BJT voltage on the emitter question

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Discussion Overview

The discussion revolves around understanding the voltage at the emitter (Ve) in a BJT circuit, particularly how it relates to the base voltage (Vb) and collector voltage (Vc). Participants explore the relationships between these voltages and the operation of the transistor, focusing on the calculation of Ve and the role of Vcc and Vbb in the circuit.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant questions why Vc is ignored in the calculation of Ve, suggesting it should be included in the formula as (Vc + Vb) - 0.7V.
  • Another participant clarifies that Vcc is not in series with Vbb in the circuit diagram, indicating that the actual value of Vcc is not critical as long as it can supply the necessary current.
  • A further explanation is provided about how the base current influences the collector current and the resulting voltage across the emitter resistor, leading to a steady state where the sum of the emitter voltage and the base-emitter voltage equals Vbb.
  • One participant expresses understanding after the explanation, indicating that the clarification helped resolve their confusion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the role of Vc in the calculation of Ve, with differing views on its relevance. The discussion remains unresolved regarding the inclusion of Vc in the voltage calculations.

Contextual Notes

There are assumptions about the circuit configuration that may not be fully articulated, particularly regarding the relationships between Vcc, Vbb, and the transistor operation. The discussion does not resolve the mathematical steps involved in the voltage calculations.

ttsky
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i got this example from our lecture notes, there is one thing i don't understand,

When calculating Ve, he takes Vb and minus the Voltage drop across transistor, so 2V - 0.7V and gets 1.3 V for Ve,

my confusion is why did he ignore Vc? should it not be (Vc + Vb)-0.7V ,so (10V+2V)-0.7V=11.3V on Ve ?

note,we haven't covered BJT yet, just reading ahead, hence the confusion.

thanks

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What you have done is fine. There is another voltage Vce which you have not been asked to calculate.
This will be Vcc - Ve = 10 - 1.3 = 8.7V
 
Vcc isn't in series with Vbb in this diagram.

Try to think of it as two loops with the current in the left loop controlling the current in the right loop. The actual value of Vcc doesn't matter much as long as it can supply the necessary current for the right loop.

When the circuit is turned on, base current starts to flow from Vbb. This turns the transistor on and collector current starts to flow in the right loop.

This (and a small contribution from the base current) produces a voltage across the emitter resistor which rises until the sum of this voltage and the base-emitter voltage equals Vbb.

At this point, the base current becomes steady and a predictable collector current is flowing.
 
vk6kro said:
Vcc isn't in series with Vbb in this diagram.

Try to think of it as two loops with the current in the left loop controlling the current in the right loop. The actual value of Vcc doesn't matter much as long as it can supply the necessary current for the right loop.

When the circuit is turned on, base current starts to flow from Vbb. This turns the transistor on and collector current starts to flow in the right loop.

This (and a small contribution from the base current) produces a voltage across the emitter resistor which rises until the sum of this voltage and the base-emitter voltage equals Vbb.

At this point, the base current becomes steady and a predictable collector current is flowing.
aha! that makes a lot of sense, thanks a lot!
 

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