Marissa's+AM+Radio+Lab

= AM Radio Lab Report =

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 * Yip Yip Martians discover a sweet radio ! **


 * Date**: March 3, 2010
 * Participants**: Marissa Marton

The purpose of this lab was to explore, build, and learn about the different components of circuits within a radio. In addition, we had to focus on how these different components work separately as well as together in amplifying an AC signal.
 * Purpose**:

In this experiment, we learned about the different components of circuits within a radio (such as an inductor, capacitor, resistor, diode, etc). In addition, we had to focus on how these different components work separately as well as together in amplifying an AC signal. Our goal was to design a capacitor / inductor pair that gave a resonant frequency of 900 K Hz (900,000 Hz à 900,000 times per second). We began this process by calculating values for these components (such as the capacitance and inductance) in order to reach a value for the resonant frequency. The resonant frequency is what identifies a signal for a specific radio station (AM radio station in this case). In addition, this was a great process of trial and error where we had to alter our area of capacitance, the distance between our two sheets of tin foil (our capacitor), as well as the amount of coils of wire for our inductor. Once we had all of our calculations, we put together our radio and began to test it out. We were then able to see what parts of our radio needed adjusting in order to reach a specific resonant frequency. As a result, we could control what radio station our radio picked up. However, prior to all of this, the class partook in online simulations of various circuits as well as note taking in order to better our understanding. This allowed us to draw many connections when we built our radios and it helped us to understand how our radios functioned as a whole.
 * Brief Description of Experiment**:



Formula//__ || The capacitance (measured in farads) is calculated by using a formula that multiplies the capacitance constant by the area of the capacitor. That is then divided by the distance between the two pieces of tin foil (the capacitor).
 * Sample Calculations:**
 * __//Capacitor

//C = ε0 x (A (m²) / d (m)) ε0 = 8.85 x 10^ -12// || Formula//__ || The inductance (measured in henries) is calculated by using a formula that multiplies the inductance constant by the number of loops / coils of the wire squared as well as by the area of the loops / coils. This is then divided by the distance between the two pieces of tin foil (capacitor).
 * __//Inductor

//L = ( μ0 x N² x A (m²))// / ℓ //(m) μ0 =4π x 10^ -7// || Formula//__ || The resonant frequency is calculated by using a formula that multiplies one divided by two pi, then by one divided by the square root of the capacitance, which is then multiplied by the inductance.
 * __//Resonant Frequency

//fres = (1 / 2 π// // ) x (1 / square root(LC)) // ||

While we may not have picked up the correct radio station with our resonant frequency of 607,479.3145 Hz, we were definitely able to draw connections between the work of precious labs to this lab. In addition, our understanding of circuits is much greater because of making these radios and learning about how the different components of the RCL circuit functioned as a whole. Also, we were able to make some changes in hopes of amplifying the AC signal. Overall, the lab was very interesting and it was nice to be able to explore the circuits in a new and interesting way.
 * Results**:


 * Lab Questions**:

1. Explain what a resonant circuit is, what circuit elements it conains, and how those elements work together to amplify an AC signal.

Resonance in RLC circuits consists of a unique frequency that is determined by the calculated values of resistance, capacitance, and inductance (1). For making our radio, we had to use a more complex circuit otherwise known as a paralell resonant circuit (rather than a series resonant circuit). Our circuit consisted of a capacitor, a resonator, a diode, an inductor, an antenna, and of course electric and audio (headphones) as well as ground to attach to. In trying to explain the AC signal, AC circuits utilize a certain adaptation of Ohm's Law, which defines impedance (electric resistnace measured in Ohms). This definiton of impedance consists of the combination of resistance, capacitance, and inductance (1). Essentially, all components of the circuit must have the most accurate values that one could reach together in amplifying the AC signal successfully and appropriately. The antenna has to be in the right position for recieving the best signal it can. Also, the goal was to construct a radio with a resonant frequency of 900 K Hz. The closer one comes to this number, the better. The signal starts out very slow and weak and originates within the antenna. With the correct resonant frequency, the signal will build and become much more intense. This is the way that these components amplify the AC signal. All calculations have to come out as close as possible to the target resonant frequency and this includes the capacitance and inductance. It also includes all parts of the radio circuit and they have to be put together correctly and much function correctly as well. If our radio was exactly 900 K Hz, it would have recieved the optimum amplified AC signal. The further one is away from that target, the weaker the AC signal. Our resonant frequency was 607,479.3145 Hz. This is not great but we were still able to decipher the words of our AM radio via our weak signal. (1) Work Cited: []

Good site: []

2. Describe (pictures, calculations, etc) your particular resonant circuit and its elements. What is its resonant frequency and how do you know? *The coils on the far left of the circuit represent an inductor
 * The two lines parallel to each other (just to the right of the coils) represent a capacitor

// fres = (1 / 2 π // // ) x (1 / square root(LC)) //

1 / 2// π = 0.159 //

// 0.159 x 1/X = 900,000 Hz //

// (1.745 x 10^ -6) // = square root LC

( // 1.745 x 10^ -6) // ² = square root LC² = 3.13 x 10^ -14

LC = 3.13 x 10^ -14

The goal was to design a capicator / inductor pair that would give off a resonant frequency of 900 K Hz (900,000 Hz). Our particular resianat circuit consisted of a capacitor, a resonator, a diode, an inductor, an antenna, and of course electric and audio (headphones) as well as ground to attach to. After testing out our calculations multiple times, we figured out that the capacitor of our radio would consist of a value of .04 meters squared. In addition we found the distance between the two sheets of tin foil to be 6.67 x 10^ -5, which is about the thickness of a piece of paper. We then took the capacitance constant and multiplied it by the .04 meters squared and then divided that product by the distance between the two sheets to get a quotient of 5.28 x 10^ -9. Equation: 8.85 x 10^ -12 x .04 m² / 6.67 x 10^ -5 = 5.28 x 10^ -9 This number (5.28 x 10^ -9) is our capacitance.

For the inductor, we came up with the value of five loops / coils of wire around a water bottle. The length of the loops / coils was about .015 meters. We then took the inductance constant (// μ0 =4π x 10^ -7) and multiplied that by the number of loops / coils squared, which was then multiplied by 7.85 x 10^ -5 (A). // Equation: // 4π x 10^ -7 = (1.257 x 10^ -6 x (5) //²) x 7.85 x 10^ -5 = 1.3 x 10^ -5 This number ( 1.3 x 10^ -5) is our inductance. Finally, in order to determine if our calculations held true to our LC value of 3.13 x 10^ -14, we had to multiply our value for the capacitance by the value of the inductance. Equation: 5.28 x 10^ -9 x (1.3 x 10^ -5) = 6.864 x 10^ -14 This depicts that we have a calculation that is very close to the target resonant frequency of 3.13 x to^ -14, and this is the final calculation for LC of our radio that we decided on.

For the resonant frequency, we used the following equation: fres = (1 / 2 π // ) x (1 / square root(LC)) // We the plugged in the values that we calculated above and found our resonant frequency of 607,479.3145 Hz (607 K Hz). We know this is our resonant frequency because we used the fres equation and we checked our answer.

3. Address one of the following (or more for extra credit): - Effect of resistance on resonance * - Head phones and electric to audio conversion * - Antennas and conversion of EM waves to currents
 * -** Diodes and demodulation *

__Head phones and electric to audio conversion__: Headphones consist of a pair of transducers. These transducers receive electrical signals from a a specific source like a receiver or a media player. The speaker in each headphone converts these electrical signals to sound waves, and this allows us to hear sound via headphones. Audio transformers are magnetic devices that are designed to be used in audio circuits (hence the name). They are capable of blocking radio frequency interferences; however they can also combine audio signals or they can come up with impedance that matches high and low impedance circuits. They were originally designed to connect various telephone systems together, but keeping the power supply separate at the same time. Today, they are often used for audio systems and their components. Due to the fact that they are magnetic devices, they are prone to coming in contact with external magnetic fields—just like those of an AC current-carrying conductor. Overall, the audio transformer provides for shielding of specific components in order to allow for audio to come through in the headphones. Work Cited: []



__Effect of resistance on resonance__: Resonance is when a system oscillates at larger amplitudes at spesific frequencies in comparison to others. It occurs when a system has the capability to store and transfer energy between two or more different storage models. Resistance is when a material opposes the flow of electrical current and in circuits, it turns electrical energy into heat. All inductors consist of a specific amount of random capacitance. This is because of certain gaps called insulation gaps. In addition, circuit conductors also have the power to generate random capacitance. As we saw in our radios, it is important to have a capacitance close to the target capacitance because there will always be a capacitance somewhere that one cannot get rid of. This can definitely cause some resonance issues. However, in order to counteract this, adding resistance can greatly help with correcting any resonance problem. This process of getting rid of unwanted resonance by adding resistance is often utilized within mechanical system designs. Work Cited: []

__Diodes and demodulation__: When looking at an AM Radio as a whole, there are many different compoenets that work together as one. When sound is multiplied by a carrier, one will get a product of the amplitude modulated signal. Sound consists of many frequencies that are constantly changing and the carrier is one constant frequency and it is usually at a high freuency. For the amplitude modulated signal, it contains characteristics of both the sound and the carrier. There is an envelope that surrounds the high frequency and the envelope changes slowly according to the sound wave (see the first picture on the far left to get an idea of what the envelope looks like). In order to hear the sound being transmitted, one must demodulate the sound waves. In order to do this, the carrier wave must be extracted. This is done by using a diode. The diode has an effect of cutting the alternating current in half. As a result, the top part of the amplitude modulated signal is remains and the lower half is discarded. In other words, the envelope is saved and the carrier is thrown away. Overall, this is the process of the diode demodulating the sound waves. In addition to this, the carrier switches much more rapidly between positive and negative in comparison to the envelope. Interestingly enough, as the current flows from left to right across the diode, there is a large resistance and as a result, very little current will flow. This means that when the current flows from right to left, there is a small resistnace and a greater amount of current will flow.

While our radio had a few issues, it still worked perfectly fine. The strength of the sound that our radio produced was not so weak that we couldn't decipher the words, but it was faint. I think there could have been other additional things that could have gone wrong, which would have caused the results to be difficult to reproduce. For example. errors within calculations could have caused the radio to function incorrectly and we then would have had to go over our calculations again until they were reasonable. However, our calculations did prove to be valid for the most part in that our radio did in fact work. The only issue with our radio was that our capacitor was too large. In other words we used too much tin foil and as a result, the resistance was too big. In order for the radio to work more accurately, we should have designed our radio and capacitor so that we used as little foil as possible. However, with the amount that we used, the large amounts of foil degraded our radio. Because our capacitor was ultimately useless, the antenna is what allowed us to receive the radio signal. The capacitance that the antenna naturally contains was enough for our radio to work. For this experiment, the antenna was moved around in different locations. Each location makes the signal received a little different or in other words, the different locations make the antenna function differently. One wants to setup the antenna in an optimal location so it can receive the radio waves produced by a broadcast tower in the best possible way. Overall, we could have gone over our calculations for the capacitor again and made a much better capacitance for the radio to function better.
 * Conclusion**: