How can I get base current in this simple transistor switch?

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Discussion Overview

The discussion revolves around determining the base current (IB) in a transistor switch circuit, specifically focusing on the base-emitter voltage (VB) and its typical value of 0.6 V. Participants explore the implications of this voltage in various operating modes of the transistor, including active, saturation, and cut-off modes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how the base-emitter voltage (VB) can be assumed to be 0.6 V across different circuits, suggesting that voltage drops in passive elements may affect this value.
  • Another participant asserts that the base-emitter voltage is approximately 0.6 V when the junction is forward biased, noting that it can vary between 0.5 V and 0.7 V depending on current levels.
  • A later reply emphasizes that while 0.6 V is typical for active mode, it may not hold in cut-off mode where there is no base drive.
  • One participant expresses a need for a mathematical or graphical relationship between IB and VB, indicating that they have not found a clear equation for this in the provided resources.
  • Another participant mentions that the exact value of VBE may not significantly affect the calculation of IB, given the tolerances in transistor parameters.
  • There is a suggestion to explore the exponential relationship between base current and base voltage for a theoretical calculation of VB, which would require numerical methods to solve.

Areas of Agreement / Disagreement

Participants generally agree that 0.6 V is a typical forward bias voltage for the base-emitter junction in active mode, but there is disagreement regarding its applicability in other modes such as cut-off. The discussion remains unresolved regarding the exact relationship between IB and VB.

Contextual Notes

Participants note that variations in transistor parameters, such as gain, can affect circuit behavior, and that the relationship between IB and VB may depend on additional factors not fully explored in the discussion.

goodphy
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Hello.

Please look at the attached figure first, which is cut image from transistor section of HyperPhysics.

In this circuit, I only need to know base current or base voltage, IB. In order to know this VB must be known but it is already given as 0.6 V. How this value is obtained? I don't think this is given parameter for each transistor and it seems enough external parameters to fully describe this circuit is already specified.

Could you please figure out What I've missed in circuit analysis?
 

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  • BJT transistor switch.png
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The base-emitter is a silicon P-N junction, so exhibits the V-I relationship of the typical P-N junction. For the range of currents over which a small signal transistor will be operated the B-E voltage is approximately 0.6v. (At the low end of expected currents, VBE will be around 0.5V, while at the upper end it may be 0.7V.) So you can assume that whileever that junction is forward biassed and is conducting base current, the B-E voltage is close to 0.6V.
 
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NascentOxygen said:
The base-emitter is a silicon P-N junction, so exhibits the V-I relationship of the typical P-N junction. For the range of currents over which a small signal transistor will be operated the B-E voltage is approximately 0.6v. (At the low end of expected currents, VBE will be around 0.5V, while at the upper end it may be 0.7V.) So you can assume that whileever that junction is forward biassed and is conducting base current, the B-E voltage is close to 0.6V.

Oh thanks for quick reply! But I still don't get one point. Can I really assume that forward bias voltage for base-emitter is always 0.6 V in whatever circuit is made with this transistor? I mean...every element of circuit has its own voltage drop along it when the current is going through. When current is going high, voltage drop is high. This is passive element example thus it may be not relevant to transistor but my point is that the voltage drop (here voltage base-emitter) should be externally dependent parameter unless it has voltage regulation. How can I have guaranty of such 0.6 V bias?
 
You can assume 0.6V - 0.7V whenever this transistor is given forward base current.

Do a google search for the V-I curve of a silicon PN junction (diode).
 
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Note that in this problem..

Ib = (10-Vbe)/1000

So if

Vbe = 0.5 then Ib = 9.5mA
Vbe = 0.7 then Ib = 9.3mA

So an "error" in the value you use or assume for Vbe only has a small effect on the value of Ib. You might consider that the exact value of Vbe doesn't matter in this circuit.

Many transistor parameters are subject to manufacturing tolerance. For example the gain of a transistor might vary from 80 to 200. Good engineering design ensures that the circuit will work for all values between 80 and 200.
 
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All right. It looks like that I need to accept 0.6 V is typical forward biasing voltage in base-emitter junction in at least active mode. However, If it is possible, can I get some mathematical expression or graphical form of relation between IB and VB? As suggested, I've found figure like the attached image but even this figure doesn't explicitly tell this relationship.

In addition, 0.6 V even holds for other mode such as saturation and cut-off mode?
 

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  • Transistor load line and characteristic curve.gif
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Actually for my last reply in this post, I've found that silicon PN junction depletion layer voltage (due to space charge) or built-in voltage is about 0.6 V. That is why this number is given as firstly suggested by NascentOxygen since this voltage must be overcome to current flows. Although I still don't have clear equation to get VB in circuit.
 
In cut-off there is no base drive, so you mustn't expect the 0.6V figure to still hold.

If you really wanted to theoretically calculate VB then you would use the exponential relation between base current and base voltage, including its temperature dependence. You would have to employ a numerical method to solve the mathematics. For normal conditions, with these small signal BJTs, the result you end up with won't differ much from 0.6V.
 
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