PNP Junction Diode Problems and Solutions

This assignment covers a range of topics related to PNP junction diodes, including Knee Voltage, V-I Characteristic Curve, Temperature Effect, and more. It provides detailed solutions to problems such as calculating base current, understanding the significance of Knee Point Voltage, and analyzing the effect of temperature on PN junction diode characteristics. The document is suitable for students studying electronics or semiconductor physics.

Problem 1:
Solution:
Condition 1: Given data: R1=12KΩ, R2=3.3KΩ, RC=1.2KΩ, RE=330KΩ, β=300, VCC=10V
Assume VBE, VBC =0.7V, VCE(SAT)=0.2V
To Find: VC, VB, VE
Situation 1: The resistor R1 is opened
Circuit diagram:
VCC –ICRC - VBC – IBRB =0 (Apply Kirchhoff's voltage law)
VCC –βIBRC - VBC – IBRB =0
Hence, IB =( VCC - VBC ) / (RB + βRC ) ------ Equation 1(Here RB is R2)
Applying the values to Equation 1 we get,
IB = 2.55×10-5 A
Hence, IC = βIB = 7.65 mA
IE ≈ IC
Ans: VE = IERE =2.508 V, VB = VE + VBE =3.208V, VC = VCC – ICRC = 0.88V

Situation 2: The resistor R2 is opened
Circuit diagram:
VCC - VBE – IBRB – IERE = 0 (Apply Kirchhoff's voltage law) ------------------------- Equation 2
IE = IB + IC = IB + βIB = (β +1) IB (Since IC = βIB) ------------------------------------- Equation 3
Substitute Equation 3 in Equation 2 we get,
VCC - VBE – IBRB - (β +1) IBRE = 0 (Here RB is R1)
Hence IB = (VCC - VBE) / (RB + (β +1) IBRE) ----------------------------------------------------------- Equation 4
From Equation 4 we get
IB = 8.35 × 10-5 A
IC = βIB = 0.025A
IE ≈ IC
Ans: VE = IERE = 8.25 V, VB = VE + VBE = 8.95V, VC = VCC – ICRC = -20V.
Situation 3: The resistor RC is opened
Circuit diagram:

The thevenin voltage is given by the formula
Vth = (VCC × R2) / (R1 + R2)
Hence Vth = 7.84V
Rth = (R1×R2) / (R1+R2)
Hence Rth = 2588 Ω
Now, from the thevenin’s circuit we can get
Vth – IBRth –VBE –IERE = 0(Apply Kirchhoff's voltage law) ------------------------- Equation 5
IE =IB + IC (Here IC = 0)
Therefore Equation 5 becomes
Vth – IBRth –VBE – IBRE = 0
IB = (Vth –VBE) / (Rth + RE) ---------------------------------------------------------------- Equation 6
Substituting the given values in Equation 6 we get
IB = 2.4mA
IE = IC + IB = 2.4 mA( since IC = 0)
Ans: VE = IERE = 0.8074 V, VB = VE + VBE = 1.507V, VC = 0V.

Situation 4: The resistor RE is opened
Circuit diagram:
The values of Vth and Rth can be taken from the situation 3
Vth – IBRth –VBE = 0(Apply Kirchhoff's voltage law)
IB= (Vth–VBE) / Rth ----------------------------------------------------------------- Equation 7
Substituting the given values in Equation 7 we get
IB = 2.75mA
IC = βIB = 0.827A
IE = 0
Ans: VE = IERE = 0 V, VB = VE + VBE = 0.7 V, VC = VCC – ICRC = -982.4V.
Problem 2:
Solution:

The condition of different voltage levels of PNP junction diode to operate in the active
mode is that the Emitter-base junction should be forward biased and Collector-base junction
should be reverse biased. The voltage condition is V3 > V2 > V1. This condition is applicable to
the forward active region. For reverse active region the condition is exactly the opposite i.e., V3
< V2 < V1.
Problem 3:
Solution:
For the first circuit, when the diode is forward biased the diode acts as a short circuit and at the
voltage side of the voltage the diode is reverse biased and it acts as a open circuit.
Resistance=50Ω Voltage=±5 V Current(Amps) I=V/R
50 0 0
50 0.7 0.014
50 2 0.04
50 5 0.1
50 -5 0
OUTPUT WAVEFORM:
For the Second circuit the diode is in the opposite position hence during the positive half voltage
the diode is reverse biased and hence it acts as an open circuit. At negative voltage the diode acts
as a forward bias and hence the diode acts as a short circuit.

Resistance=2Ω Voltage=±20 V Current(Amps) I=V/R
2 0 0
2 20 0
2 -1 0.5
2 -10 5
2 -15 7.5
2 -20 10
Output Waveform:
Problem 4:
Solution:
a) KNEE VOLTAGE:
The diode consists of holes at P junction and electrons at the N junction. Due to
recombination process the electron hole pair will appear. When a moving electron combines with
a hole, both holes and electrons vanishes and leaving behind an immobile positively charged
donor on the N side and negatively charged acceptor on the P side. This tends to occur rapidly
and finally we get a depletion layer which will at last stop this recombination. When a voltage
source is connected in forward bias, a minimum voltage of 0.7V for Silicon diode and 0.3V for
Germanium diode is rEquationuired to cross the depletion layer. At this minimum voltage the
diode starts to conduct and the current increases exponentially which is termed as CUT IN

VOLTAGE or KNEE VOLTAGE. The graph of the exponentially increasing diode look like
the structure of our leg at knee so called this point as knee voltage. The significance of Knee
Point Voltage can be seen in Current Transformer which is mainly used for protection purposes.
b) V-I CHARACTERISTIC CURVE:
c) TEMPERATURE EFFECT:
The increase in temperature on the PN junction diode tends to decrease the width of the
depletion layer. Due to this the barrier potential will be decreased at an approximate rate of 2 mV
per degree Celsius.
Problem 5:
Solution:
Given data: VCC=5 V, VBB=2 V, RB= 250KOhm, RC= 4KOhm. Assume VBE = 0.7V (for silicon)
To Find: Base current

Applying KVL for the first loop, we get
VBB – IBRB – VBE =0
Hence, IB = VBB - VBE / RB
= 2 – 0.7 / (250000)
Ans: IB = 5.2μA
REFERENCES:
1) S. Huseinbegovic, B. Perunicic-Draenovic, N. Hadimejlic, Discrete-time sliding mode direct power
control for grid connected inverter with comparative study, IEEE 23rd International Symposium
on Industrial Electronics (ISIE) (2014).
2) Nam Sung Kim A. D, Ann Arbor S. Hu Mary Jane Irwin, “Leakage Current: Moore‟s Law Meets
Static Power,” Published by the IEEE Computer Society, 2003.