A uniformly doped silicon pnp transistor is biased in the forwardactive mode. The doping profile is N_{E} = 10^{8} cm^{3}, N_{B }= 5.2 x 10^{16} cm^{3} and N_{C} = 10^{15 } cm^{3} .For V_{EB} = 0.6 V, the p_{B} at x =0 is
An npn bipolar transistor having uniform doping of N_{E }= 10^{18} cm^{3 } N_{B }= 10^{16} cm^{3} and N_{c} = 6 x 10^{15} cm^{3} is operating in the inverseactive mode with V_{BE} =  2 V and V_{BC} = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x = x_{B} is
An npn bipolar transistor having uniform doping of N_{E }= 10^{18} cm^{3 } N_{B }= 10^{16} cm^{3} and N_{c} = 6 x 10^{15} cm^{3} is operating in the inverseactive mode with V_{BE} =  2 V and V_{BC} = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x" = 0 is
An pnp bipolar transistor has uniform doping of N_{E} = 6 x 10^{17 }cm^{3}, N_{B }= 2 x 10^{16} cm^{3} and N_{C }= 5 x 10^{14 }cm^{3}. The transistor is operating is inverseactive mode. The maximum V_{CB} voltage, so that the low injection condition applies, is
Low injection limit is reached when
The following currents are measured in a uniformly doped npn bipolar transistor:
I_{nE} = 1.20 mA, I_{pE } = 0.10 mA, I_{nC} = 1.18 mA
I_{R} = 0.20 mA, I_{G} = 1 μA, I_{pC0} = 1 μA
Q.
The α is
The following currents are measured in a uniformly doped npn bipolar transistor:
I_{nE} = 1.20 mA, I_{pE } = 0.10 mA, I_{nC} = 1.18 mA
I_{R} = 0.20 mA, I_{G} = 1 μA, I_{pC0} = 1 μA
Q.
The β is
The following currents are measured in a uniformly doped npn bipolar transistor:
I_{nE} = 1.20 mA, I_{pE } = 0.10 mA, I_{nC} = 1.18 mA
I_{R} = 0.20 mA, I_{G} = 1 μA, I_{pC0} = 1 μA
Q.
The γ is
A silicon npn bipolar transistor has doping concentration of N_{E} = 2 x 10^{18}cm^{3}, N_{B} =10^{17}cm^{3} and N_{C } = 15 x 10^{16} cm^{3}. The area is 10^{3 } cm^{2} and neutral base width is 1 μm. The transistor is biased in the active region at V_{BE} = 0.5 V. The collector current is
(D_{B} = 20 cm^{2}/s)
A uniformly doped npn bipolar transistor has following parameters:
N_{E} = 10^{18} cm^{3} N_{B} = 5 x 10^{16} cm^{3},
N_{c} = 2 x 10^{19} cm^{3},
D_{E} = 8 cm^{2} /s , D_{B} = 15 cm^{2} /s , D_{c} = 14 cm^{2} /s
x_{E} = 0.8 μm, x_{B} = 0.7 μm
The emitter injection efficiency γ is
In bipolar transistor biased in the forwardactive region the base current is I_{B} = 50 μA. and the collector currents is I_{C }= 27 μA. The α is
For the transistor in fig., I_{S} = 10^{15} A, β_{F} = 100, β_{R} = 1. The current I_{CBO} is
= 0
Determine the region of operation for the transistor shown in circuit in question
V_{BE} = 0 , V_{BC} < 0, Thus both junction are in reverse bias. Hence cut off region.
Determine the region of operation for the transistor shown in circuit in question
V_{BE }> 0, V_{BC} = 0, BaseEmitter junction forward bais, Basecollector junction reverse bias, Hence forwardactive region.
Determine the region of operation for the transistor shown in circuit in question.
V_{BE }= 0, V_{BC}> 0, BaseEmitter junction reverse bais , Basecollector junction forward bias, Hence reverseactive region.
Determine the region of operation for the transistor shown in circuit in question.
V_{BE }= 6V, V_{BC =} 3V Both junction are forward biase, Hence saturation region.
For the circuit shown in fig., let the value of β_{R }=0.5 and β_{F} = 50. The saturation current is 10^{16 }A
Q. The baseemitter voltage is
The current source will forward bias the baseemitter junction and the collector base junction will then be reverse biased. Therefore the transistor is in the forward active region
For the circuit shown in fig., let the value of β_{R }=0.5 and β_{F} = 50. The saturation current is 10^{16 }A
Q. The current I_{1} is
The leakage current of a transistor are I_{CBO} = 5μA and I_{CEO} = 0.4 mA, and I_{B} =30 μA
Q. The value of β is
The leakage current of a transistor are I_{CBO} = 5μA and I_{CEO} = 0.4 mA, and I_{B} =30 μA
Q. The value of I_{C} is
I_{c} = βI_{B} + I_{CEO} = 79(30μ) + 0.4m = 2.77 mA
For a BJT, I_{C} = 5 mA, I_{B} = 50 μA and I_{CBO} = 0.5μA.
Q. The value of β is







