* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Methods Part A (Diode without Transformer)
Nanofluidic circuitry wikipedia , lookup
Spark-gap transmitter wikipedia , lookup
Standing wave ratio wikipedia , lookup
Oscilloscope history wikipedia , lookup
Analog-to-digital converter wikipedia , lookup
Radio transmitter design wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Valve audio amplifier technical specification wikipedia , lookup
Integrating ADC wikipedia , lookup
Two-port network wikipedia , lookup
Wilson current mirror wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Transistor–transistor logic wikipedia , lookup
Power MOSFET wikipedia , lookup
Surge protector wikipedia , lookup
Current source wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Power electronics wikipedia , lookup
Operational amplifier wikipedia , lookup
Schmitt trigger wikipedia , lookup
Voltage regulator wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Current mirror wikipedia , lookup
Jacob Kazam Lab Partner: Joey Lawlor 2/24/04 Diodes and Transistors Introduction: Impurities can be added to semi-conductors to change conduction characteristics. This process is called doping. Semi conductors generally have four valance electrons, resulting in 2 bands. Added impurities can have one more electron, 5 or one less, 3. If a semi-conductor is doped with a 5 valance electron atom such as Sb or As, the extra electron is forced to fill an empty band. It becomes free to move and leaves behind a fixed positive atom. This type of conductor is now called an n-type. If a semi-conductor is doped with a 3 valance electron atom such as Al, the missing electron leaves a hole. The hole can be thought of as a moving positive charge. It leaves behind a fixed negative atom. This type of conductor is now called a p-type. Putting together N and P type conductors (called a diode) results in a net positive charge on the n-type side (due to movement of electrons to the p-type side) and a net negative chare on the p-type side (due to movement of positive holes to the n-type side). This results in a build up of potential difference between the two ends which induces an electric field. It discourages charge drifting and reaches equilibrium. If the positive terminal of a battery is connected to the p side of the diode and the n side to the negative terminal the current flows with relatively no resistance. This is called Forward Bias. If the diode is flipped around, the result is no current. The diode then acts as a component of infinite impedance. This is called Reverse Bias. Transistors are mixed semi-conductors with 2 junctions, resulting in three parts: an emitter, a base and a collector. They could be PNP or NPN. Charge carriers flow out of the emitter, 99% will stay in the collector and very few will stay in the base. The transistors can also be connected in forward and reverse bias. This lab deals with inserting diodes and transistors into simple circuits and analyzing them. Methods Part A (Diode without Transformer): Diodes allow current to flow in one direction with relatively no resistance (forward-bias) and in the opposite direction with almost infinite resistance (reverse-bias). This property can be used to change an AC signal into a DC signal. In order to illustrate this property, the circuit illustrated in figure 1 below was built: Figure 1: Figure 1: An AC circuit where Vin is a sine wave of 1KHz, the diode is diode 1N4001 and R=1000 Ω. Where Vin is an AC sine wave of 1KHz, the diode is diode 1N4001 and R=1000 Ω. Since the input voltage is AC, we should expect to only see the output as the positive parts of the sine wave and zero for the negative parts. This circuit is thus called a half wave rectifier in that it should only produce the positive half of the sine wave. To test this expectation, the input voltage and output voltage vs. time were connected to an oscilloscope and the resulting display was measured and sketched. Results Part A (Diode without Transformer): (Q1) One completed period of the output waveform as a function of time is sketched below in Figure 2: Figure 2: Figure 2: Input voltage (full sine wave) and output voltage vs. time for the half wave rectifier. As predicted, only the positive half of the sine wave is produced as output voltage. There seems to be some sort of a delay between the input and output voltage. The output voltage does not begin to rise until approximately 0.04ms after the input voltage becomes positive and drops to zero approximately 0.04 ms before the input voltage does. A logical explanation for this behavior is that the diode has some threshold voltage where until it is reached, no current is allowed to pass through. This threshold voltage can be approximated by taking the difference between the peaks of the input and output voltages since at forward bias there is almost no resistance. This difference is approximately 0.8V. (Q2) The amplitude of the output voltage is 4.8V. It is not half the input voltage nor should it be. (Q3) It is not half the input voltage because the diode is connected in forward bias. When the current is positive, the diode should act as if it isn’t even there and the current across the resistor should be close to the input voltage. Methods Part A (Diodes with Transformer): Two diodes, a symmetric transformer (TC016) and a resistor of 20 KΩ were connected to create the full wave rectifier illustrated in Figure 3, below: Figure3: Figure 3: A Full wave rectifier with symmetric transformer (TC016) with center tap and resistor R=20kOhms The center tap allows the creation of a full wave rectifier with only two diodes. The center tap splits the voltage across the transformer, sending it in opposite directions with equal magnitude. This should insure that negative or positive, there will always be a signal for Vout. The diodes insure that the signal will always be positive and since the signal is split into equal halves, Vout should equal approximately half of Vin. Next, a 5000 pF capacitor was added in parallel with the resistor. The circuit was analyzed with the oscilloscope. Next, a 0.1 μF capacitor was inserted in place of the previous capacitor. The circuit was analyzed once more using the scope. Results Part A (Diodes with Transformer): (Q4) One completed period of the output waveform as a function of time of the full wave rectifier (without capacitor) is sketched below in Figure 4: Figure 4: Figure 4: Full wave rectifier without Capacitor. Output voltage vs. time. As predicted, the output voltage across the resistor results in a positive hump for both positive and negative input voltages. It has a peak of 2.6V, half of the previous output voltage, as predicted. There seems to also be a threshold voltage for the diodes in this case as well as there is a slight offset and the output voltage is not exactly half of the input voltage. (Q5) A sketch of output voltage vs. time for the full wave rectifier with a 5000 pF capacitor in parallel with the output resistor is shown in Figure 5, below: Figure 5: Figure 5: Output voltage vs. time for a full wave rectifier with 5000 pF capacitor in parallel with output resistor. The output voltage is no longer a well defined hump. Rather, it begins as a well defined hump and once it reaches its peak it seems to stay at a higher voltage than it would have it remained the same shape. The output voltage also no longer touches zero. This is because by the peak, the capacitor is fully charged. Once the input voltage drops, the capacitor begins to lose its charge. It can only lose its charge to the resistor, resulting in an increase in the output voltage. There is still 0.8V across the resistor at its lowest point. (Q6) The capacitance was then increased. A sketch of output voltage vs. time for the full wave rectifier with a 0.1 μF capacitor in parallel with the output resistor is shown in Figure 6, below: Figure 6: Figure 6: Output voltage vs. time for the full wave rectifier with a 0.1 μF capacitor in parallel with the output resistor. The figure above shows an even more drastic effect than Figure 5. The larger capacitor can store more charge and decays over a longer period of time. (Q7) It has become obvious that a large capacitor will result in almost no change in current and will result in a DC. But how large a capacitor is needed to keep the variation in voltage to less than 1%? This means that over the time of 1 period, the capacitor must cause a decay of only 1% in the voltage across the resistor. Vout across the resistor, (VR), is given by equation 1, below: VR Ve t Where V is the voltage across the resistor at the peak of the wave. The period was plugged in as t=0.42 ms and τ is given by, τ=RC, since R is known, C can be found if τ is found. We want VR 0.99V and thus plug in 0.99V for VR in equation 1 and solve for τ. The result is C=2.1 μF. Methods Part B (Transistors): Two simple circuits involving transistors were constructed to look at basic transistor behavior. The first demonstrates how a transistor can act to transform a poor voltage source into a relatively good one. A voltage divider was set up with each resistor, R1 and R2 equal to 1 KΩ. A DC input of +11.8V was used. It is already known that a voltage divider is a bad voltage source. The voltage across the input, (Vout) and one of the resistors (R2) was measured as well as the current through R2. This is Vout without a load. Next, a 500 Ω load resister was added in parallel with R2 and the output voltage and current through was measured once more. The two points were plotted to make a line of output voltage vs. output current. The output impedance is the slope of that line and was calculated. Next a voltage divider with a transistor was created. It is illustrated below in Figure 7: Figure 7: Figure 7: A voltage divider with NPN transistor. R1=R2=Re=1000 Ohms A voltage divider with NPN transistor was constructed as shown above, with R1=R2=Re=1000 Ohms. The output voltage across Re was measured. The current through the base, collector, emitter, R1 and R2 were measured. Next a 500Ω load resistor was added in parallel with Re and the output voltage was measured once more. The output voltage vs. the output current was graphed once more and plotted on the same graph as the voltage divider. The output impedance of this circuit was calculated from the slope of the line. Results Part B (Transistors): (Q8) The voltage across the input and across R2 was found to be 11.8 and 6V respectively. The output current was measured to be 5.9mA. (Q9) After the 500Ω resistor was added, the output voltage and current were found to be 3.0 and 8.7 respectively. The output impedance, given by the slope of the line, was found to be 1100Ω. (Q10) The circuit shown in Figure 7 was constructed. The output voltage across Re was measured to be 5.23V. (Q11) The currents into the base, through R1, R2, the collector and emitter can be found below in table one: Table 1: Table of circuit components and current for voltage divider with transistor Circuit Component Base R1 R2 Collector Emitter Current 42 μA 5.97 mA 5.95 mA 5.28 5.2 (Q12) A 500Ω resistor was added in parallel to Re and the output voltage was measured again. Output voltage vs. output current was plotted on the same graph as the data given above. This plot can be found in Figure 8 below: Figure 8: Figure 8: output voltage vs. output current for a voltage divider with an without transistor. It can be seen from the above graph that inserting the transistor results in an almost horizontal line with a small slope, while the voltage divider alone produces a steep line with large slope. (Q13) The output impedance of the voltage divider with transistor was calculated to be 25Ω, very small in comparison to 1100Ω. (Q14) Good voltage sources produce an output voltage independent of the load resistance and very low output impedance. Adding the transistor to the voltage divider resulted in a voltage source that varied very little with the load resistance and low output impedance in comparison with the ordinary voltage divider. Methods Part B (Transistors continued): Next the circuit shown below was connected, which acts as a simple current source. The load resistance RL was set equal to 100, 1000 and 5000Ω. The circuit is shown below in Figure 9: Figure 9: Figure 9: Simple Current Source with an NPN transistor, where R1=R2=Re=1000 Ohms and RL is variable R1=R2=Re=1000 Ohms and RL is variable. The output voltage and current through RL were measured for each of the three above cases. Results Part B (Transistors continued): (Q15) Given below is a table of measured values for the circuit illustrated in Figure 9. Table 2: Table for Resistance, output voltage and RL current values for the Figure 9 circuit Resistance RL (Ω) 515 1000 5000 Vout (V) 5.28 5.28 4.01 IRL (mA) 5.27 5.27 1.54 (Q16) At first, the change in resistance has no affect on the output current, but changing the resistance to 5000Ω results in a large variation in current. Increasing the resistance RL, decreases the voltage, Vcb. At first the transistor is in its linear operating region. With a large enough resistance, the transistor moves to a saturated region. Conclusion: Diodes allow current to flow in only one direction. Transistors can be used to turn a relatively bad voltage source into a relatively good one.