BJT voltage on the emitter question

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    Bjt Voltage
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The discussion revolves around calculating the emitter voltage (Ve) in a BJT circuit. The original confusion stemmed from whether to include the collector voltage (Vc) in the calculation, but it was clarified that Vc does not factor into the Ve calculation in this context. Instead, the focus is on the base voltage (Vb) and the base-emitter voltage drop (0.7V), leading to Ve being 1.3V. It was emphasized that Vcc is not in series with Vbb, and the circuit operates with two loops where the left loop controls the right. Understanding this relationship helps in predicting the behavior of the circuit once the transistor is activated.
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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 alot!
 
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