Understanding KVL Loop in BJT Biasing and Calculating Ve

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Discussion Overview

The discussion revolves around the calculation of the emitter voltage (Ve) in a BJT biasing circuit, specifically focusing on the application of Kirchhoff's Voltage Law (KVL) and the interpretation of circuit elements. Participants explore the definitions and assumptions related to Ve and Vcc, as well as the implications of circuit configurations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes that Ve can be calculated using the formula Ve = -Vee + IeRe, questioning how Ve can be considered positive.
  • Another participant clarifies that Ve is defined as the emitter voltage with respect to ground and suggests that it can range between -VEE and +VCC, emphasizing the importance of current direction and voltage drops.
  • A participant expresses confusion regarding the equations related to Vcc, seeking further assistance.
  • Another participant requests clarification on a specific line of a solution, indicating a lack of understanding.
  • One participant questions the application of KVL to what they perceive as an open loop, asserting that there is no open loop present in the circuit.
  • A later reply elaborates on the circuit configuration, explaining that the base current is supplied from VCC and discussing the implications of the transistor operating in the active region.

Areas of Agreement / Disagreement

Participants exhibit varying interpretations of the circuit elements and the application of KVL, indicating that multiple competing views remain. The discussion does not reach a consensus on the correct approach to calculating Ve or the validity of the open loop concept.

Contextual Notes

Participants express uncertainty regarding the definitions of Ve and Vcc, as well as the assumptions underlying the application of KVL in the context of the circuit. There are unresolved questions about the circuit configuration and the implications for voltage calculations.

karandeepdps
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thread moved into homework forum, so comes without template
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Screen Shot 2015-10-02 at 2.48.17 pm.png
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Please help how Ve is calculated what i think is that it should be:
-Ve=-Vee+IeRe;
How Ve is taken positive please someone draw equivalent circuit for same
 
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VE is defined as the emitter voltage wrt ground. Nothing about this says that VE must turn out to be a positive voltage, though. You can see that VE will be somewhere between -VEE and +VCC.

If NPN emitter current is the current out of the transistor emitter, then the upper end of RE is the more positive. Draw arrows to show current direction, and voltage drop across resistances, and apply Kirchoff's Voltage Law.
 
Thanks will you please help me in this question also i am confused in Vcc equations
Screen Shot 2015-10-02 at 6.47.15 pm.png
 
Explain which line in the solution you are having trouble with.
 
Last line,how they have applied kvl to an open loop.
 
karandeepdps said:
Last line,how they have applied kvl to an open loop.
There is no open loop.

RB is supplying base current from the positive rail, VCC. Because the transistor is operating in the active region, its base is forward biased and its B-E potential is labelled VBE.

The wire from the base ending in mid-air simply indicates where the amplifier's AC input will be impressed when the circuit is later completed to function as an audio amplifier, for example.
 

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