Engineering BJT Circuit - Determine region of operation

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The discussion focuses on determining the region of operation for a BJT circuit, specifically whether it is in forward active or cutoff mode. The user initially concludes that the circuit is in forward active operation based on voltage relationships but questions the direction of the base current, which contradicts their notes on PNP transistors. Further analysis leads to the realization that if the base current is assumed to flow into the base, it aligns with the circuit being in cutoff mode instead. The key takeaway is that without base current, there can be no emitter or collector current, confirming the cutoff state. A methodical approach to analyzing such circuits is emphasized for exam preparation.
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Homework Statement



See the jpeg image for the circuit. Question is simply determine the region of operation.

BJT_Circuit.jpg


The Attempt at a Solution



The answer is cut off, but i don't really get how it is.

Looking at the circuit, emitter voltage Ve should be higher than base voltage Vb (simply because the base current has to go through a resistor Rb). Through this reasoning i say that Veb = Ve-Vb>0 . This means emitter-base junction is forward biased. Voltage at the collector Vc must be higher than Vb because there is a voltage drop between the base and collector terminal. Hence Vbc = Vb-Vc<0 and collector-base junction is reverse biased

That information suggests the circuit is in forward active operation. However i noticed one strange thing - the direction of the base current is going into the base (due to Ve > Vb). However from my notes it says a PNP transistor in forward active operation should have conventional current going out of the base.

So i must be analyzing the circuit wrong...help please?
 
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Blehs said:

Homework Statement



See the jpeg image for the circuit. Question is simply determine the region of operation.

View attachment 52778

The Attempt at a Solution



The answer is cut off, but i don't really get how it is.

Looking at the circuit, emitter voltage Ve should be higher than base voltage Vb (simply because the base current has to go through a resistor Rb). Through this reasoning i say that Veb = Ve-Vb>0 . This means emitter-base junction is forward biased. Voltage at the collector Vc must be higher than Vb because there is a voltage drop between the base and collector terminal. Hence Vbc = Vb-Vc<0 and collector-base junction is reverse biased

That information suggests the circuit is in forward active operation. However i noticed one strange thing - the direction of the base current is going into the base (due to Ve > Vb). However from my notes it says a PNP transistor in forward active operation should have conventional current going out of the base.

So i must be analyzing the circuit wrong...help please?

I believe the part of your quote that I bolded is the key.
 
Ah ok so when i assumed that conventional current was going OUT OF the base, the results seem to make sense and point towards the BJT being in cut off.

However my real concern is - how am i meant to determine that current is in fact going into the base? Was it really just as simple as

1) assume forward region of operation
2) in forward active region current goes INTO base, so its incorrect
3) assume cut off
4) everything adds up so therefore its cut off

Got an exam coming up soon and id like to prepare a good methodical way to approach questions like this
 
The point berkeman was alluding to is that the circuit does not PROVIDE for (conventional) current flowing out of the base. No base current, no emitter or collector current (ignoring reverse-biased b-c junction leakage current).
 
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