Find Max VB for BJT Active Region in Fig. P5.69

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SUMMARY

The discussion centers on determining the maximum base voltage (VB) for a BJT transistor operating in the active region, specifically referencing the circuit in Fig. P5.69 from the Sedra/Smith 5th edition textbook. The key condition is that the collector voltage (VC) must be greater than the base voltage (VB), with a specific focus on the relationship VC - VB = -0.4V. Participants clarify that saturation occurs when VC is less than VB, and that for low collector currents, VCE can drop to 0.1V or lower, allowing VB to exceed VC. The confusion arises from the interpretation of saturation conditions and the voltage relationships in the circuit.

PREREQUISITES
  • Understanding of BJT operation and active region characteristics
  • Familiarity with the Sedra/Smith 5th edition textbook
  • Knowledge of transistor saturation conditions
  • Ability to interpret voltage relationships in transistor circuits
NEXT STEPS
  • Review BJT transistor operation in active and saturation regions
  • Study the voltage relationships in transistor circuits, focusing on VC, VB, and VE
  • Examine data sheets for common small signal transistors to understand saturation voltages
  • Explore practical examples of BJT circuits to reinforce theoretical concepts
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Electrical engineering students, circuit designers, and anyone studying BJT transistor behavior in active and saturation regions.

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Homework Statement


Capture.PNG


VBE=0.7V

The transistor in the circuit of Fig. P5.69 has a very
high 0. Find the highest value of VB for which the transistor
still operates in the active mode.

Homework Equations


VC>VB

IC=IE (since IB is approximated to zero due to high beta)

EDIT: IE= [VE-0]/1k (VB - VE = VBE = 0.7V ---> VE = VB-0.7V)

The Attempt at a Solution



This question is taken from sedra/smith 5th edition, there's some worked solution floating around the internet, however the solution provided is confusing. The condition given is that VC-VB=-0.4V

I can't agree with the solution given because that equation would mean that the voltage at base is higher than that of the voltage of the collector since the potential difference is negative.

What I was thinking is that VC - VB = 0.

Am I correct, or did I missed anything?
 
Last edited:
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I think you're correct. Saturation is defined as Vc < Vb by any amount. Which is when the b-c junction becomes forward-biased.

However, fact remains that saturation voltages well below Vb are typically attained. Well below 0.3V which would roughly correspond to their answer. But you don't have the information you need to determine that level.
 
Kurokari said:
This question is taken from sedra/smith 5th edition, there's some worked solution floating around the internet, however the solution provided is confusing. The condition given is that VC-VB=-0.4V

I can't agree with the solution given because that equation would mean that the voltage at base is higher than that of the voltage of the collector since the potential difference is negative.

What I was thinking is that VC - VB = 0.
The textbook is correct in making that reasonable assumption. For very low collector currents, VCE can go down to 0.1V or less for some transistors, so this means that (for NPN) the base can be at higher potential than the collector. Take a look at the data sheet for some common small signal transistors.

Saturation occurs when the base current loses control of the collector current, i.e., when an increase in IB fails to elicit a corresponding increase in IC. Under these conditions, IC is being controlled by the external collector/emitter circuit, and not by the base.
 

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