BJT voltage on the emitter question

In summary, when calculating Ve, the voltage drop across the transistor is subtracted from Vb to get the value of Ve. Vc is not included in this calculation. Vcc, which is not in series with Vbb, is used to supply the necessary current for the circuit. The base current controls the collector current and when the circuit is turned on, the voltage across the emitter resistor rises until it equals Vbb. At this point, the base current becomes steady and a predictable collector current is flowing.
  • #1
ttsky
20
0
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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  • #2
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
 
  • #3
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.
 
  • #4
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 alot!
 

1. What is a BJT voltage on the emitter?

A BJT (Bipolar Junction Transistor) is a type of transistor that consists of three layers of doped semiconductor material. The voltage on the emitter refers to the potential difference between the emitter and the base of the transistor.

2. Why is the BJT voltage on the emitter important?

The BJT voltage on the emitter is important because it affects the current flow through the transistor. In a BJT, the emitter current is directly proportional to the emitter-base voltage, so any changes in the emitter voltage can significantly impact the performance of the transistor.

3. How is BJT voltage on the emitter calculated?

The BJT voltage on the emitter can be calculated using Ohm's Law, which states that the voltage across a resistor is equal to the product of the current and the resistance. In a BJT, the emitter voltage can be calculated by multiplying the emitter current by the emitter resistance.

4. What factors can affect the BJT voltage on the emitter?

The BJT voltage on the emitter can be affected by various factors such as temperature, base current, and collector-emitter voltage. Changes in these factors can cause variations in the emitter voltage, which can impact the performance of the transistor.

5. How does the BJT voltage on the emitter relate to other parameters in a transistor circuit?

The BJT voltage on the emitter is closely related to other parameters in a transistor circuit, such as the base voltage and the collector voltage. These parameters, along with the emitter voltage, determine the operating point of the transistor and ultimately affect its performance and behavior in a circuit.

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