Download Lab 2 - Full wave rectifier

Building the full wave rectifier circuit and passive filters.
Make sure the CRO is on DC coupling
So after last weeks lab, where you prototyped a circuit and understood what mechanism
was behind rectifying a time varying signal today we shall build our own circuit.
If we examine the circuit diagram in Fig. 1 we shall notice some peculiar features not
present in last weeks lab.
+
+
Vout 1
-
Figure 1: Schematic of
the full wave rectifier.
-
Vout 2
+
The input port on the left hand side is going to take the three prong power AC
transformer plug pack and rectify the wave such that we get a useful ±15V DC power
supply. The transformer has two output pins which are out of phase by 180º degrees.
Draw the current flow and hence explain how this enables a negative voltage.
The input port is labeled with a + and – to make the sign of the polarized capacitors
sensible. The polarized capacitors must be biased such that the pin labeled ‘negative’
must be placed at a point of lowest potential. This means that since the negative rail is
more negative than the ground pin we must account for this.
Part II
Last week we solved the differential equation describing the charging and discharging of
a capacitor through a resistive load. We made the implicit assumption that with the
voltage source removed we had a DC circuit. For today’s lab we shall investigate the
capacitors behavior when the placed in an AC circuit.
Background theory.
We begin with the equation for the charge stored on a capacitor (see Fig. 2.1),
Q(t)=CV(t),
where V(t) is the time varying voltage applied.
2.1
Figure 2.1: Current i(t)
+
flowing through the
V(t) capacitor due to the
applied voltage V(t).
i(t)
The current through the capacitor is just the change in the amount of charge stored with
respect to time,
dQ(t )
i (t ) 
.
2.2
dt
Hence the current through the capacitor is,
dV (t )
i (t )  C
.
2.3
dt
Let us now consider the case where V(t)=Vs cos(ωt+φ) such that Vs is the peak value of
the voltage wave, ω is the angular frequency of the wave in radians and φ is the phase
angle. As the cosine function completes one cycle when the angle increases by 2π we
have
and
f=1/T.
2.4
T  2
V (t)
i(t)
+
-
Figure 2.2: Current i(t)
flowing through the
capacitor due to the
applied voltage V(t).
Evaluating the current through the capacitor using Eq’s. 2.3 and 2.4 we have,
i(t )  CVs (t ) sin( t   )
We note that the current lags the voltage by 90º hence we can recast Eq. 2.5 as
i(t )  CVs (t ) cos(t    90)
Using V=IZ we evaluate the impedance of the capacitor as,
2.5
2.6
Z
cos(t   )
cos(t   )

 C sin( t   )  C cos(t    90)
2.6
Using phasors (don’t worry if you haven’t seen this yet as it will come up in class) we
find that the impedance of a capacitor is,
Z
1
.
jC
2.7
The significance of this equation is that it tells us that the capacitor has frequency
dependent impedance at low frequencies (i.e. DC current) the impedance of the capacitor
is infinitely large.
Laboratory:
For this part of the laboratory we shall build and evaluate the frequency response of a RC
circuit, in effect we shall construct a low pass filter. Consider the circuit in Fig. 2.3
Vin
R1
+
C1
Vout
Figure 2.3: Circuit
diagram for the low pass
filter.
Task 1
The transfer function for a filter is given as a ratio of Vout/Vin, calculate the transfer
function of the filter above. (Hint: The impedance of the capacitor can be though of as the
resistance of the capacitor. Knowing this use KVL.)
Once you find the transfer function we can calculate at what frequency the output voltage
will drop to half the input value. Using this fact prove that this frequency is,
1
fb 
.
2.8
2 RC
Task 2
Setup two 47kΩ resistors in series with the signal generator and measure voltage drop
across one of the resistors as a function of frequency. This is known as a Bode plot.
This is constructed by plotting (log) amplitude vs (log) frequency. Set the signal
generator output amplitude to 10V peak to peak. Use Log-log scales. Measure
amplitude at 10Hz, 20Hz, 50Hz, 100Hz,...., 100kHz. Draw a smooth curve
through the measured points.
Task 3
Change one resistor to a 2.7nF capacitor and measure the voltage drop across the
capacitor as a function of frequency (i.e. measure Vout). Predict using Eq 2.8 what the 3dB point, or Half power point also known as the "corner frequency" is and how does
this compare with your measured result?
Task 4 (optional)
Plot the R-C circuit phase vs frequency over the same range 10Hz to 100kHz.
Setup the CRO to plot input Amplitude on the X axis and Output amplitude on the Y
axis. What angle on the CRO is represented as 0 degrees? Find the frequency where the
phase is 15 degrees? 30 degrees? 45 degrees? 60 degrees? 75 degrees? Plot a smooth line
through the measured points.
Appendix
Vero-boards are a cheap way to fabricate circuits. They consist of a series of pre drilled
holes with copper tracks underplayed. They act as a kind of breadboard with permanent
connections. How you choose your component layout is very important for a neat clean
circuit. Shown in Fig. A1 is the plain underside of a vero-board with the connections
clearly evident. The component layout done by student in the 2007 class, and their
soldered connections.
Figure A1: Various vero-board construction stages.