For a single stage BJT common base amplifier.
∴ Voltage gain can be greater than unity but current gain is always less than unity.
A bipolar transistor has an emitter current of 1 mA. The emitter injection efficiency is 0.99, base transit factor is 0.995 and depletion region recombination factor of 0.998. The base current flowing through the transistor is _____μA.
The common emitter current gain is given by:
α = γ ∗ β ∗ δ
= 0.99 × 0.995 × 0.998 = 0.983
Collector current
I_{c} = αI_{E} = 0.983 mA
Base current
I_{B} = I_{E} – I_{C}
= 17 μA
In a Bipolar junction transistor the base width is 0.54 μm and base diffusion constant is D_{B} = 25 cm^{2}/sec. The Base transit time time is ________ × 10^{10} sec
The Base transit time
⇒ 0.5832 × 10^{10}
Find the value of bias resistor (in kΩ) if quiescent collector current and voltage value are 4.6 mA and 2.2 V. The transistor has DC gain 110, V_{BE} = 0.7 V and V_{CC} = 4.5 V.
Applying KVL in collector to emitter loop
V_{CC} = R_{L}I_{C} + V_{CE
}
Now applying KVL in collector to base loop
I_{B}R_{B} + V_{BE} = V_{CC }
Consider two pnp bipolar junction transistors. For the first transistor when emitter to collector voltage is 5 V, V_{EB} is 0.85 V and emitter current is 10 A. (The β for this transistor is 15). Second BJT conducts with a collector current of 1 mA and V_{EB} = 0.70. The ratio of emitterbase junction area of the first transistor to the second transistor is ______.
(Assume KT = 26 mV)
Given: V_{EC} = 5V^{}, this means pnp transistor is operating in the active node
The leakage current of a transistor with usual notation are I_{CEO} = 410 μA, I_{CBO} = 5 μA, and I_{B} = 30μA. Calculate the I_{C} ________mA
I_{CEO} = 410 μA, I_{CBO} = 5 μA, I_{B} = 30 μA
I_{C }= βI_{B} + (1 + β) I_{CBO}
I_{CEO} = (1 + β) I_{CBO}
410 = (1 + β)5
β = 81
I_{C} = 81 × 30 + 82 × 5
I_{C} = 2.84 mA
For what value of current gain β , the given transistor will be in saturation
(Assume V_{in} = 5V, V_{BE(SAT)} = 0.8 V , V_{CE(SAT)} = 0.2V)
Apply kVL in base emitter loop
at saturation V_{BE(sat)} = 0.8
substituting the value of V_{BE}
I_{B} = 0.0525 mA
Apply Kvl , Form the collector to emitter
5 × 10^{3} × I_{C }+ V_{CE} = 12 V
= 2.36 mA
At the edge of saturation
Hence for β value greater than 45 the transistor will be in saturation
An npn bipolar transistor having uniform doping of N_{E} = 10^{18} cm^{3} N_{B} = 10^{16} cm^{3} and N_{C} = 6 × 10^{15} cm^{3} is operating in the inverseactive mode with V_{BE} = 2V and V_{BC} = 0.6 V. The geometry of transistor is shown
The minority carrier concentration at x = x_{B} is _____ × 10^{14} cm^{3}
(Assume n_{i } = 1.5 × 10^{10}/cm^{3}, V_{t }= 25 mV)
= 5.96 × 10^{14} cm^{3}
The common emitter forward current gain of the transistor shown is β = 100
The transistor is operating in
Assume Transistor in Active region
20 – 2 (I_{B} + I_{C}) – 0.7 – 250 I_{B} – 10 = 0
20 – 2 × 101 I_{B} – 0.7 – 250 I_{B} – 10 = 0
10 – 452 I_{B} – 0.7 = 0
20 – 2 (I_{B} + I_{C})  V_{EC} – 2I_{C} = 0
20 – 202 I_{B}  V_{EC} = 0
V_{EC} = 11.68 V
V_{CB} = V_{CE} + V_{EB }= V_{EC }+ V_{EB }=  11.68 + .7
V_{CB} = 10.98
∴ Collector base junction is Reverse Biased
∴ Transistor is operating in forward active region
Second Method
Assume transistor in saturation region
I_{B} min ≤ I_{B}
i) V_{EB (sat)} = 0.8 V
ii) V_{EC (sat) }= 0.2 V
I_{C} ≠ βI_{B}  Because it is valid only active region
20 – 2 (I_{B} + I_{C}) – 0.8  250 I_{B} – 10 = 0
20 – 2 I_{B} – 2I_{C} – 0.8 – 250 I_{B} – 10 = 0
10 – 252 I_{B} – 2I_{C} – 0.8 = 0
252 I_{B} + 2I_{C }= 9.2  (1)
20 – 2 (I_{B} + I_{C}) – V_{EC(sat) }– 2I_{C} = 0
20 – 2I_{B} – 2I_{C} – 0.2 – 2I_{C} = 19.8  (2)
By solving equation (1) and (2)
I_{B} = 0.00278 mA
I_{C} = 4.95 mA
∴ I_{B(min) }> I_{B}
∴ Transistor is operating in forward active region
In a silicon PNP transistor the mobility of charge carries is μ_{4} = 1300 cm^{2}/Vs and μ_{p} = 450 cm^{2}/Vs and carrier life time τ_{p} = 0.10 μs. The most appropriate base width for effective transistor function is (Take T = 300° k)
Given μ_{n} = 1300 cm^{2}/Vs and μ_{p} = 450 cm^{2}/Vs
D_{p} = 450 × 0.0259
= 11.655 cm^{2}/s
Base diffusion length = 1.08 mm
For injected holes from emitter to reach the collector they should not be recombined in base. Hence the base width should be very less compared to base diffusion length.
W ≪ L_{P}
In all options the base width is significant compared to base diffusion length.
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