Download Methods Part A (Diode without Transformer)

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.