0
\$\begingroup\$

What is the significance of the early effect? We increase the voltage at the collector, increase Ic and Ie increases. How does

What im getting from this is that, there is a point where Ic = 0. So possibly there is some threshold voltage we need to increase the collector voltage to in order to have some current flowing at the Emitter regardless of the set base current?

Is this correct?

\$\endgroup\$
  • \$\begingroup\$ It gives an indication of the slopes of \$I_C\$ vs \$V_{CE}\$. \$\endgroup\$ – StainlessSteelRat Jan 9 at 0:57
  • 1
    \$\begingroup\$ The Early Effect doesn't have just one significance and it's unlikely there's a book on the topic on all the possible ways one might use a BJT where the Early Effect is significant. But it causes distortion in the common CE amplifier configuration and the effect can be mitigated in some cases using a cascode configuration. So you could look up those two cases, for example, to see a problem and a solution that relates to the Early Effect. But you won't understand everything about it by just staring at a chart. You need to learn where it matters by designing and testing stuff. \$\endgroup\$ – jonk Jan 9 at 3:48
  • \$\begingroup\$ The Early Effect is one cause of finite gain, for a single-bipolar-transistor gain stage. \$\endgroup\$ – analogsystemsrf Jan 9 at 4:24
1
\$\begingroup\$

"What im getting from this is that, there is a point where Ic = 0. So possibly there is some threshold voltage we need to increase the collector voltage to in order to have some current flowing at the Emitter regardless of the set base current?"

No - this is not correct. The set of curves as shown exhibits a relatively small positive slope (the current Ic slightly increases with rising Vce). This increase is identical with an internal output resistance of the transistor in the range of some tenth of kohms. For this reason, the BJT is not an ideal but only a real "current source" with a finit output resistance.

It can be shown that - THEORETICALLY - an artificial continuation of the slopes (as shown in the simplified graph) will meet in a common point called Early voltage -VA. That means: This point -VA offers a simple way to describe the slopes of the curves - NOTHING ELSE!.

Of course, it is not a real voltage (no "threshold voltage" within the transistor) and you cannot measure it with a voltmeter! It is just a theoretical parameter that helps to model and to describe some BJT properties.

Comment: To me, it is not quite clear if the EARLY voltage is defined for the case IB=const or VBE=const. Both cases are possible, but lead to slightly different voltages VA. Surprisingly, both cases do exist in the literature - some books and university papers define VA for IB=const. and some other for VBE=const. (In some contributions the curves are even not labelled with IB or VBE). Of course, due to tolerances the difference is not very important - however, it is an unsatisfying situation that two definitions exist in parallel.

\$\endgroup\$
1
\$\begingroup\$

You need voltage for current to flow across the junction, but that's not the significance. The abrupt drop shown in the diagram isn't quite accurate either- there is a smooth curve downward to zero as the transistor saturates.

The extrapolation of the active-region Vce/Ic curve out to Ic = 0, ignoring the reality of saturation, yields the Early voltage Va.

If you think about a model of the transistor where Ic is a function of base-emitter voltage (or base current) then Ic should be constant regardless of changes in Vce (assuming it's not saturated or doesn't break down). In fact, it isn't exactly constant, it increases as the collector voltage increases.

Small-signal hybrid-pi model from Wikipedia :

enter image description here

The Early voltage is the Vce (or sometimes Vcb) at which the collector current doubles, so you can also think of it in the small-signal domain (where collector current changes a bit around a bias point Ic) as a resistor of value Va/Ic (ro in the above model). An ideal current sink would have infinite output resistance.

The forward and reverse Early voltage parameters appear in models such as Ebers-Moll and Gummel-Poon (used in SPICE).

\$\endgroup\$
  • \$\begingroup\$ Thanks but how can you get a negative voltage of Vce? Vce = Vcb - Vbe Assuming Vbe = 0.7V Increasing Vcb will increase Vce Vce will only be negative if Vcb < 0.7V I have read that Early voltage is around 10 to 150V So when Vce = -10 to -150V How to get to these negative values? \$\endgroup\$ – user209109 Jan 9 at 7:32
  • 1
    \$\begingroup\$ The Early voltage is just a theoretical tool that helps to describe some transistor properties - you cannot measure it and it does not "exist" in reality (see my detailed answer). \$\endgroup\$ – LvW Jan 9 at 8:58
  • \$\begingroup\$ As LvW said, it's a theoretical value. The output resistance is a more intuitive number that directly affects the performance, however it depends on the bias current. The product Va of bias current and output resistance is a device characteristic. So a transistor with Va of 100V and bias current of 10uA has a small-signal output resistance of about 10M ohm. \$\endgroup\$ – Spehro Pefhany Jan 9 at 12:35
  • \$\begingroup\$ @SpehroPefhany-typing error? The mentioned output resistance is by far to large. The slope of the Ic=f(Vce) curve is m=Ic/(VA+Vce)~Ic/VA (for VA>>Vce). Hence, the small-signal output resistance at the bias point (Ic, Vce) is app. ro=1/m=VA/Ic. Example: VA=100V and Ic=1mA leads to ro=100/1E-3=100 kohms. This value does not depend on the actual base current. \$\endgroup\$ – LvW Jan 9 at 14:07
  • \$\begingroup\$ Ohh sorry, only now I have recognized that you wrote "bias" current, which I have interpreted as "base" current. My fault. On the other hand my (false) assumption was supported (enabled) because in your example you have chosen a very low collector (bias) current of 10µA only. \$\endgroup\$ – LvW Jan 9 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy