Solve This Question: Ve of PNP Transistor - Why is it 1.68?

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Discussion Overview

The discussion centers around understanding the emitter voltage (Ve) of a PNP transistor circuit, specifically why it is calculated as 1.68 volts instead of the expected +9 volts. Participants explore the implications of the current source and its voltage drop in relation to the circuit components.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant believes that the emitter voltage Ve should equal +9 volts, arguing that the current source does not drop any voltage.
  • Another participant asserts that the current source does have a voltage drop, and that voltages at the base and collector are determined using Ohm's Law and the known currents through the resistors.
  • A participant questions how to find the voltage drop across the current source, suggesting it might be calculated as 9 - Ve.
  • It is mentioned that the voltage drops across the current source, the resistor, and Veb must sum to 9V.
  • One participant states that the current source has the same voltage drop as the resistor, emphasizing that the resistor carries current, thus the voltage drop is not zero.
  • A later reply discusses the implications of removing the resistance from the circuit, suggesting that it leads to a theoretical problem regarding the voltage drop across the current source.
  • Another participant explains that for an ideal current supply, the voltage drop would be infinite if the terminals of a voltage source are connected directly together, while a real current supply would have a maximum voltage drop based on its specifications.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the current source and its voltage drop, indicating that there is no consensus on how to approach the calculation of Ve or the implications of the current source in this context.

Contextual Notes

Participants discuss the assumptions related to the ideal versus real behavior of current sources, as well as the dependencies on circuit components and configurations, which remain unresolved.

munna007
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solvethis question please

hello to all...

i m having some doubts in solution of this question...


what i know about this ckt is that this is pnp transistor. +9 volt is connected to emitter (where arrow made) and -9 volt is connected to collector.

i think voltage emitter Voltage Ve should be equal to +9 volt , because current source is not dropping any voltage . but in solution part its found something else i.e 1.68 ...

why is it ? please clear my doubt
 

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The current source does have a voltage drop.

Voltages at the base and collector are calculated using Ohm's Law and the known currents through the resistors. The emitter voltage is assumed to be 0.7V above the base.

Here's a more obvious, convincing example of a current source that has a voltage drop:

LEC3_2.GIF

(from http://www.labsanywhere.net/circuit/lectures/lect3/lecture3.php )[/size]
 
Last edited:


Redbelly98 said:
The current source does have a voltage drop.
[/size]

thanks for help fnd.

but how to find voltage drop across current source. would it be like 9-Ve ?

i m unable to understand how is your diagram showing voltage drop across current source ?

please clarify
 


munna007 said:
how to find voltage drop across current source. would it be like 9-Ve ?
You don't need the voltage drop across the current source to solve this problem. But if you wish to know it, use the fact that the voltage drops across the current source, the 50k resistor, and Veb must sum to 9V.

i m unable to understand how is your diagram showing voltage drop across current source ?

please clarify
The current source has the same voltage drop as the resistor. And the resistor has a current, therefore the voltage drop is not zero.
 


Redbelly98 said:
The current source has the same voltage drop as the resistor. And the resistor has a current, therefore the voltage drop is not zero.

thanks for the reply ...if we remove the resistance from the circuit shown by you , then will there be a voltage drop across current source ?
 


That becomes a problem. Similar to asking what is the current when you connect the terminals of a voltage source directly together.

For a theoretically ideal current supply, the voltage drop becomes infinite since the current is being driven through an infinite resistance (V=IR, and R=∞ for an open circuit).

But in a real current supply, the voltage drop would be whatever the maximum voltage of the current supply is.
 

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