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The Early Effect seems to play a role when output voltage is increased in a transistor.

1)In the Common Base config, it causes the emitter current to increase,which in turn causes the collector current to increase. What happens to the base current in such a situation and why?

2)In the Common Emitter mode, my textbook says the emitter current increases while the base current decreases. It also says the Emitter current is constant because input voltage is constant. I am not able to follow this and seem to think otherwise.

Any help would be appreciated.

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  • \$\begingroup\$ You say you "think otherwise." It would help if you'd discuss your thoughts, in detail. What do you think should occur? \$\endgroup\$ – jonk May 6 '17 at 7:44
  • \$\begingroup\$ According to me, the Emitter current should increase, same as in the case of CB config. This is because technically we can still assume the Emitter-Base region to be Forward biased and the Collector-Base region to be reversed biased in CE config. According to me, the Early effect should play the same role because the input voltage is constant in the CB config as well. \$\endgroup\$ – b_boundary May 6 '17 at 8:03
  • \$\begingroup\$ Hmm. Does your textbook actually simultaneously argue that in the CE mode, the emitter current both increases and is constant? Did I read you correctly there? \$\endgroup\$ – jonk May 6 '17 at 8:28
  • \$\begingroup\$ I'll post up a short note. See if it helps or if I'm totally missing the mark. \$\endgroup\$ – jonk May 6 '17 at 8:43
  • \$\begingroup\$ The text book only says that the emitter current remains constant in CE mode. Their explanation is that the input voltage is constant, as well as the voltage drop across the base-emitter junctions (~0.7 for Si) \$\endgroup\$ – b_boundary May 6 '17 at 9:25
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As far as the mentioned Early effect is concerned, there is no difference between CB and CE configuration. The effect is as follows (modulation of the base width Wb) :

As the reverse bias across collector-base junction increases (due to a rising collector-emitter voltage) the width of the collector-base depletion layer increases - and the base width Wb decreases correspondingly. This effect has - in principle - the following consequences:

1) When VBE is held constant, the emitter current IE=f(VBE) remains constant but the collector current increases and the base current is reduced correspondingly (due to IE=IB+IC=const).

2) When IB=const, both IE and IC increase at the same time (B value is larger).

In reality, neither of the cases 1) or 2) will be exactly met - instead, we will have a mixture of both cases.

Remark: As can be derived from the above (and as shown in the corresponding data sheets for constant IB and VBE, respectively) the slopes of the curves Ic=f(VCE) are not equal in both cases. Hence, the derived values for the EARLY voltage VA are different.

Thus, do we have two EARLY voltages? Surprisingly, this question has not been answered in textbooks (as far as I know). Some sources define the EARLY voltage for VBE=const, some others for IB=const. and some make no statement at all.

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  • \$\begingroup\$ Regarding the remark: I believe it has been addressed and was discussed in some of the earlier papers going quite far back (Shockley-Read-Hall recombination, for example.) See: ecee.colorado.edu/~bart/book/recomb.htm for an expanded list. The VBIC model incorporates 2.11.4, 2.11.5, and 2.11.6 on that page, memory serving. \$\endgroup\$ – jonk May 6 '17 at 17:21
  • \$\begingroup\$ Jonk, thanks for the hint. However, what is the outcome of the contribution regarding the EARLY voltage (IB or VBE constant)? \$\endgroup\$ – LvW May 7 '17 at 9:00
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The following only applies if the BJT is in active mode and not saturated, obviously.

Let's say you measure \$I_{C_{CAL}}\$ and \$I_{B_{CAL}}\$ at the operating point \$V_{CE_{CAL}}=2\:\textrm{V}\$. If you hold \$I_B=I_{B_{CAL}}\$ you can predict the collector current as a function of \$V_{CE}\$ as:

$$I_C\left(V_{CE}\right)=I_{C_{CAL}}\cdot\frac{V_A+V_{CE}}{V_A+V_{CE_{CAL}}}$$

Conversely, if you hold \$I_C=I_{C_{CAL}}\$ then you can predict the base current as a function of \$V_{CE}\$ as:

$$I_B\left(V_{CE}\right)=I_{B_{CAL}}\cdot\frac{V_A+V_{CE_{CAL}}}{V_A+V_{CE}}$$


However, if you have circuit elements where neither \$I_B\$ nor \$I_C\$ is held constant, then you can still compute:

$$\beta\left(V_{CE}\right)=\frac{I_{C_{CAL}}}{I_{B_{CAL}}}\cdot\frac{V_A+V_{CE}}{V_A+V_{CE_{CAL}}}$$

And then use this computed value of \$\beta\$.


There are plenty of other factors that affect \$\beta\$. I'm sure you've heard of the "late effect", as well. There are effects at (1) low currents caused by carrier recombination effects at the surface and in the emitter-base space-charge layer and due to the formation of emitter-base surface channels; and (2) high currents where the minority carrier injection becomes significant and, because the space-charge must be neutral in the base region, the total majority carrier concentration must be similarly increased (by the same amount) and ultimately causing the collector current to asymptote. So there are other parameters to deal with these effects on \$\beta\$ at low and high collector current levels, which are not included in the above equations.


Note that I haven't mentioned \$V_{BE}\$ above.

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  • \$\begingroup\$ Thanks for that, but i still don't understand why emitter current remains constant in CE with varying base-collector voltage \$\endgroup\$ – b_boundary May 6 '17 at 9:42
  • \$\begingroup\$ @b_boundary Who says it does??? When I read your words, I couldn't even tell what exactly you were saying there. Perhaps you are confused? \$\endgroup\$ – jonk May 6 '17 at 16:28
  • \$\begingroup\$ My text book says it does, apparently because the 'input voltage is constant'. That was what confused me, thanks for your help. \$\endgroup\$ – b_boundary May 6 '17 at 17:26
  • \$\begingroup\$ @b_boundary No problem. \$\endgroup\$ – jonk May 6 '17 at 17:27

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