Joseph Walsh
Partner: Elizabeth Tozour
750:327:02
February 24, 2004
DIODES AND TRANSISTORS
Introduction:In this lab, we were supposed to analyze the outputs of different circuits containing diodes and transistors. A diode is a combination of two semiconductors, one p-type and onen-type. When these two semiconductors are initially placed next to each other, the holes, and free electrons rearrange to give p-type semiconductor a net negative charge, the n-type semiconductor a net positive charge, and create an electric field heading from the n-type to the p-type semiconductor. Hence, there is a potential difference, orpotential barrier, across the point where the two semiconductors meet. When this diode is put into a circuit, it allows current traveling from the p-type to the n-type, but not the reverse. This is due to the size and effect of the potential barrier between the two semiconductors. For instance, if a positive current is trying to travel from the p-type tothe n-type, since the p-type is already negative, the current will decrease the potential barrier and will flow freely. A circuit wired in this fashion is said to have a forward bias. However, if the current is trying to pass through the n-type to the p-type, since the n-type is already positively charged, the current will increase the potential barrier and be blocked. A circuit wired this way is said to have a reverse bias. In theory, an ideal resistor would have 0 impedance to a forward bias current and infinite impedance to a reverse bias current. In practice, however, the current through the diode increasesexponentially with the voltage drop across it. Diodes are extremely useful in creating rectifier circuits, circuits that turn AC voltage into a DC voltage source; this is due to the fact that they generally only allow current to travel in one direction.
A transistor is basically a semiconductor, either n-type or p-type, with a smaller semiconductor of opposite type attached to either side. In this lab, we will be studying npn transistors, those transistors that have a p-type semiconductor with n-type semiconductors attached to each side. In this case, the p-type semiconductor is called the “base” of the transistor, for obvious reasons. The transistor is attached to a circuit such that one of the np regions has a forward bias and the other has a reverse bias. The one with a forward bias is called the “emitter” because it emits current into the base. Most of the current then flows out of the other np region, called the “collector”, rather than bleed out through the base connection. This is because the base-emitter junction is much closer (by a factor of hundreds) to the base-collector junction than it is to the base connection. Therefore, the current through the collector, Ic, is very close in magnitude to the current through the emitter, Ie.
Part A:
Methods:In the first part of the lab, we were to create a simple voltage divider consisting of a diode and a resistor connected in series with an AC voltage source. We fixed the input voltage and frequency, used a sinusoidal waveform, and used the oscilloscope to measure the output voltage across the resistor. The values we took for the resistance of the resistor, input voltage, and frequency can be found in the results section. We recorded from the oscilloscope the output voltage for one period and plotted it in Figure 1. The data can also be found in the results section.
Diagram 1: Schematic of Simple Voltage Divider Circuit.
Next, we built a rectifier as shown in Diagram 2 using an alternating voltage source, two diodes, a resistor, and a transformer. The data for these circuit elements is in the results section. We used a transformer with a center tap because it provided a way to hook up the two diodes in such a fashion that when one was biased forward, the other had a reverse bias. A symmetric transformer was used so the current generated in the second coil is equal to the first. We again measured the output voltage across the resistor with the oscilloscope; the results are plotted in Figure 2. We then wired different capacitors in parallel with the resistor and again used the oscilloscope to show the output voltage.
Diagram 2: Schematic of Voltage Rectifier
Results:
Table 1: Element Data of Simple Voltage Divider
Resistance of Resistor / 1007 ΩType of Diode / 1N4001
Input Voltage / 4 V
Input Frequency / 1000 Hz
Period of Oscillation = 1/f = 1 ms
Table 2: Output Voltage versus Time for One Period
Time (ms) / Output Voltage (V)0 / 0
.1 / 1.6
.2 / 2.4
.25 / 3
.3 / 2.4
.4 / 1.6
.5 / 0
.6 / 0
.7 / 0
.8 / 0
.9 / 0
1.0 / 0
Table 3: Element Data for Voltage Rectifier
Type of Transformer / TC016Type of Diode / 1N4001
Resistance of Resistor / 19870 Ω
Input Voltage / 4 V
Input Frequency / 1000 Hz
Table 4: Output Voltage versus Time Data for Rectifier Circuit
Time (ms) / Output Voltage (V)0 / 0
.1 / .9
.2 / 1.4
.3 / 1.0
.4 / .3
.5 / 0
.6 / .9
.7 / 1.4
.8 / 1.1
.9 / .5
1.0 / 0
List of Capacitors Used in Rectifier Circuit:
5000 pF
.1 μF
1 μF
2 μF
Analysis and Conclusions:Figure 1 is the graph of the output voltage against time for the length of a period. It shows that when the AC input voltage is positive, current passes through the diode and most of the voltage is lost across the resistor. However, when the AC voltage is negative, there is no voltage drop across the resistor, or equivalently, no current through the circuit. These results are in agreement with the theory since the way we had the circuit set up created a forward bias over the diode when the current was positive and a reverse bias when the current was negative. The amplitude of the output was 3 V, or a gain of .75. This is much greater than half the amplitude. We only would expect the output voltage to have an amplitude which is half that of the input voltage in the special case that the impedance through the diode is equal to that of the resistor. However, we know that the impedance of the diode—besides its natural internal resistance—decreases exponentially as a function of voltage. Therefore, it is highly unlikely that we would see a voltage drop across the resistor that was roughly half that of the input.
Figure 2 shows the output voltage measured across the resistor of the rectifier circuit before any capacitors were added. The length of time corresponds to one period of the input voltage. The amplitude of each peak is 1.4 V, much less than the 3 volts as measured before. This is probably mostly due to a loss of voltage in the transformer. The first peak is cause when the current is biased forward through one diode, but reverse biased through the other. The second peak comes when the current switches direction, since it is an alternating current, and the biases switch on both diodes.
When the 5000 pF capacitor was placed in parallel with the resistor, we noticed that the peaks smoothed out and began to mesh together out the bottom, and the downward slope of the sinusoid gained a concavity common to exponential decay. For the .1 μF capacitor, we found that the peaks were not as high, nor did they travel all the way back down to the baseline. Rather, there was only around a 40% drop in the voltage between peaks. For the 1 and 2 μF capacitors, the oscilloscope showed a straight line; we could not discern any drop in voltage although we knew there must have been. This is a voltage rectifier; it had taken an AC input and transformed it into a DC output. The theory behind this is simple. As the input voltage rises, it not only causes a voltage drop across the resistor, but also stores energy in the capacitor as well. After the voltage peaks and begins to decline, the capacitor starts to discharge its stored voltage across the resistor at an exponential decay rate. It stops discharging when it hits the next peak in voltage, caused by the other diode. Then it charges up again and repeats the process. How quickly it discharges depends on the time constant of the RC loop, and since the resistor is fixed, it depends entirely on the capacitance of the capacitor. The 5000 pF capacitor was too small to make any major change in the waveform, while the 1 μF capacitor was so large that no drop between peaks was noticeable.
Hence, our results fit the theory very well.
Part B:
Methods:In this part of the lab, we first had to set up a simple DC voltage divider using two resistors in series, as done previously in lab 1. We used the DC power source to generate a constant input voltage. A multimeter was used to measure the voltage drop across R2. We took the current through the load to be zero amps since there was no load. We then attached a load resistor in parallel with R2. Again, we used the multimeter to measure the voltage drop across R2, but we also measured the current through the load resistor. These values were plotted and connected by a straight line, since the relationship is linear, and we calculated the slope of the line, which corresponded to the output impedance.
Diagram 3: Simple Voltage Divider
We then made a more complicated voltage divider by attaching the base of an npn transistor to the node between R1 and R2. A resistor, Re, was placed leading from the emitter terminal to the ground, while the collector was simply connected to the input line. In this more complicated divider, the output voltage was measured across Re, and again the current through the load was taken to be 0, initially. We thenattached a load resistor in parallel with Re, and again measured the voltage drop across it and the current through it. We graphed this data on the same plot as the graph of the simple voltage divider, and compared them. Again, since the form is known to be linear, we simply connected the two points. We found the output impedance in the same manner as before.
Diagram 4: Voltage Divider with Transistor
Next, rather than attach the load resistor across Re, we placed different load resistors in the connection between the collector and the voltage source, as shown in Diagram 5, and used the multimeter to determine the output voltage through it. This configuration should act as a simple current source through the load; hence, we calculated and plotted the current through the resistor in each case.
Diagram 5: Schematic for a Simple Current Source through the Load
Results:
Table 5: Element Data for Simple Voltage Divider
Input Voltage / 12.03 VR1 / 1011 Ω
R2 / 1001 Ω
Load Resistor / 517 Ω
Table 6: Current and Voltage for Simple Voltage Divider
Current (mA) / Output Voltage (V)0 / 5.98
5.82 / 3.03
Output Resistance = -dV/dI = - ΔV/ ΔI = -(-2.95 V/5.82 mA) = 506 Ω
Table 7: Element Data for Transistor Voltage Divider
Input Voltage / 12.03 VR1 / 1011 Ω
R2 / 1001 Ω
Re / 1023 Ω
Transistor Type / 2N3904
Load Resistor / 517 Ω
Table 8: Recorded Currents through Various Elements
Element / Current through It (mA)R1 / 5.96
R2 / 5.94
Base / .0413
Emitter / 5.15
Collector / 5.15
Table 9:Current and Voltage for Transistor Voltage Divider
Current (mA) / Output Voltage (V)0 / 5.31
9.98 / 5.25
Output Resistance = -dV/dI = - ΔV/ ΔI = -(-.06 V/9.98 mA) = 6 Ω
Table 10: Element Data for Current Source
Input Voltage / 12.03 VR1 / 1011Ω
R2 / 1001 Ω
Re / 1023 Ω
Transistor Type / 2N3904
Table 11: Current through Load Resistor in Current Source
Load Resistance (Ω) / Voltage Drop across Load (V) / Current through Load (mA)517 / 2.65 / 5.1
1006 / 5.15 / 5.1
5080 / 7.96 / 1.6
10000 / 8.21 / .8
Analysis and Conclusions:As you can see from the data in Table 8, there is very little current flowing through the base connection, while the emitter and collector currents are nearly identical. This matches up perfectly with the theory. Figure 3 shows the graphs of output voltage vs. output current for both the simple voltage divider and the transistor voltage divider. This data can be found in Tables 6 and 9, respectively. As you can see, as the current through the load increases, the output voltage decreases quite dramatically for the simple circuit compared to only very slightly for the transistor circuit. In other words, the output resistance is much greater in simple divider than the output resistance in the transistor circuit. Hence, the transistor circuit is a much better voltage divider than the simple divider since there is very little dependence of the output voltage on the load resistance.
Figure 4 shows the graph of Current vs. Resistance of the load resistor when the load was placed in the collector connection. Theory states that the current through this connection is supposed to be roughly the same as the current through the emitter. Hence, it should be a simple constant current source. As you can see from the graph and the data in Table 11, there is a near constant current flowing when the resistance is small. However, it drops drastically when the resistance is increased from 1000 to 5000 Ω. To explain this apparent problem, we need only to apply Ohm’s Law and the Law of Conservation of Energy. By Ohm’s Law, if the current had kept its constant value of 5.1 mA, the voltage drop across the load with a resistance of 5080 Ω would have been 25.9 V. However, this is impossible since the input voltage of the circuit was only 12 V, a violation of conservation of energy. Hence, the constant current can only be applied to small resistances.