Principle of Electronics Coursework I
Daniel Ruiz Blanco k20055658
November 25, 2021
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Problem 1
This problem involves the understanding of the baseline theory behind diodes.
For the circuit showing in Fig.1 below let us assume that the diode is ideal.
Draw the current-voltage characteristic (V-I) curve assuming that R1 = 25kΩ
and R2 = 25kΩ.
The circuit above can be simplified to a single resistor and a diode. The new
resistor will have a value of:
R3 =
R1 ∗ R2
= 12.5kΩ
R1 + R2
Now the circuit looks like:
In this circuit the diode acts as a switch (as we assumed it is ideal). Therefore
the I-V curve in this circuit will follow the Reverse Bias trend up to a voltage
level of approximately 0.7V, meaning that there will be no current flowing until
the voltage reaches that level. After it reaches around 0.7V, it will enter the
Forward Bias zone, in which the current spikes, turning the diode into a “wire”,
meaning current will flow freely without getting any resistance or block.
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Therefore the I-V curve for the diode will be:
Current won’t start flowing until the voltage across the diode goes over the
threshold value. The I-V curve of the resistor is just a straight line, where
the current is directly proportional to the voltage. The gradient of the line is
1
R.
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Problem 2
Consider the circuit configuration as the one shown in Figure 2 below. Assuming
that the diodes are ideal find the states of the diodes (i.e., if they are ON or
OFF). Assume that VA = 5V , VB = VC = 3V , R1 = 10kΩ and R2 = 10kΩ.
To analyse this circuit, we need to find the states of the diodes, whether
they’re on or off. To find it out, we must do assumptions about the circuit.
We know that when a diode is ON, there is current flowing through it.
Therefore, we can say that if the voltage across VA and VC is bigger than 0,
then the diode is ON. Otherwise, the diode is OFF.
We need to analyse the circuit when each of the diodes is OFF, so I will first
assume D1 is off, and then I will procceed with D2 .
For D1 : VC can be safely ignored, as the cathode blocks the incoming voltage
due to the forward bias. Therefore in the equation I will only consider VA and
VB .
Potential across the now open circuit space, which would be D1 is VA − VB ,
as the batteries’ anodes are facing the same direction. We obtain a voltage
across D1 of 2V , which is greater than 0V , therefore confirming us that the first
diode is indeed in an ON state.
For D2 : Voltages VA and VB add up, as they are both going towards D2 . Now
the currents in these first two lanes are:
iA =
5V
= 0.5mA
10kΩ
iB =
3V
= 0.3mA
10kΩ
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Then the current delivered to D2 is of 0.8mA. The total resistance is the
sum of the two parallel resistors R1 and R2 :
Req =
R1 ∗ R2
= 5000Ω
R1 + R2
Now, following Ohm’s law:
V = iR −→ V = 4V
As VC is in the opposite direction, the voltage across D2 will be
V = 4V − 3V
Which means that D2 is also on, therefore doing the assumptions of both
being OFF is not neecsary.
We can conclude saying that the state of the diodes is ON for both of
them.
Excercise 3
Consider the circuit configuration as shown in Fig. 3 below and an input
sawtooth-like voltage that we apply as an input signal. Assume that the peak
voltage of the input signal (Vp ) is equal to 9V . Calculate and plot that output
voltage for this configuration. Please provide full explanations.
The circuit displayed in the left side of Figure 3 shows us a capacitor and
a diode. As we can see, during the phase where the sawtooth waveform has a
positive voltage, the diode blocks the current, as it is going towards the cathode.
This means that durign the time where the voltage is positive, the capacitor
is getting charged (Capacitor does not get discharged as . When discharged,
both plates of the capacitor have an equal amount of positive and negative
charge. Once the positive voltage phase starts, the left plate starts to get charged
positively, therefore pushing the electrons towards the right plate. This means
that the polarity of the capacitor will add up with the voltage of the source once
it enters into negative voltage.
We can consider the capacitor as a “doubler” of the voltage whenver voltage’s
negative. Once voltage is positive again, the capacitor gets charged, ready to
release the energy in the next cycle.
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As the voltage source has a peak of 9V , then the capacitor gets charged up
to 9V , which are not discharged as it faces the cathode. When the source turns
into negative voltage, it outputs a max of 9V , but this value is doubled as the
capacitor also discharges its energy. Therefore we obtain a sawtooth like output
waveform that will look like:
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