A
Diode Rectifier Problem
Problem
Diode4P00Int
This is a guided problem to help you with diode rectifiers.
-
Click
here for the circuit simulator for a rectifier using a zero-threshold
diode.
-
Click
here for the circuit simulator for a rectifier using a diode with a
threshold.
There are some fundamental behaviors of rectifier circuits that you will
need to understand. This interactive problem is designed to let you
see and understand how some simple rectifier circuits work.
Problems
Start the first filter simulator - using a zero-threshold diode.
In the simulator, we assume the following.
-
The circuit is the one
shown below.
-
In this circuit, note
the following: (Click
here to go to a point in the electronic lessons where this circuit
is discussed in more detail.)
-
The input is assumed to
be a sinusoidal signal. If you want a square wave signal or a triangular
signal, you'll have to look elsewhere.
-
Initially, when the input
signal is positive, current flows through the diode, D, and some of the
current flows into the capacitor, C, charging it, and some of the current
flows through the resistor, R.
-
When the input voltage
signal drops below the voltage on the capacitor current ceases to flow
through the diode. At that point, however, current flows out
of the capacitor, through the resistor, discharging the capacitor, reducing
the voltage across the capacitor, i.e. reducing the output voltage.
During the discharge part of the cycle, the discharge follows a decaying
exponential since the resistor-capacitor circuit is effective de-coupled
from the input circuitry because the diode is not conducting.
-
As the sine wave cycles
through its cycle, it eventually becomes larger than the capacitor voltage,
and current again flows through the diode, recharging the capacitor.
-
The simulator shows the
operation - as described above - starting from zero voltage across the
capacitor. The simulator runs once after you click the Start
button.. To re-run, click the Clear
button, then click the Start
button.
The output of the circuit fluctuates, but it has a non-zero average value.
That's what this kind of circuit is used for. It gives you a DC source
when you have an AC source. There are two things to remember.
-
The value of the DC level
is just the average value of the output. If there were no discharging
of the capacitor, then the average value of the output would be the peak
level of the input.
-
The DC signal that comes
out of this circuit is not "pure". It has a fluctuating component
riding on top of the DC level. One way to specify the level of impurity
is to give the Peak-to-Peak value of the fluctuation, i.e. the total range
of the fluctuation from the minimum point to the maximum point.
Now, answer the following question.
Question
1
Run the first simulator (the zero threshold diode) and observe the output.
Before you run the simulator, set the time constant to 50 seconds.
Is the output constant at the peak level of the sine wave?
Question 2
Now, run the first
simulator again, but change the time constant to 5 seconds. Is the
output constant at the peak level of the sine wave?
Question 3
Now, answer this question.
What is the peak-to-peak variation in the first signal?
Problem 1
For the second simulation,
estimate the peak-to-peak value of the signal. Don't get fancy, just
aim for the closest half of a volt.
There are some important points to note, so don't close the simulator yet.
-
When the current flows
through the diode - in the charging portion of the curve - the output voltage
is equal to the sinusoidal input voltage because we have assumed that there
is no voltage across the diode. (We assumed a zero
threshold.)
-
You can estimate
the DC voltage output (i.e. the average voltage) by taking the maximum
voltage and the minimum voltage and averaging. This would be the
average if both the charging and discharging portions of the output signal
curve were straight lines. However. they are not straight, but they
are close enough to being straight that it is a good approximation for
what we have done above.
Now, let's examine a slightly different situation.
Question
4
Re-run the simulator with a time constant of 0.5 seconds keeping everything
the same as before. Examine the curve to determine if a straight
line approximation is appropriate, then click on the button of your choice
below.
Problem 2
Now, re-run the simulator with a time constant of 5 seconds and estimate
the DC value (average value) of the output signal. You might
want to check what this button has to say.
Now we are going to consider another question. For this question,
use the second simulator, the
one that simulates a diode with a threshold - more like a typical, real
diode.
Question
5
Run the simulator with a time constant of 20 seconds keeping the default
values for everything. How does the peak value change?
You should also note the following:
-
The output voltage does
not begin to change (the capacitor does not begin to charge) until the
input voltage exceeds the capacitor voltage by the amount of the threshold
- which is 0.6 volts in the simulator.
-
When the input voltage
drops below the capacitor voltage, the capacitor begins to discharge -
as before.
Finally, we have one last thing for you to do - one final observation.
If the capacitor is small, then the time constant becomes small.
Run the first simulator with a
very low value of the time constant. Let's arbitrarily choose a time
constant of 0.01 seconds. Run the simulator.
Question 6
Which of these statements is true?
Question
7
Run the simulator with a time constant of 0.1 seconds keeping the
default values for everything else. Now, do you see exponential decay?
Question
8
Run the second simulator with
a time constant of 0.01 seconds keeping the default values for everything
else. Do you see the half sine you saw with the no-threshold simulator?
Again, there are some conclusions to be drawn.
-
The time constant can
be critical. The time constant needs to be substantially longer than
the period of the sine wave in order for there to be little "droop" in
the output curve. In practical power supplies that might mean pretty
big capacitors, depending upon the load resistor.
-
If the time constant is
too small, in the limit it might as well not be there since you get the
output curve you would get without a capacitor. The capacitor is
necessary for good filtering, i.e. getting a good DC signal out of the
circuit.
That's it for this problem. We hope that you have learned a little
bit about this basic power supply circuit, and that you enjoyed answering
these questions and using the simulators.