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So the parameter beta for a Common Emitter Amplifier is called the Current Gain and is defined as

\$\beta =\dfrac{I_\mathrm{collector}}{I_\mathrm{base}}\$

\$\alpha =\dfrac{I_\mathrm {collector}}{I_\mathrm{emitter}}\$

What is this \$\alpha\$ called and what is its significance?

Like for example \$\beta\$ can be treated as the amount by which a signal is getting modified.

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    \$\begingroup\$ On Mathjax: Because engineering often involves how much something costs. Also, Mathjax wasn't added to the site until a few hundred questions had already been asked and answered. \$\endgroup\$
    – The Photon
    Jan 2, 2019 at 19:47
  • \$\begingroup\$ Note: You can still do display equations with $$. \$\endgroup\$
    – The Photon
    Jan 2, 2019 at 19:49
  • \$\begingroup\$ @ThePhoton So? How does that relate? \$\endgroup\$ Jan 2, 2019 at 19:49
  • \$\begingroup\$ If they had added Mathjax using just $ as a delimiter, it would have broken existing questions that talked about costs. \$\endgroup\$
    – The Photon
    Jan 2, 2019 at 19:50

2 Answers 2

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\$\beta\$ is a ratio between the collector and base currents whereas \$\alpha\$ is also a ratio but between the collector and emitter currents. There is no special name for \$\alpha\$ and \$\beta\$ as they are both unit-less.

However, you may call \$\alpha\$ as the common-base gain and \$\beta\$ as the common-emitter gain.

Either way, you will usually see \$\beta\$ in the spec sheets of BJT chips.

\$\alpha\$ is always less that 1.0 because of carrier generation and recombination going through the base region of the transistor, thus, the base and collector current are always less than the emitter current.

There is a relationship between \$\alpha\$ and \$\beta\$...

\$\displaystyle \beta= \frac{\alpha}{1-\alpha}\$

(Also, I asked a question about why Mathjax is different on the EE.SE than other websites.)

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  • \$\begingroup\$ No, \$\alpha\$ is always less than 1 because collector current is always less than emitter current -- the difference is the base current. This is a simple application of Kirchhoff's current law -- the sum of all three currents must be 0. \$\endgroup\$
    – Dave Tweed
    Jan 2, 2019 at 20:01
  • \$\begingroup\$ @DaveTweed We were taught this using a CB pnp transistor that a fraction of the holes from the emitter go to the collector. This fraction is called \$\alpha\$. The rest go through the base. Is that correct? \$\endgroup\$ Jan 2, 2019 at 20:08
  • \$\begingroup\$ @harshit54: Yes, that's correct. It's one measure of the "efficiency" of the transistor. \$\endgroup\$
    – Dave Tweed
    Jan 2, 2019 at 20:16
  • \$\begingroup\$ @harshit54 In reply to what you asked me: Yes there exists a common-collector characteristic. I was about to type it out but user G36 answered that in their answer. \$\endgroup\$
    – user103380
    Jan 2, 2019 at 20:18
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Yes, we have:

1 - "beta" \$\beta = \frac{I_C}{I_B}\$ for common-emitter current gain.

2 - "alpha" \$\alpha = \frac{I_C}{I_E} =\frac{I_C}{I_B +I_C}= \frac{\beta \cdot I_B}{I_B +\beta \cdot I_B}= \frac{\beta}{\beta + 1}\$ for common-base current gain.

3 - "gamma" \$\gamma = \frac{I_E}{I_B} = \frac{I_B + I_C}{I_B} = \frac{I_B +\beta \cdot I_B}{I_B} =\beta + 1 = \frac{1}{1-\alpha} \$ for common-collector current gain.

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