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The BJT diagram is shown below:

BJT diagram

The voltage source at the base side is increased incrementally from 1V to 10V, with the voltage source at the collector side being constant.

The Beta values are recorded in OrCAD PSPICE tool:

  • 1V - 147
  • 2V - 168
  • 3V - 174
  • 4V - 176
  • 5V - 177
  • 6V - 176.9
  • 7V - 176
  • 8V - 174
  • 9V - 173
  • 10V - 158

The beta value increases from 1V and reaches its peak around 5V, and it starts dropping from there till 10V. I want to find the most appropriate DC amplification factor from these values, which will mainly be used in doing the DC analysis of a common emitter BJT amplifier circuit. Do I assume that the most appropriate value of a DC amplification factor is the mean value of all the beta values from 1V to 10V, or? Am a bit confused here.

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    \$\begingroup\$ Most appropriate for what? \$\endgroup\$
    – user76844
    Sep 2, 2018 at 5:11
  • \$\begingroup\$ Hi, welcome to EE Stack Exchange. Are you asking what value can you assume for design purposes? Transistor are messy devices whose parameters vary alot \$\endgroup\$
    – jramsay42
    Sep 2, 2018 at 5:31
  • \$\begingroup\$ @GregoryKornblum Hi, I want to determine the most appropriate DC amplification factor (beta value) based on the values mentioned in the question for doing the DC analysis in a typical common-emitter BJT amplifier circuit. \$\endgroup\$
    – jd043
    Sep 2, 2018 at 5:40
  • \$\begingroup\$ @jramsay42 Thank you! As mentioned in the above comment, I want to determine the most appropriate DC amplification factor based on the values mentioned in the question, so that it can be used in doing the DC analysis on a typical common emitter BJT amplifier circuit. Hope that clarified your query. \$\endgroup\$
    – jd043
    Sep 2, 2018 at 5:43
  • \$\begingroup\$ @jd043 If you take a look at the 2n2222's datasheet there are different DC Current Gain (Hfe) values listed for different Ic, and Vce. In design, I would be designing for a particular Ic and Vce, and use the appropriate Hfe as a starting point, while also investigating the effect of variance of Hfe, as we can never rely on it being an exact value. For your analysis have a think about what you think Vce/Ic may be and use that as your starting point. It would also be good to use the largest and smallest values you have in your analysis, to see the impact of the variance. \$\endgroup\$
    – jramsay42
    Sep 2, 2018 at 6:05

2 Answers 2

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The answer is that using a fixed beta (gain) for analysis is not the right way to do BJT analysis. They are given as a range because they vary for all sorts of reasons. That's why BJT circuits that work are designed to be very insensitive to the BJT's gain - they need to still work over the whole range.

Often a BJT circuit is designed to work for a gain of at least, say, 50 or 100. Then you just make sure the gain of the BJT you choose can't be less than that value, and you're done.

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  • \$\begingroup\$ I think I get your point. But for this case, would choosing a fixed beta (gain) for DC analysis still matter if I were to choose a suitable gain from this circuit for a typical common emitter BJT amplifier circuit that has a supply voltage of 0.1V AC? Or would it make more sense to choose any of the beta values from 1V to 10V, because all of them are <100? \$\endgroup\$
    – jd043
    Sep 2, 2018 at 10:30
  • \$\begingroup\$ Jd043 - at first, select a suitable DC collector current (with respect to some requirements and other constraints (voltage gain, supply voltages, available resistors, power consumption,). As asecond step, select a typical beta value in the middle of the given range and e this value for all further calculations (bias network). \$\endgroup\$
    – LvW
    Sep 2, 2018 at 15:27
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Jd043 - If I understand your question right, you are asking for a certain beta-value that can satisfy specific voltage gain requirements, correct? In this case, I consider it as important to know how a bipolar transistor really works.

Please note that there is one single parameter that really matters - as far as voltage amplification is concerned: The transconductance gm=d(Ic)/d(Vbe). This parameter is identical to the slope of the steering characteristics Ic=f(Vbe).

The actual value of gm depends on the chosen DC collector current only (gm=Ic/Vth) and does NOT depend on the beta-values. The beta value (called "current gain") determines the base current and, hence, the input resistance) only.

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    \$\begingroup\$ While this may be technically correct, thinking of a BJT as voltage-driven is confusing to newcomers, and rarely useful in a real circuit anyway. The just wants to understand how to characterize the transistor he has according to the current gains he measured. Confusing him with voltage-driven characteristics is not useful. \$\endgroup\$ Sep 2, 2018 at 13:29
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    \$\begingroup\$ I do not want to start again a long discussion on this subject, on the other hand - how can a "technically correct" explanation confuse somebody. I have more than 25 years experience in teaching electronics - and, very often, some students were of the false opinion that a large current gain would be necessary for a large voltage gain. I think, it is best from the beginning to teach the "truth". Do you think that voltage control (reality) is much harder to understand than current control (only a model, which cannot explain all transistor properties)? \$\endgroup\$
    – LvW
    Sep 2, 2018 at 15:16
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    \$\begingroup\$ Continue: In contrary - if one starts thinking about what`s really going on, he will reach the point where the question comes up: Can a small current DIRECTLY control the flow of a larger current? And the answer, of course, will be: NO! This is impossible from the energy point of view! \$\endgroup\$
    – LvW
    Sep 2, 2018 at 15:21
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    \$\begingroup\$ Actually BJTs do work as current amplifiers at the low level. For a NPN, for example, to get electrons out the base, lots more electrons have to be injected via the emitter, since so many of them diffuse across the junction and get to the collector before they can go out the base lead. That's where current gain comes from. A BJT as a current amplifier is also the model that is most useful for designing circuits. This site is about electrical engineering, not as much about semiconductor physics. \$\endgroup\$ Sep 2, 2018 at 16:28
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    \$\begingroup\$ In my contribution I did not go into semiconductor physics (that`s what you did in your last comment). Instead I have used the input-output steering function Ic=f(Vbe) (to be found in the data sheets) for defining the transconductance gm which appears in each formula for the voltage gain of an amplifier stage. And - it is not true that BJTs do work as current amplifiers. I know that this misconception can even be found in some textbooks. It is not a problem to prove and to explain why the BJT works as a voltage-controlled current source. I am sure you have heard about Shockleys equation \$\endgroup\$
    – LvW
    Sep 2, 2018 at 16:38

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