NPN Transistor Circuit Analysis

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SUMMARY

The discussion focuses on analyzing an NPN transistor circuit to determine the voltage level at V3 under two cases: both V1 and V2 at 0V, and V1 at 0V while V2 is at 5V. For Case #1, the calculations show that Q3 is on and in the saturated region with a voltage V3 of 4.51V. The user also discusses the calculation of the base current (Ib) and collector current (Ic) using the formulas Ic=β*Ib and the application of KVL and KCL. The confusion arises from the interpretation of β when Ib is zero, leading to a misunderstanding of transistor states.

PREREQUISITES
  • NPN transistor operation principles
  • Basic circuit analysis techniques (KVL, KCL)
  • Understanding of transistor parameters (β, Vce, Vbe)
  • Ohm's Law and resistor calculations
NEXT STEPS
  • Study the impact of varying V1 and V2 on transistor states in NPN circuits
  • Learn about transistor saturation and active regions in detail
  • Explore advanced circuit analysis techniques using simulation tools like LTspice
  • Investigate the role of β in transistor operation and its practical implications
USEFUL FOR

Electronics students, hobbyists, and engineers interested in understanding NPN transistor circuit behavior and analysis techniques.

bschwartz
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Homework Statement


Determine the voltage level at V3. For each case also determine which transistors are on and saturated, on and active, and which are off. Let R1=R2=R3=2.2kΩ and let Rc=240Ω and Vcc=5V. Calculate β for each transistor as well.

We are to assume that if the transistor is on, then it is in the saturated state so Vce=0.2V. If β>20 then the transistor is in active and Vce=2V+. Also, Vbe=0.6V if the transistor is on

Case #1: V1=V2=0V
Case #2: V1=0V V2=5V

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Homework Equations


Ic=β*Ib
KVL, KCL

The Attempt at a Solution


Ok, so here's what I did for the first case where both V1 and V2 are 0V. Since they are 0V, Ib=0A and therefore β=∞ for both Q1 and Q2, right? If so, then I'm basically left with the "upper" circuit containing Q3 and V3.

I then found Ib=Vcc/(240+2.2k)
Ib = 0.002A

I did the same approach for Ic=Vcc/(240)
Ic = 0.0208A

β=10.1 so Q3 is then on and in the saturated region. To find V3 I simply subtracted the voltage drop of the 240Ω resistor

V3 = 5 - (240*.002A) = 4.51V

I don't know if this is right at all since I've never done any circuit analysis with transistors before. If it's right, I'm still completely lost on how to start Case #2 where V2=5V and V1=0V. Any help is greatly appreciated.
 
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Your method looks right, though I haven't checked the arithmetic. It is unusual to use β in the two ways you have. Writing β=∞ is not right. Just say IB=0 and therefore IC=β*0=0.

When V2=5v there will be current into the base of Q2, hence collector current, too.
 
excuse me. , where are you find this question. I also need this similar question to finish my homework
 

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