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MA2009: Tutorial 9 (Diodes and its applications)- Part II
(Forward: b, d, e; reverse: a, c)
Answer:
(a) VD = 0-5V = -5V < 0, reverse biased.
(b) VD = 10V-5V = 5V > 0, forward biased.
(c) VD = -10V-5V = -15V < 0, reverse biased.
(d) VD = -5V-(-12V) = 7V > 0, forward biased.
(e) VD = 0-(-10V) = 10V > 0, forward biased.
(a: 0.318A; b: -70.7V)
Answer:
(a) When forward biased, maximum voltage from the source will be:
Vmax = 50 × sqrt(2) = 70.7 (V) 70.7 V – 0.7 V = Imax 220Ω , Imax = (70.7-0.7)/220 = 0.318 (A)
(b) When reverse biased, the diode is closed, there is no current passing through the circuit,
then all the voltage will drop on the diode, so the peak reverse voltage will be:
Vmax = -70.7 V
(no need to consider Zenner breakdown)
Question 3. The diode shown in Figure 3 has a piecewise linear characteristic that passes
through the points (0 V, 0 mA), (2 V, 1 mA), (5 V, 10 mA), (8V, 25 mA), and (10 V, 50 mA). (Exam
question of
2012S2)
(i)
(ii)
(iii)
Draw I-V characteristic of diode using these points;
Determine the load line equation;
Find the values of i and υ (Q-point).
Figure 3
(iii: 18mA and 6.3V)
(2006S1 exam question)
Answer:
(i)
Positive cycle upper diode ON, negative cycle lower diode ON. In both cases,
current flows downwards, and I = Vs/RL
(ii)
The efficiency is doubled.
(813Ω)
Answer:
IS = (VS-VZ)/1800 = (18-5.6)/1800 = 6.89 mA
IL = VZ/RL = 5.6/RL
The current passing through Z should be:
IZ = IS – IL = 6.89 × 10-3 – 5.6/RL
This current must be larger than 0
6.89 × 10-3 – 5.6/RL >0
RL > 5.6/(6.89 × 10-3) = 813 Ω
Optional Questions:
(a: Vin > 0; b: iD1= 0 (when Vin < 0V); iD1 = Vin / 2k (when 0<Vin <10V); iD1= 5mA (when Vin >=10V))
Answer:
(a) As long as Vin > 0, VD1 will be larger than 0 (diode D1 will be forward-biased) no
matter D2 is forward or reverse biased.
(b) The current flowing D1 not only depends on Vin, but also depends on the on/off state of
D2.
When VA<5V (or Vin <10V) D2 is off, iD1 = Vin / 2k
When Vin >=10V, D2 is on, and VA will be locked at 5V, so iD1= 5V/1k=5mA.
Of course iD1= 0 when Vin < 0V
(hints: 1. Voltage transfer characteristic means the output voltage waveform in responding the given
input voltage; 2. Output is 4V when diode is on or Vs<8V, otherwise equal to Vs/2)
Answer:
The diode has two states: on and off.
When it is on, output voltage will always be 4 V, ie., υo(t) = 4 V
When D is off, the circuit is composed of one voltage source and two resistors, and the
output voltage will be: υo(t) = υS(t) 10/(10+10) = 5 sin(2.000πt)
Now we look the conditions for on/off of the diode. The voltage across the diode is:
VD = 4 V - υo(t)
When VD < 0, the diode will be off, and at this moment,
VD = 4 V - υo(t) = 4 - υS (t) 10/(10+10) = 4 - υS (t)/2
Since VD < 0 now, so 4 - υS(t)/2<0, which means υS(t) > 8 V
Conclusion:
when υS(t) > 8V, diode will be off, and the output υo(t) = υS(t) 10/(10+10) = 5 sin(2.000πt)
All other conditions, the diode will be on, and the output will be υ o(t) = 4 V.
(hint: in part a it could be considered as using large signal model for offset diode)
(a: 0.6 V and 0.5 mA; b: 0.975 V and 0.46 mA)
Answer:
The diode is made of silicon, so the offset voltage is about Vγ = 0.6 V
Q-point is DC operation point, AC could be neglected. Also the peak value of AC component
of Vs is 0.11V, smaller than 0.6V. It will have no impact on the ON/OFF of diode.
The Vd in diode equation should be the voltage on PN junction, which is apparent diode voltage
deducting Vγ
(a) Using offset model, diode’s I-V curve is a straight vertical line at VD = Vγ = 0.6 V.
Load line: VS-VD = iD R, or iD = VS/R – VD/R
iD = 6.0× 10-4 –VD/7000 (A)
So the load line is a line passing through points (0 V, 0.6 mA) and (4.2 V, 0 mA)
Q-point for this diode will be VD = 0.6 V, and iD ~ 0.5 mA
(b) Using iterative method: by changing the VD, we calculate the current for the diode using
diode equation and the load current using load equation. Increase VD if calculated load current
is larger than diode current, while decrease VD if calculated load current is smaller than diode
current. Repeat the calculation until these two current equal to each other. For example:
Step 1: we arbitrarily take VD =1.1 V, then
Load current iL = 6.0× 10-4 –VD/7000 = 6.0× 10-4 –1.1/7000 = 0.443 mA
Diode current iD = 250 × 10-12 × {exp[(1.1-0.6)/0.026]-1} = 56.2 (mA)
Load current is smaller than diode current, so we need decrease VD
Step 2: we now take VD =0.7 V, then
Load current iL = 6.0× 10-4 –0.7/7000 = 0.5 (mA)
Diode current iD = 250×10-12 × {exp[(0.7-0.6)/0.026]-1} = 1.14×10-5 (mA)
Load current is larger than diode current, we need increase VD
Step 3: repeating….
Step 4: we now take VD =0.975 V, then
Load current iL = 6.0× 10-4 –0.975/7000 = 0.461 (mA)
Diode current iD = 250×10-12 × {exp[(0.975-0.6)/0.026]-1} = 0.459 mA
Therefore the Q-point of the diode is: VD = 0.975 V and iD = 0.46 mA
(hint: don’t forget Vγ)
Answer:
VS(rms) = 12 V suppose it is a sin curve,
vs(t) = 12 sqrt(2) sin(2πft) = 17 sin(377t)
When |νs(t)| < 1.2 V, there is no diode in ON state, so the output VL(t) = 0
When |νs(t)| > 1.2 V:
For the positive half-cycle, D1 and D4 ON, and VL(t) = |νs(t)|-1.2
For the negative half-cycle, D2 and D3 ON, and VL(t) = |νs(t)|-1.2