write-up tells us this offset will be 1.75V, however when we drive the circuit with a 20mVpp 1kHz sine wave from the function generator and gradually raise the DC offset by 10mV at a time we find that the output saturates and hits a rail at only a DC offset of approximately 70mV. This is because the DC offset is being routed through the op amp and also receiving a gain of 100.

To prevent the DC offset portion of the signal from being amplified we want the op amp to act as a follower at DC, but an amplifier at AC. This is what a high-pass filter does, so we accomplished this by further modifying CKT#1 to include a high-pass filter. We added a capacitor between the 5kΩ resistor and the rest of the circuit. To select the value of the capacitor we used the formula 1/(w*C) = Zcap < 5kΩ and the fact that we need 20Hz to pass through the amplifier to calculate the capacitance from the following: 1/(2*pi*20*C) = 5000Ω. We end up with C = 1.5 microFarads and since we didn't have a 1.6 microFarad capacitor we used one 1.2 microFarad capacitor and three 0.1 microFarad capacitors in parallel to achieve that value. You can see the resulting circuit (and the several capacitors) in the photo below:

We confirmed that the DC offset at the output is equal to the DC offset at the input while the AC signal is amplified and you can see this in the photo below: 

However, the amplifier does not have the same gain at all frequencies and thus the amplifier has distortion. The change in gain between 20Hz and 20kHz was 2.16V to 780mV and thus we know that the amplifier is distorted by a factor of 3.

Adding a Microphone

Now that we have a circuit that will take in a signal and output a volume controlled audio output through the speaker that is not clipped and has "space" to swing as needed, we need to change the circuit so that the input signal comes from the microphone instead of the function generator. We added the microphone to a separate section of the breadboard and set it up as shown in the photo below:

We confirmed that the microphone worked by connecting the Vout of the circuit shown above to the oscilloscope. We could see amplitude changes on the oscilloscope when we spoke into the microphone, but as we have no clue what the signal pattern of our voice should look like we also tested it with a tuning fork. Tuning forks produce a sine wave signal and you can see that test in the video below:

It's important to note that even though the signal in the video is a sine wave it did not come from the signal generator, but from the microphone. We just wanted to make sure everything worked as expected and thus used a tuning fork so that we would know what the output should look like (a sine wave). We tried several different tuning forks and we also raised and lowered our voices in pitch while talking in to the microphone and the higher the pitch the faster/higher the frequency of the signal output. 

Connecting the Microphone to Amplifier

Now that we have confirmed that the microphone is working as expected, we need to connected to the amplifier and speaker. However, we cannot just directly them because the microphone's output is a +5V DC signal with a very small AC signal of roughly 20mV on top. This DC signal is too much for our amplifier so we need to remove it from the microphone signal before it reaches the amplifier. However, as we know from the "Avoiding Clipping" section of this lab, we do need some DC offset, specifically 1.75V. To accomplish this we need to design another circuit, CKT#2, to connect the microphone to the rest of the circuit via the (+) input of the amplifier.

First, in order to cut the +5V DC offset off of the microphone signal and isolate the AC signal we need to use a high-pass filter. Then to add back in a 1.75V portion of a DC voltage we need to use a voltage divider made of two resistors. We can combine the high-pass filter and the voltage divider so that the the bottom resistor of both is the same. A sketch of this circuit can be seen below:

Now we just need to determine the values of the components in the sketch above. We will use a +5V voltage source since we have the option of 5 or 12V. The two resistors need to have a ratio of 0.65 in order to pass 1.75V from the 5V voltage source and we want them to be very high in order to have lower power dissipation (P = V^2/R). Thus we chose the values of 1MΩ and 560kΩ (closest to 538kΩ which is the calculated value) for the resistors. Now for the capacitor, we know we want our high-pass filter to pass 20Hz so we use the formula 1/(2*pi*R*20) = C to get a capacitance value of 0.113 microFarads which we rounded to 0.1 microFarads.

Now that all of our components have values, we build the circuit! We lengthened the distance between the microphone and speaker using more of the cables to connect the pins of the microphone and the speaker to the correct spots on the bread board. The circuit worked!!! And it was very fun to play with :).

Unfortunately I don't have any good videos of the microphone and speaker working because there was so much ambient noise in the room and the microphone on my phone isn't very good.

During my playing with the speaker though, I noticed that high frequencies came through more clearly and more loudly than low frequencies. This was especially noticeable when I used the child's xylophone that was in the blue cabinet to produce an octave of frequencies. I think the clarity of the high tones versus the low tones has to do with the properties of our high pass-filter, but it might also be because of the properties of the buzzer itself or even human hearing which is logarithmic. It would be very interesting to explore this further and see what is the cause and how best to compensate. I wonder if this is why speakers for phones and smaller devices often sound "tinny" and don't produce bass notes well.

Session 8 - Tuesday, February 24

Op Amp Current Source

If you have a variable resistor as R2 in the configuration shown below:

We can apply the Op Amp Golden Rules and Ohm's Law to show the following:

1. V- = V+ = Vin
2. IR1 = V- / R1 = Vin / R1
3. IR2 = IR1 = Vin / R1

Thus we know that the current flowing through R2 is independent of the value of R2 over a wide range of resistances and therefore this circuit is an excellent current source.

Op Amp Current Source, experimentally

Student Manual Section 8-5

In this lab we create an op amp current source by connecting a 15kΩ resistor from a +12V voltage source to a 1kΩ resistor which is then connected to ground. This is a voltage divider which we will connect the middle of to the (+) input pin of the op amp. The (-) input pin is connected to a 180Ω resistor in series with ground. The output of the op amp is connected to a 10k "load" potentiometer and digital multimeter in series and then connected between the (-) input and the 180Ω resistor.

As we vary the current source with the potentiometer we measure currents between 1.08mA and 4.16mA. Because of our 15kΩ and 1kΩ voltage divider we know that the voltage input is +0.75V or 1/16th of the +12V supply. Since we know from the golden rules that V- = V+, V- is also 0.75V. This 0.75V is connected to ground by a 180Ω resistor and using the law V=IR we know that the current should be at most 4.2mA and this is very near our measurement of 4.16mA showing that the results are as expected.

This current source is much more precise and stable than the transistor current source, but it unfortunately needs a "floating load" meaning that the load is not connected to ground on either side. Another disadvantage of this current source is that its speed limitations are not very good with either output current of load impedance varying at microsecond speeds.

Photodiodes
We are already familiar with LEDs from a previous lab and we know that one photon of light is emitted for every electron of current through the LED. A photodiode is a reverse of this process with an electron of current produced for every photon of light absorbed. Since the current produced is approximately linearly proportional to the intensity of light, photodiodes serve as light detectors.

For us to measure the induced photo-current from a photodiode we need to convert the photocurrent into a voltage. Usually a resistor is the standard current to voltage converter, but this will not work well for a photodiode because it is not a very good current source and thus the biggest output voltage is can sustain is about 0.5V and also as voltage develops across the photodiode the linearity of the device is diminished because of the varying response.

Op Amp Current-to-Voltage Converter
In order to have a better current source than our photodiode, we will be using a phototransistor which is essentially a photodiode with gain, making it a much more sensitive light detector.

It is important to note that the current for a photodiode flows towards the voltage source which is backwards from what we would expect, but for phototransistors the current flows in the more logical direction towards the next circuit element.

Current-to-Voltage Converter, experimentally
Lab 8-6 in the Student Manual
For this lab we constructed the circuit shown below with the +15V source changed to +12V and the 100kΩ resistor changed to a 10kΩ.

We measured Vout to be 0.7V and thus the average input photocurrent in this circuit is I = V/R = (0.7V)/10kΩ = 70 micro-amps. The output as seen on the oscilloscope can be seen in the photo below:

We then found the percentage modulation of this circuit by taking the change in Vpp which was 620mV and the change in Vbase which was 800mV. Putting these values into the formula (Vpp/Vbase)*100 = percent-modulation gave us a percent modulation value of 77.5%

Summing Amplifier
Op amps get there name from being "operational amplifiers" meaning that they can be configured to do mathematical operations. To do the mathematical operation of addition you can configure the op amp to take in two voltages sources by getting a current from applying that voltage to a resistor and adding the resulting currents together at a "summing junction" so that the sum of the two currents is fed into the feedback resistor. This summing junction is a virtual ground so we know that if the two resistors between the voltage sources and the summing junction are equal then Vout = (-Rfeedback / Rsource) * (V1 + V2)

Question: Can you design a circuit using op amps that subtracts two voltages?
To create an op amp that would subtract two voltage one would just need to invert one of the voltages using an inverting follower op amp and then feed the inverted voltage and the other voltage into a summing amplifier.

Summing Amplifier as a Digital to Analog Converter
Another use of the summing amplifier is as a digital to analog converter. If we say 6V represents a binary  "1" and 0V represents a binary "0" then by proportionally increasing the resistance by a factor of two from each power source for each additional digit we can create a digital to analog converter. To explain further the most significant digit will have the smallest resistor , let's say a 1kΩ, and the next digit will have a 2kΩ resistor, and the next a 4kΩ, and so on. Using the law V = IR this means that the smaller the resistor the larger the current at the summing junction leading to a weighted sum. Since the number coming in is in binary and the resistors are bases of two, the weighted sum is equivalent to the analog base ten current and thus the circuit acts as a digital to analog converter.

Summing Amplifier, experimentally
Student Manual Section 8-7
In this lab we constructed the circuit shown in the diagram below, but with the + and - 15V sources changed to + and - 12V and also the 15kΩ resistor changed to a 12kΩ.

This circuit sums a DC level with the input signal allowing you to add a DC offset to the signal. We applied a 5Vpp sine wave at 1kHz to the input and then adjusted the offset voltage. As we approached the end points of our offset range we noticed that the voltage peaks were being clipped. We tried this again with smaller input signal amplitudes and found that with a small enough Vpp we could avoid clipping all together because the Vout never got near the 12V limit. We also experimented with the AC coupling option and found that if the oscilloscope was AC coupled the signal would "bounce" briefly but then settled back to 0V. 

Op Amp Limitations
As has been mentioned in prior blog posts, real op amps are not ideal. To deal with the reality of op amps instead of the ideal version we can use the following two modified golden rules:

• Modified Golden Rule IWith negative feedback in place, the output of the op amp will try to do whatever is necessary to keep the voltage difference between the inputs equal to a very small voltage difference, called the offset voltage, Vos.
• Modified Golden Rule IIDue to their very high input impedance, the inputs of an op amp will sink a very small current, called the input bias current, Ibias.

Typical input bias currents for a 411 op amp are on the order of 3 pico-amps and a 411 op amp can supply at most 25mA. Another very important limitation is that at high frequencies the open loop gain of the op amp will "roll off"and decrease as frequency increases because of the slew rate. The slew rate for the 411 op amp is about 15V per micro-second. We will explore the effects of the slew rate in lab 9-1.
Op Amp Limitations, experimentally
Lab 9-1 in the Student Manual
For this lab we constructed a circuit with the (-) input of the op amp connected directly to Vout and the (+) input connected via a 10kΩ to the input signal. Driving this circuit with a square input wave of Vpp = 10V we found the falling slope to be 12.8mV per nanosecond and the rising slope to be 10.4mV per nanosecond. These measurements give us a slew rate for slewing down to be 12.8V per microsecond and the slew rate for slewing up to be 10.4V per microsecond.

We then changed the input to be a sine wave with an amplitude of 10V and measured the frequency at which the output waveform begins to distort. We found that frequency to be approximately 500kHz and the output appeared to look like a sawtooth wave. Using the equation 2*pi*f*A = 2*pi*500kHz*10 = 31.4V per microsecond. This did not match the expected value of  approximately 10 to 13V per microsecond we had just measured with the square wave.

Because sine waves were difficult for us to notice distortions on until the distortions were fairly significant, we changed back to a square wave and again measured the frequency at which distortions began. With a square wave we found this frequency to be 10kHz and the new calculated slew rate to be 6.3 V per microsecond, much closer than 31.4 to our expected values of 10-13V. I expect that if we had a more accurate method of measuring than just estimating with our eyes that we could get an even closer value.

Op Amp Integrator
As mentioned before, op amps can perform several mathematical operations. We have explored summation and now we will consider integration. An op amp integrator consists of a capacitor with the (+) input grounded, the (-) input connected via a resistor to Vin and a feedback loop containing a capacitor connecting (-) input to Vout.

Using an op amp to integrate, instead of the RC integrator circuit we explored earlier in the semester, allows us to remove the restriction of Vout << Vin. Using the Golden Rules and the definition of capacitance we can derive the expression Vout(t) = -(1/(R*C))*INTEGRAL(Vin(t))dt and this result will work for a wide range of frequencies.

This op amp integrator works very well which can cause a problem if there is even a small DC offset on the input voltage. Over time the offset is accumulated on the capacitor until the Vout "drifts to one of the supply rails".

We can correct this drifting by adding a large by-pass resistor to the feedback loop in parallel to the capacitor. This will give the DC current another path through the resistor and thus prevent it from accumulating on the capacitor.

Op Amp Integrator, experimentally
Lab 9-2 int he Student Manual
In this lab we constructed the circuit shown in the image below:

We then drove this circuit with a 1kHz square wave. As the lab write-up tells us, this circuit is sensitive to small DC offsets which causes the output to go into saturation near the 12V supplies or to not be symmetric about 0V. We used the function generator's offset controls to prevent these problems.

Next we used the component values to predict the triangle wave amplitude produced from an input of 2Vpp, 500Hz square wave. A 2Vpp means a value of +-1V for Vin and thus a current of (1V/100kΩ) = 1*10^(-5)A which we can use in the formula deltaV = (I/C)*delta(t) with the values t = 1ms and C = 0.01 microFarad. This gives us a change in voltage of -1V as the gain is negative.

The 10MΩ resistor serves to keep the output signal from railing. You can see the results of removing the 10MΩ resistor in the video below:

As you can see, without the resistor the DC current doesn't have another path and is forced to build up on the capacitor causing the output to drift to the rail.

Session 7 - Friday, February 20

Operational Amplifiers
Amplification and Gain

We have previously explored the use of transistors as electronic switches, but in this lab we explore their uses for amplifying electronic signals.

If you want to connect a microphone "source" to a speaker "load", it won't work well at all if you just directly connect them. This is for two main reasons. The first reason is that microphones don't produce very large voltage signals and there wouldn't be enough voltage to power the speaker sufficiently. The second reason is that microphone output impedance is usually very high (~1kΩ), while the input impedance of speakers is usually very low (~8Ω). This means that the signal would be very strongly attenuated as it moved from the microphone to the speaker and since microphones already produce very small voltages, this would leave almost no voltage for the speaker.

This problem is a common one in electronics (needing to amplify a small electronic signal), and we can learn to solve it with the help of the following two statements:

1. The output impedance of a device can be thought of as a measure of its ability to provide current to a load. (smaller Zout = better current supply to load)
2. When connecting a "load" to a "source" the output impedance of the source must be small compared to the input impedance of the load. (Zout << Zin)

Differential Amplifiers

Measuring the difference between two input signals is important in order to minimize unwanted pick-up.

For differential amplifiers, the two leads (ground and signal) are affected by the same pick-up, but in a differential amplifier this common pick-up mode is rejected.

Ideal differential amplifiers (which reject the pick-up) have the following the following characteristics:

• nearly infinite differential mode gain (Gdiff = Vout / ((V+) - (V-)))
• zero common mode gain (zero Vout when the two inputs are given identical non-zero voltages)
• infinite input impedance
• zero output impedance

As is usually the case, ideal things don't exist in the real world. However the operational amplifier, also known as "op amp" is amazingly close to this ideal. Op amps have an inverting input symbolized with (-) and a non-inverting input symbolized by (+). Op amps also have other pins, but in this lab we are only concerned with the non-inverting and inverting inputs, the positive and negative voltage sources and the output pin. We get our +12V and -12V voltage sources from a 5-pin DIN which we use to modify the wall's power supply and plug it into the bread board.

Open Loop Gain of an Op Amp, Experimentally
Lab 8-1 in the Student Manual

We began this lab by carefully setting up the circuit with 0.01 microFarad capacitors connecting the positive and negative voltage sources to ground in order to "decouple" the power supplies. Decoupling stabilizes the power supply and thus minimizes "fuzz" or noise on the oscilloscopes which will make for better results later.

We then connected our circuit to a potentiometer and when we spun the potentiometer the output reading on the oscilloscope alternated between the two rails of ~+12V and -12V. Some groups were able to delicately spin the pot so that it just lined up so that the output was ~0V, but we didn't have a very good pot (nor hand stability). The behavior of jumping between rails is as expected though because the gain is 200V/mV and thus even 1mV is multiplied so that it is 200V, this means that you would have to adjust the pot so that the voltage is between -0.06mV and +0.06mV in order to see anything other than a railed value of 12, which is very difficult to do, as you can see in the video below:

Feedback
Feedback is what we call taking some of the output and feeding it back into the input. Positive feedback is when we have a feedback designed to reinforce the input. This reinforcement of the input causes the sound to get louder and louder and the growing outputs keep getting resubmitted to the input causing the next output to be even greater. This kind of feedback is what causes the loud screeching sounds caused when you sometimes get the microphone too close to the speaker in common audio systems.

This screeching is not very useful, so let's look at the other type of feedback called negative feedback. Negative feedback pulls the output into the input in such a way that the output partially "cancels" some of the input. This stabilizes the system and corrects it when it varies too far from which is good, but unfortunately it also lowers the gain of the system, which is not so good, but is usually worth the benefits of the stabilization.

Almost all op amp circuits use some form of feedback and we will be exploring this later in this lab.

Op Amp Golden Rules

  1. With negative feedback in place, the output of the op amp will try to do whatever is necessary to keep the voltage difference between the inputs equal to zero.

  2.  Due to their very high input impedance, the inputs of an op amp will neither "source" nor "sink" appreciable currents.

Inverting Amplifier
By the application of the Op Amp Golden Rules and Ohm's law to a circuit with a negative feedback, we know the following:

1.  V- = V+ = 0
2. IR1 = Vin/R1
3. IR1 = IR2
4. Vout = 0 - IR2 * R2 = -(Vin/R1) * R2
5. Vout = (-R2/R1) * Vin (simplification of 4)

It is important to note that if R2>R1 the signal will be amplified and that for the circuit to work as an inverter the differential gain needs to be large, but the gain is independent of the differential gain of the amp. Also note that the minus sign in 4. and 5. above show that the output is 180degrees out of phase relative to the input and is thus inverted, which is why it is called an inverting amplifier.

Two other quick things to note before doing the lab 8-2; the input impedance of this circuit is Zin = Vin/Iin = R1, and the maximum output voltage will be ~1 volt less than the positive supply voltage while the minimum output voltage will be ~1 volt greater than the negative supply voltage. 

Inverting Amplifier, Experimentally
Lab 8-2 in the Student Manual
For this lab we constructed an inverting amplifier with a 1k resistor between the input voltage and the (-) input and a 10k resistor in the negative feedback loop. Also (+) input was grounded. We powered the op amp with +-12V and drove the circuit with a 1kHz sine wave. The output is shown in the photo below:

The gain is 10x as you can you see from the photo above by the fact that the scale for CH1 (output) is 5V while the scale for CH2(input) is 500mV. The maximum output swing is 1V for the input and 10V for the output and we can see that the output is 180 degrees out of phase with the input. We repeated this with a triangle wave as you can see in the photo below:

We then repeated it again with sine waves of various frequencies. As you can see in the video below, the amplifier stops working well as the frequencies get higher.

This has to do with the slew rate, but for the rest of this session we will treat the op amp as ideal with good slewing (we're on our honeymoon with the op amp :] )

Going back to driving the circuit with a 1kHz sine wave we measured the input impedance of the amplifier circuit by adding a 1kΩ resistor in series with the input. We found Vin to be 1V and Vout to be 520mV, telling us that the input impedance is 1kΩ since there is a voltage drop of roughly 50%.

We then attempted to measure the output impedance. After a long time fiddling with various small resistors and a very cool resistor box and lowering our voltage as much as possible, we could only conclude that the Zout is significantly smaller than 4Ω. This difficulty in measuring the output impedance has to do with the effects of limit on op amp output current. We hit a limit with our current output and thus couldn't conclude anything from results past this limit. You can see evidence of this limit in the clipping of the signal in the photo below:

Non-Inverting Amplifier
For this part, we explored the properties of a non-inverting amplifier which has Vin connected to the (+) input and has the (-) input connected to ground with a resistor between it and ground. It also has a feedback loop from Vout to (-) input with a resistor.

Applying the Op Amp Golden Rules and Ohm's Law to this circuit reveals the following properties:

1. V- = V+ = Vin
2. IR1 = (V-)/R1 = Vin/R1
3. IR1 = IR2 
4. Vout = V- + IR2 * R2 = Vin + (Vin/R1) * R2
5. Vout = (1 + R2/R1) * Vin (simplification of 4)

The gain for this setup is G = (1 + R2/R1). The output will be in phase with the input, which is why it is called a non-inverting amplifier. Similar to the inverting, this amplifier has a very small output impedance for small signals. Also, because the input signal is connected directly to the input of the op amp, the input impedance is very high. However, this circuit is not as stable as the inverting amplifier at high gains.

Non-Inverting Amplifier, Experimentally
Lab 8-3 in the Student Manual
For this lab we built a non-inverting amplifier with a 1kΩ resistor between (-) input and ground and a 10kΩ in the feedback loop and drove it with a 1kHz sine wave signal. We found Vin to be 1.06V and Vout to be 11.3V, giving a gain of 10.66 or ~11x. We then tried to measure the circuit's input impedance by putting a 1MΩ resistor in series with the input, and since the Vout was approximately the same we concluded that the input impedance must be >> 1MΩ.

This non-inverting amplifier maintains the low output impedance we measured in the inverting amplifier because it is essential the same circuit with the only difference being whether (+) or (-) input is connected to Vin. Since every other element is the same, we know that we maintain the low output impedance.

Op Amp Follower 
An Op Amp Follower is a special case of the non-inverting amplifier. When a non-inverting amplifier approaches the limit of R2 = 0 and R1 = infinity, the gain becomes one. Such a circuit is called a follower because the output signal "follows" the input signal.

Followers are useful because they maintain the high input impedance and very low output impedance of a non-inverting amplifier and thus can act as a buffer between a source and load without changing the output signal (because it has a gain of 1).

Op Amps are almost ideal followers, with their drawback being the inability to supply large currents.

Op Amp Follower, Experimentally
Lab 8-4 in the Student Manual
For this lab, we built the very simple circuit of a follower by connecting the Vin directly to the (+) input and connected the (-) input to Vout with no resistors. We confirmed that this follower circuit has the same Zin (very large) and Zout (very small) by briefly trying to take this measurements by using resistors in series.

Rather than having us measure the follower's input impedance, we added a 1kΩ resistor in series with the output in order to show that the feedback is producing the low output impedance. By looking at Vout with and without a 1kΩ load attached and with the feedback loop connected either before or after the 1kΩ resistor (not the load resistor). When the feedback loop is connected before the 1kΩ resistor which is in series with the 1kΩ load we see a ~50% attenuation in the signal which is expected because half the voltage drop is occurring across each 1kΩ resistor. However, when the feedback loop is connected in between the two 1kΩ, that is the 1kΩ resistor is treated as a part of the follower, we see no attenuation of the signal.

This results are not at all surprising since a follower is simply a special case of the non-inverting amplifier. As the lab writeup said, "No surprises here: Rout had better be 1kΩ."

Session 6 - Tuesday February 17

Diodes
Diodes only allow current to flow through them in one direction and have an effective resistance given by the ratio V/I. Since diodes allow current in only one direction we can think of the resistance being approximately 0 in the direction current is allowed to flow and infinite in the direction it can't flow.

Golden Rule of Diodes: When a diode with an I-V curve that looks like an exponential function is "forward-biased" and is carrying appreciable current (> 10 mA) it will have a fixed voltage drop across it: approximately 0.6 V in the case of a "regular" silicon diode and approximately 2-3 V in the case of a light-emitting diode.

Measuring the Resistance of a Diode
Ohmmeters output a constant current in order to measure the resistance of an unknown resistor. However, the amount of current output depends on the scale of the ohmmeter, so changing the sensitivity of the ohmmeter changes the current output. Since a diode's resistance is dependent on current this means that an ohmmeter will measure different resistances of the diode depending on what setting it is on. Specifically, the more sensitive the ohmmeter setting the smaller the resistance measured on the diode. To prove this to ourselves we measured a resistance of 136kΩ on our ohmmeter (not the standard ohmmeter) at a sensitive setting and 2.83 MΩ on its least sensitive setting.

Half Wave Rectifier
Rectifiers turn an AC signal into a DC signal, which is very useful for converting the AC voltage from the standard wall power supply in the DC voltage needed by most devices.

If we consider a circuit with a diode and 2.2kΩ resistor in series to create a kind of voltage divider with the diode as its top half we can see that current will only be allowed to flow in one direction. The diode is between the voltage source and the resistor and current can only flow from the voltage source to the resistor. Since the voltage drop across our diode is approximately 0.6 V, we know that if the voltage signal is below this value no current will move through the diode. We also know that the output voltage will now be 0.6V less than the input since that is how much was lost across the diode and that the diode's resistance is huge compared to the resistor.

Half Wave Rectifier, Experimentally
Lab 3-2 in the Student Manual
For this lab we constructed the circuit discussed above with a 2.2kΩ resistor and a 1N914 diode and a 6.3Vac(rms) transformer as the power source. It is important to note that since the transformer is 6.3 V rms we will actually see 9.6 V on the oscilloscope. You can see our observed output in the picture below:

This is the output expected as the output (yellow) never goes negative due to the influence of the diode which prevents any voltage less than +0.6V, giving the result a positive polarity. There are have as many "wiggles" in Vout (yellow) as in Vin(blue) because the diode blocks the current when there is not sufficient current to produce the necessary +0.6V.

We also expect to see a 0.6 V or 600 mV difference between Vout and Vin due to the voltage drop across the diode. We took measurements of this difference at the peaks of the voltages and off to the side. You can see the side measurement in the picture below:

The voltage difference measured in both locations was 640 mV which is very close to the expected value of 600 mV.

Ripple
If we add a capacitor in parallel with the load resistor in order to add a "RC time lag", our output voltage will no longer closely follow Vin on the falling edges. This is because on the rising edge the diode is allowing current and therefore has an extremely small resistance resulting in a small RC allowing the capacitor to charge very quickly and the Vout to act as normal. However, on the falling edge of Vin the capacitor can not discharge through the diode as the current is not allowed to flow in that direction. Instead the capacitor has to discharge through the resistor and therefore has a much larger RC time constant. Since the capacitor charges quickly, but takes longer to discharge and doesn't have enough time to discharge much at all(1/60 s), the resulting Vout looks like ripples as can be seen in the picture below:

The height of these ripples can be calculated by deltaV = Vmax - Vmax*e^((-1/60 s)/(Rload * C)) which is approximately deltaV = Vmax*((1/60s)/(Rload*C)). 

To test the above equation, we added a 47 microFarad capacitor in parallel with the 2.2kΩ load resistor in our half wave rectifier circuit. We calculated that the "ripple" amplitude should be 9.6V*(1/60)/(2.2k*47micro) = 1.55V. We then measured the "ripple" amplitude on the oscilloscope and got a value of 1.14V which is approximately equal to our calculated value, showing that the formula above works.

Signal Diodes, Experimentally
Lab 3-3 in the Student Manual
For this lab we constructed a rectified differentiator as seen in the diagram below, but with a 470 pF capacitor instead of the 560:

We then drove it with a 5Vpp square wave at 10kHz and the input(blue) and output(yellow) can be seen below:

This output makes sense because this is the same result as the differentiator circuit we made in the previous lab session, but cleaner because the rectifier eliminates the dip below 0V.

The 2.2k load resistor decreases the time the capacitor needs to discharge by providing another resistor for the current to discharge across. If you remove the 2.2k resistor the time constant increases and the capacitor takes longer to discharge. This longer time constant is just 560pF*1kΩ. or 560 ns. We took a measurement of this time constant using the oscilloscope and got 480 ns which is close.

With the 2.2k resistor in the circuit the time constant that we measured using the oscilloscope is 320 ns. Using the relation 320 ns = RC and knowing that the R is 2.2kΩ, we calculate the capacitance to be 145pF. This capacitance is called "parasitic capacitance" and it is inherit in the system. Roughly 30pF comes from the oscilloscope and there is another 30pF or so from each foot of coaxial cable. Knowing this, our value of 145pF makes sense because we were using a total of roughly 3 feet of coaxial cable which combined with the machine's 30pF should give a parasitic capacitance of approximately 120pF which is very near 145pF.

Diode Limiter
Large voltages can easily damage some instruments, but these instruments can be protected by using two diodes in pointing in opposite directions to limit the voltage to values less than 0.6V. We will explore how this works in lab 3-6.

Lab 3-6 in the Student Manual
We begin this lab by constructing a voltage divider that has a 1kΩ resistor on top and an 1N914 diode on the bottom which connects to +5V voltage source.Driving this circuit with a sine wave from our function generator at max output amplitude we can see that the output is cut off on the top of the sine waves. You can see this is the photo below:

This cutting off of the peak is called clamping and, as we can see in the zoomed in photo below, is not quite a flat chop. This is the effect of the diode's non-zero impedance causing the voltage to gradually cut off.

Diode Emitter, Experimentally
Lab 3-7 in the Student Manual
For this lab, we built the circuit shown in the diagram below:

We then drove it with a sine wave, triangle wave, and square wave of various amplitudes. You can see the output of these in the following order below: sine wave 5Vpp, triangle wave 5Vpp, and square wave 5Vpp.

Note that in all of these wave types, the output voltage has a max Vpp of just over 1V. To show that this 1 Vpp of the output is independent not just of the wave type, but also the amplitude of the Vin look at the pictures below:

The only difference between the two inputs in the pictures above was the amplitude of Vin, but the Vout didn't change at all. This is the perfect solution to the difficulties of needing to power sensitive circuits that can't handle large voltage from large voltage sources such as a standard wall plug.

Impedances of Test Instruments
Ideal instruments for measuring voltage will have infinite input impedance in order to have little impact on the circuit they are measuring. Of course, things are rarely ideal in the real world. Inside our oscilloscopes there is a "parasitic" or "stray" capacitance of 30pF that is in parallel with its 1MΩ resistor. The engineers who create instruments work as hard as they can to make this "stray" capacitance as small as they can and thus get their instruments closer to the ideal. This "stray" capacitance has the unfortunate affect of limiting the scope's ability to measure higher frequencies. The high frequencies make it difficult to meet the impedance requirement that Zout << Zin. Also, Zin is frequency dependent and thus the signal can be attenuated by different amounts at different components leading to distortion.

The Oscilloscope Probe
The Oscilloscope Probe is one way to help minimize these problems discussed above. The Oscilloscope Probe has a 10x setting that raises the input impedance of all frequencies for the oscilloscope by a factor of 10. This helps with the Zout << Zin requirement and also minimizes distortion. The drawback to using an Oscilloscope Probe is that its 10x setting decreases the sensitivity of the oscilloscope by a factor of 10 and therefore doesn't work well for weak signals.

In order to raise the input impedance of the oscilloscope by a factor of 10, the Zprobe has to be 9 times the impedance of the scope and cable. This is because the total impedance is Zprobe + Zscope&cable which we want to be 10*Zscope&cable. We can accomplish this by matching the oscilloscopes 1MΩ resistor in parallel with a 30pF capacitor with a 9MΩ resistor in parallel with a 3pF capacitor for the probe.

Impedances of Test Instruments, experimentally
Lab 3-8 in the Student Manual
We began this lab by measuring the internal resistance of the DVM. We made this measurement by measuring the Vin, Vout, and Rload which we made the R1 in a voltage divider with the DVM being R2. Vin was 5V, Vout was 4.95V, and Rload was 1kΩ. Using our formula for voltage dividers (Vout=(R2/(R1+R2))*Vin) we found the resistance of the DVM to be 99kΩ (our DVM is not the standard lab DVM, but a cheaper one). This value makes sense because for a DVM to be good at measuring voltage it must have a high input resistance and thus we should choose a large resistor in order to measure its impedance.

Next, we again used the voltage divider trick, but this time to measure the input resistance of the oscilloscope. This time we used a voltage source of 10Vpp sine wave at 100 Hz. We found that the voltage attenuated by half when we used a 1MΩ resistor which means the oscilloscope has an internal resistance of 1MΩ. Again, this makes sense because for the oscilloscope to be good at measuring voltages it must have a very high resistance in order to minimally affect the circuit it is measuring.

After this we measured the scope's input impedance. We did this by driving it with a signal in series with a 1MΩ to channel 2 and just the signal to channel 1. We know that the output impedance of the function generator is 1MΩ which is of course not << 1MΩ. This is important because we want most of the voltage drop to happen across the oscilloscope's 1MΩ resistor so that it can measure the voltage drop.

The attenuation of the signal is 50% at low frequencies, but as we raise the frequency of the input the output signal actually becomes larger than the input. This effect can be explained by the model in the diagram below:

Since there is a capacitor in parallel with the oscilloscope's 1MΩ the oscilloscope's attenuation value is frequency dependent. When the frequency is very high the capacitor acts like a wire and short circuits the resistor.

We know from the above that the R of the scope is 1MΩ. We can find the approximate value of C by solving the equation Vout = magnitude((-i/(w*C))/(R - i/(w*C)))*Vin with our measured values of Vout = 1.52V, Vin = 10V, delta t = 22 mircoseconds, and f = 11kHz. We get a final value of C = 150pF.

We can fix the circuit above to work as a divide-by-two signal attenuator at all frequencies by also making the load capacitance match the oscilloscope's, just like the resistor. This way, no matter what the frequency the load and oscilloscope are each losing half the voltage causing the oscilloscope to only measure half (divide-by-two) of the input voltage. Since we calculated C to be 150pF we know this is how much capacitance to add to the load. Since we don't have 150pF capacitors in the lab we used a 100pF and 50pF capacitor in parallel. You can see our circuit in the photo below:

This circuit does in fact create a divide-by-two signal attenuator for all frequencies. You can see some of our tests in the photos below:

Finally, we measured the internal resistance of the function generator. We did this by loading the generator with a known resistor and watching the output drop with a signal of 1 Vpp at 1kHz. To choose our known resistor we considered the purpose of a function generator. Function generators want to be good voltage providers and therefore want to minimize internal voltage drops and thus internal resistance. Knowing this we chose the small value of 10Ω for our known resistor. This resulted in a Vout of 208 mV from a Vin of 1.06V and by applying our voltage divider equation of Vout=(R2/(R1+R2))*Vin we find that the value of R1 is 40.9 Ω. To double check this result, we chose the closest resistor to 41Ω, the 47Ω resistor. The attenuation in the circuit with a 47Ω is approximately 50% which confirms that the internal resistance of the function generator is approximately 41Ω.

Session 5 - Friday, February 13

Section 1 - Reactance of a Capacitor

Capacitors change their charge with relation to the change in voltage over time and this relationship is shown by V(t) = q(t)/C or the differentiated version, dV/dt = (dq/dt)/C. This changing charge over time is the same as current (I = C(dV/dt)) and we can think of this as the "sloshing"(changing) charge causing current to be induced in response on the other side.

If the Voltage is a sin wave we can say that V(t) = Vo*sin (w*t) and therefore I(t) = C*w*Vo*cos(wt) = Io * cos(w*t).

As these are all sine waves, there are potential phase shifts of between 0 and 90 degrees between the V(t) and I(t). This phase shift can be modeled a Vo/Io = Vo/(C*w*Vo) = 1/(w*C). We call this 1/(w*C) term the capacitive reactance and it is sometimes written as Xc. Comparing this result to the one we found in a previous lab for resistors (Vo/Io = R) we realize that the value 1/(w*C) is analogous to resistance, but with a frequency-dependent phase shift component.

A very important concept that comes from these statements is that at high frequencies capacitors act like a wire or "short circuit", and that at low frequencies they look like an "open circuit".

Integrator Revisited
When a resistor and capacitor are in series in a voltage divider we can utilize the frequency dependence of the capacitor to insure that Vout << Vin. We can do this by making Xc << R and thus insure that this integrator circuit will have an output voltage proportional to the integral of the input signal, provided that the frequency of the input is high enough to meet the requirement that w is much less than 1/(R*C).

Differentiator
Remembering that Vin(t) = Vo*sin (w*t) and I(t) = Io * cos(w*t), we can also note that dVin/dt = Vo*w*cos(w*t). If we reverse the voltage divider we created for our integrator circuit, we now have a voltage divider with a capacitor and then a resistor. If we apply the principles mentioned in the previous sentence and Ohm's Law we can show that Vout(t) = Io * cos(w*t)*R = C*w*Vo*cos(w*t)*R = R*C*(dVin/dt). This has the fascinating result that at sufficiently low frequencies, that is RC << (1/w), this circuit works as a differentiator.

An RC Differentiator, Experimentally
Lab 2-2 in the Student Manual
For this lab, we constructed an RC differentiator with a capacitor of 100pF and a resistor of 100Ω. We then drove this circuit with a square wave of 1 Vpp amplitude and frequency of 100kHz. The output is shown below:

When you zoom in in units of time on a square wave, you can see that there is actually increases on a curve. The slope of this curve is shown by our differentiator circuit and is the yellow line. This output makes sense because as the blue curve gets steeper (the slope increases) the yellow line increases, and as the blue line starts to level off (the slope decreases until it gets to zero) the yellow line decreases but does not go below zero.

We confirmed that our circuit works as expected by trying a 100kHz triangle wave. The output can be seen below:

This output makes sense because the slope of the blue line is alternating between a positive constant value and a negative constant value, which is reflected in our flat yellow line that alternates between a positive and negative value.

Trying this one more time for a sine wave we get the output seen below:

This output also makes sense because the yellow line, which represents the slope, is zero at each peak of the blue sine wave and the yellow wave has a peak each time the slope of the blue line is greatest as the blue line passes through zero. This further confirms that the circuit (yellow) is a differentiator of the input voltage (blue).

Input Impedance
As a final exercise in lab 2-2 we did a quick worst-case impedance calculation for a DC input and a input at infinite frequency. Due to the capacitor, we know that the input impedance at DC would be infinite since the capacitor acts like an open circuit, and that at infinite frequency it would be 0 since the capacitor acts like a wire.

Lab 2-3 Part 2 in the Student Manual
For this lab we are looking at the behavior of the differentiator circuit when w<< 1/(R*C) is violated. We can see the effects of breaking this condition in the outputs below for the square wave (picture) and triangle wave (video). The effect for the square wave happens because the frequency is so high that the capacitor does not have time to sufficiently charge and therefore instead of the brief peak, it is a stretched out peak that decreases gradually.

The effect for the triangle wave can best be seen when you vary the frequency of the triangle wave. In the video we manually varied the frequency from 500Hz to 100kHz, adjusting the scale of the display as needed. Just like for the square wave, these odd effects occur because the capacitor doesn't have enough time. For the triangle wave, the capacitor isn't able to charge fully to then hold its constant value for a while. It never reaches this constant value that models the slope of the input and thus it increases and decreases similarly to the input wave. You can see as this happens because the yellow line goes from the expected stair stepping, to a rounded sawtooth, to finally a triangle wave matching the input.

Filters:

Low-Pass Filter

For a circuit of two resistors set up as a voltage divider we know that if R2>> R1 then Vout ≈ Vin and  conversely, if R2 << R2 then Vout << Vin. This translates to an RC circuit set up as a voltage divider with the capacitor as the bottom half of the divider as if 1/(w*C)>> R then Vout ≈ Vin and  conversely, if 1/(w*C) < R2 then Vout << Vin. From these we know that if a frequency is low Vout ≈ Vin and if a frequency is high Vout << Vin. Therefore this circuit only passes through low frequency inputs and "kills" high frequency inputs. Thus, we call this circuit a "low-pass filter".

Complex Impedance

Since the reactance of a capacitor is similar to resistance, but isn't the same, we need a way to describe reactance that can be used in the voltage divider equation of Vout = (R2/(R1 + R2))*Vin, but also keeps track of both the amplitude and phase. The answer to this dilemma is to use complex numbers. Instead of 1/(w*C) as we were using earlier, Xc = -i/(w*C).

Now, the magnitude of the complex factor (Xc/(R + Xc)) gives the relative amplitudes of the signals for a low-pass filter voltage divider. If we graph the relative amplitudes as a function of frequency we find that the cutoff frequency which is the boundary between the frequencies that do and don't pass through the filter(approximately 70.7%) is wo = 1/(R*C) which makes sense as this is also the limiting factor of the capacitor. We will verify this later in lab 2-4.

Phase Shifts

The complex number we used for capacitance also carries information about the phase shift between Vout and Vin. Looking at just the complex part of the term gives us Vout = (1/(1+i))*Vin = ((1-i)/2)*Vin. Plotting the complex number on a complex plot shows there should be a 45 degree phase shift with Vout lagging 45 degrees behind Vin. 

Based on the fact that at low frequencies the "impedance" of the capacitor is larger than the impedance of the resistor, we know that there will be little contribution from the imaginary part of complex term and thus Vout will nearly be in phase with Vin. That is, the phase shift will be close to 0 degrees.

On the other end of this spectrum, at high frequencies the capacitor will have a much smaller impedance than the resistor. This means that the imaginary term dominates and the Vout will be near 90 degrees out of phase with Vin. Again, we will verify this with Lab 2-4.

Decibels

Decibels is the a unit for measuring the ratio of two voltages. It is defined as dB = 20*log(10, (A2/A1)). We can use decibels as another way to describe the cutoff frequency point of the low-pass filter. The cutoff frequency point has a ratio of voltages such that the signal is being attenuated by a factor of 1/(2^0.5). Expressed in decibels this is known as the 3dB point because 20*log(10, (1/(2^0.5))) = -3.01 dB

When people say things like 'beyond the cutoff the response of a low pass-filter drops at a rate of 6dB per octave', they mean that since an octave is a factor of 2 change in frequency the attenuation changes to 20*log(10, 2). This evaluates to 6.02 and thus is called a 6dB drop rate per octave.

Low-Pass Filters, Experimentally

Lab 2-4 in the Student Manual

For this lab we constructed a low-pass filter out of a 15kΩ resistor and a 0.01 microFarad capacitor. We calculated the filter's 3dB frequency to be 1061.03 Hz (w = 1/RC = 1/(15k * 0.1 micro) = 6666.6, f = w/(2*pi) = 6666.6/(2*pi) = 1061.03). We then drove this circuit with a sine wave ranging from 1kHz and 10kHz to observe how the high frequencies were cut out by the low-pass filter and the signal became heavily attenuated. 

We then found the 3dB frequency point experimentally by finding the frequency at which the filter  attenuated the Vout down to 70.7% of its full amplitude. We found this value to be 1.2 kHz which is very close to our calculated value of 1.06 kHz. 

Then we found the limiting phase shift at very low frequencies to be 0 degrees, and the phase shift at high frequencies to be approximately 90 degrees, confirming the findings above in Phase Shifts.

We also measured the output, frequency, and phase shift at 2, 4, and 10 times f3dB to make sure that the low-pass filter was attenuating 6dB per octave. The input voltage was 1 V. The results are below:

Frequency in f3dB	Frequency in kHz	Amplitude in mV	Attenuation in dB (20*log(10, Vout/Vin)	Phase shift in microseconds	Phase shift in degrees (phase shift in seconds/period)*360
2	2.4	448	6.9	78	67
4	4.8	252	12.	49	85
10	12	110	19	21	90.7

These results confirm that the attenuation is approximately 6dB per octave (note that 4.8kHz to 12kHz is not an octave jump) and that the phase shift approaches 90 as we increase the frequency (going over 90 is probably an error in measurement).

High-Pass Filter

A high-pass filter is the opposite of a low-pass filter. The positions of the capacitor and resistor are swapped and the high-pass filter attenuates low frequency signals while passing along high frequency signals and the cutoff frequency is still 1/(R*C).

High-Pass Filters, Experimentally

Lab 2-5 in the Student Manual

For this lab we constructed a high-pass filter out of a 15kΩ resistor and a 0.01 microFarad capacitor. We calculated the filter's 3dB frequency to be 1061.03 Hz, same as the previous low pass circuit. Although this time we experimentally found the 3dB frequency to be 1kHz. We then drove this circuit with a sine wave ranging from 500Hz and 10kHz to observe how this circuit passed along the high frequencies, while the low frequency signal became heavily attenuated. 

We then checked that the output frequencies at low frequencies well below the 3dB point (1/2, 1/4, and 1/10 times f3dB) were proportional to frequency. The input voltage was 1 V and the results are below:

Frequency in f3dB	Frequency in Hz	Amplitude in mV
1/2	500	448
1/4	250	240
1/10	100	112

We can see that the output is in fact proportional to frequency as the amplitude in mV is the same fraction of the original 1 V as the frequency fraction.

The limiting phase shift for this circuit is 90 degrees at very low frequencies and 0 degrees at very high frequencies.

Garbage Detector

Lab 2-6 in the Student Manual

This lab combines a high-pass filter with with a transformer to show you the high frequency "garbage" that comes with the 110-volt power line. Since this is the same circuit as the lab before, but now with an added transformer, we know that the calculated 3dB frequency is 1061Hz, and we confirmed this with a measured 3dB of 1kHz. To find the filter's attenuation at 60 Hz we counted octaves to make the arithmetic easier. 1kHz -> 500Hz -> 250 Hz -> 125 Hz -> 62.5Hz is 4 octave changes. The voltage at 1kHz will be 70.7% of the total 6.3V provided by the transformer, since it is the cutoff frequency. A 4 octave change means that the fraction of the original voltage is 1/(2^4) or 1/16. Thus, the attenuation at 60 Hz is (0.707 * 6.3)/16 = 0.28 V.

Blocking Capacitor

Since capacitors act as an open circuit for direct current and are closer to a short circuit for alternating current, a capacitor can be used to take in a varying voltage and block the DC component.

Blocking Capacitor, Experimentally

Lab 2-8 in the Student Manual

In this lab, we added a 10kΩ resistor and +5V source from our Arduino chip, to a high-pass filter consisting of a 4.7 microFarad capacitor and a 4.7kΩ resistor. Driving the high-pass filter with the function generator and looking at the output on the oscilloscope on the DC coupling setting, we can see the AC signal "riding" on +1.6V which comes from the voltage divider giving a third of the 5 volts from the Arduino.

Now we add another high-pass filter to the output of our previous circuit. This time the high-pass filter added consists of a 4.7 microFarad capacitor and a 100kΩ resistor. The low frequency limit for this blocking circuit is 0.34 Hz (wo = 1/(R*C) = 1/(100k * 4.7 micro) = 2.13, wo/(2*pi) = f = 2.13/(2*pi) = 0.34 Hz).

We will explore the applications of this circuit in the next lab.