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Full wave rectifier circuit
In today’s laboratory, you will design and build a rectifier circuit. This rectifier circuit
will be tested on a bread board, then it will be reassembled on a vero board and soldered
together so that it can be used to power future laboratory exercises.
The power that we receive from the power lines is transmitted via a 240V AC 50 Hz
signal. Unfortunately most electronics that we will wish to use requires that we have a
DC supply, clearly if we want a cheap way to power our electronics we will need to
convert the AC signal to DC. This is what our rectifier circuit does. The basic building
block for this task will be the trusty diode, the diode acts as a valve, only allowing current
to flow in one direction (see Fig. 1).
Figure 1: The circuit diagram
symbol for a diode. Current flows
when it is positively biased.
Task (1)
Connect the diode in series with a resistor and a signal generator (set to 10 V p-p and
frequency 1000 Hz sinusoidal wave). Draw the circuit diagram and display the input and
output simultaneously on the CRO, sketch this display labeling all axes. (Hint: you want
to measure the voltage across the resistor and the signal generator.)
If all went well you will notice that you have two different signals, however you will also
notice that they aren’t that useful on their own, you have lost half the wave form and it
still resembles a sinusoid. What we need to do now is smooth out the wave, how we do
this is to use a charge storage device in short a capacitor.
Task (2)
The voltage we measured was across the resistor so that should tell you the most logical
place to insert a capacitor. Draw the circuit diagram for a half wave rectifier and predict
the voltage across the load resistor.
Pen and paper time now folks. In order to keep cost and size down manufacturers would
use the smallest capacitor possible to smooth out the power ripple on their rectified
circuit. We also want to follow the same principle so we will want to evaluate what time
constant we need for the capacitor in our circuit and still keep the ripple current small.
Here we now analyze a capacitor; the current across a capacitor is given by the following
differential equation
where t is the time and Q the charge stored on the capacitor. The charge stored on a
capacitor is given by
Q  CVc ,
where C is the capacitors capacitance and Vc is the voltage across the capacitor. Thus for
the circuit shown in Fig. 2 and using Kirchoff’s voltage law (KVL) we have,
dQ Q
Vs  R
 .
dt C
Task (3)
Explain each term in Eq 3. (i.e. identify what component each term is associated with)
and solve the differential equation.
Help with Task (3)
Step (1)
Step (2)
Step (3)
Step (4)
Step (5)
Transpose equation (3) to isolate the differential term.
Multiply both sides by dt
Group all terms with the variable Q on one side and the
remainders on the other
Integrate both sides with respect to t use the identity
 (a  bx) dx  b log e (a  bx) and evaluate from t to t0
Transpose to obtain Eq. (4)
Figure 2: Capacitor
charging circuit.
Solving Eq. (3) yields the equation,
Q  CVs 1  e  ,
which describes the amount of charge Q stored on a capacitor for an applied voltage Vs.
If we assume that a fully charged capacitor begins to discharge when the diode stops
conducting thus ‘filling’ the missing voltage. So to begin we remove the voltage source
Vs , and solve Eq. (3) again but with some extra a-priori knowledge that the general
solution to this Equation is,
Vc (t )  Ae pt  B
Task (4)
Obtain the discharge equation by substuting in Eq. 5 into your modified Eq. 3 (Hint: as
t→∞, Vc→0, hence you have a value for B. It will also be easier if you use the identity
provided in Eq 2 to make your modified Eq 3 a function of Vc rather than Q. This allows
us to find an expression for p). Finally, verbally justify what value A should be.
Figure 3: Capacitor
discharging circuit.
If we sketch the result of Task (4) we can use the standard definition for the time constant
which is the time taken for the voltage across the capacitor to drop to 1/3 its initial value
and an indication of the capacitance value we should use for our circuit. Here we state it
is just given by the equation τ=RC (however it can be calculated from the result of Task
Task (5)
Chose a capacitance for your capacitor with justifications and construct the amended
rectifier circuit. Measure the voltage across the load resistor of the amended rectifier
circuit and sketch the output displayed on the CRO.
Being environmentally friendly we see that the input has a negative component which
when we use a half wave rectifier would imply that we are wasting 50% of the output
power of our transformer! We now require the use of a diode bridge (remember 1st year
labs anyone?).
Task (6)
Add in extra diodes to make the diode bridge with the smoothing capacitor.
Here is the clincher, when you buy ‘off the shelf’ electronics i.e. transistors; they actually
require a positive and negative voltage rail as well as a ground or earth point. Ask your
demonstrator for a transformer pack which has three output pins and deduce which pin is
what (extra marks if you can identify why this is redundant for two of the pins).
Task (7)
Using the transformer and your previous designs, construct a ±V power supply with a
suitable ground reference point. Draw the circuit diagram FIRST (Hint: you will need to
add in an extra capacitor).
Task (8) (almost there)
Get a vero board from the demonstrator, build the circuit and solder away.
Breadboard connections
The solid lines indicate the connections. Please note that the boards you are using have a
break in the middle.