though transistor is reverse biased , it produces a large amount of current inspite of a diode which when reverse biased do not allow the current to flow.why?
Asked by vasturushi | 19th Nov, 2017, 11:13: AM
- The thin layer region on both sides of a p-n junction which has immobile ions and is devoid of any charge carrier is called depletion region or depletion layer.
- A p-n juction is said to be reverse biased if the positive terminal of the battery is connected to n-side and the negative terminal to the p-side of the p-n junction.
- In reverse biasing the applied voltage of battery mostly drops across the depletion region of the p-n junction and its direction of voltage is same as that of the potential barrier.Due to it, the reverse bias voltage supports the potential barrier.
i..e during reverse biasing the applied dc voltage aids the fictitious battery developed across the junction.
- Due to this the potential drop across the junction increases and as a result the diffusion of holes and electrons across the junction decreases. It makes the depletion layer thick and the juction diode offers high resistance during reverse bias.
- It is nothing but a switch where the base current controls the flow of current from collector to emitter.
- In NPN transistor when operating in active region the base to collector junction is reverse biased (positive to negative , negative to positive)
- A large number of electrons in emitter layer is repelled by negative terminal and they flow towards b-e junction.
- They cross the junction and enter into small base layer.- In this region, some electrons combine with holes (in the base) and are attracted by positive terminal and remaining maximum number of electrons flow into collector layer, crossing the second junction i.e. collector-base junction.
- The electrons in the collector region are repelled by these newly arrived electrons and thus, then all the electrons are present in collector layer are attracted by positive terminal.
- Thus, all these electrons complete their journey back into emitter layer and produce conventional currents.
Answered by Abhijeet Mishra | 8th Dec, 2017, 04:32: PM
- How do we bias the junctions of a transistor in common emitter configuration?
- What are Transistors?
- A P-N transistor is used in common - emitter mode in an amplifier circuit. A change of 40μA in the base current brings a change of 2mA in collector current and 0.04V in base -emitter voltage, Find (i) input resistance (ii) current amplification factor β. If a load resistance of 6 k oh m is used, then find voltage gain?
- Draw a labeled circuit diagram of a common emitter transistor amplifier. Draw its input and the output characteristics. .
- A transistor has a current gain of 30. If the collector resistance is 6kΩ, input resistance is 1kΩ, calculate its voltage gain?
- The input resistance of a silicon transistor is Ω Its base current is changed by 15μ A, which results in the change in collector current by 2mA. This transistor is used as a common emitter amplifier with a load resistance of 5k Ω. Calculate current gain (βac) transconductance and voltage gain.
- The base of a transistor is lightly doped. Explain why?
- In the given figure, is (i) The emitter (ii) collector forward or reverse biased? Justify.
- Define current amplification factor in a common emitter mode of transistor?
- Calculate emitter current for which β = 100 IB = 20μA?
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number