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

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The discussion centers on understanding the emitter voltage (Ve) of a PNP transistor circuit, where the expected voltage is +9 volts but is calculated as 1.68 volts. Participants clarify that the current source does have a voltage drop, which must be considered along with the voltage across resistors and the base-emitter junction. The voltage drop across the current source can be determined by the equation that sums the voltage drops across the current source, the resistor, and the base-emitter junction to equal 9V. It is emphasized that without resistance in the circuit, the voltage drop across an ideal current source would be infinite, while a real current supply would have a finite maximum voltage drop. Understanding these principles is crucial for accurately calculating Ve in the circuit.
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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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