4
\$\begingroup\$

I have always thought that for a P-N junction, forward-biased simply means that \$ V_{PN} > 0 \$ and reverse biased means that \$ V_{PN} < 0 \$.

However, in this article about NPN BJTs, there is the following sentence (emphasis by me):

Vbe < 0.7V(emitter-base junction is not forward biased)

So it defines forward bias of a P-N junction as \$ V_{PN} > V_{th} \$, where \$ V_{th} \$ is the threshold voltage for the junction to conduct.

So which one is the correct definition of forward-bias? When I see an expression like: "Base-emitter is forward biased" should I understand:

  • \$ V_{BE} > 0 \$

Or

  • \$ V_{BE} > V_{th} \$
|improve this question|||||
\$\endgroup\$
4
\$\begingroup\$

It's a bit of handwaving. What they really mean is that the B-E junction is not carrying any significant forward current. If VBE is between 0.0 V and 0.65 V, it is technically forward biased, but the bias is not enough to overcome the internal barrier voltage of the junction, so it does not conduct.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ So let's say in an article talking about BJT NPNs, I encounter an expression like: "In order to operate on saturation mode, both base-emitter and base-collector must be forward biased." Does this expression mean \$ V_{BE} > V_{th(be)} \; AND \; V_{BC} > V_{th(bc)} \$ or does it mean \$ V_{BE} > 0 \; AND \; V_{BC} > 0 \$ or something else? \$\endgroup\$ – Utku Jan 23 '16 at 15:33
  • \$\begingroup\$ It means \$ V_{BE} > V_{th(be)} \$ (i.e., the base is conducting current) and \$ V_{BC} > 0 \$ (i.e., the collector voltage is less than the base voltage). There is no barrier voltage for the B-C junction, because the conduction there is caused by the active injection of carriers into it. \$\endgroup\$ – Dave Tweed Jan 23 '16 at 15:49
  • \$\begingroup\$ By the way post #7 in this forum says: "The collector voltage of a saturated transistor is a diode voltage drop less than its base voltage." This contradicts with what you say. Is that false? Isn't it required \$ V_C \$ to be one diode voltage drop less than \$ V_B \$ to be in saturation? \$\endgroup\$ – Utku Jan 23 '16 at 16:40
  • \$\begingroup\$ No, that isn't possible. The collector voltage is always slightly higher than the emitter voltage (\$V_{CE(SAT)}\$), usually by 200-300 mV or so. \$\endgroup\$ – Dave Tweed Jan 23 '16 at 17:03
  • \$\begingroup\$ But AFAIK, this doesn't contradict with what is written on that forum. If the diode voltage drop over BE is say 0.7V and over BC is say 0.4V (due to different doping concentrations), \$ V_{CE} \$ can be around 0.2-0.3V. Am I missing something here? \$\endgroup\$ – Utku Jan 23 '16 at 17:14
1
\$\begingroup\$

It depends on the context, i think. If that statement is made in an article/textbook trying to model real-life behaviour of a transistor or diode (or when dealing with small signals in general), then it probably means: \$V_{BE} > 0.7\$

But in some cases this may not matter, since it may not be the point of the lesson e.g. an article about diode clippers for 220 V AC. In this case, the relatively small voltage drop across the diode could be considered irrelevant, though its still there, of course. In order to not get caught in details, you may simply assume that it means: \$V_{BE} > 0\$

|improve this answer|||||
\$\endgroup\$
1
\$\begingroup\$

The definition is rather interesting, because its based on the models of the transistor, not the transistor itself. Obviously a real transistor has smooth responses to all inputs, with no magic "points" where its behavior suddenly changes. However, that response is complicated. We observe that, once Vbe goes above a certain voltage (0.7V for many transistors), we can use drastically simpler equations to model the behavior and still achieve very accurate results. Thus, datasheets typically give a "forward bias" Vbe which is the bias needed to get the transistor into that mode.

Under the hood, transistors are closely related to diodes, so we can use the diode equations to understand where some of these behaviors come from. The Shockley equation for a diode is \$ I=I_s(e^\frac{V_D}{nV{T}}-1) \$ As we can see, we have an exponential minus a constant, so for small or negative \$V_D\$, the exponential will be very close to 1, so the subtraction will result in almost zero current across the diode. However, at larger \$V_D\$, the subtraction becomes almost unimportant, and we see an exponential rise in current with voltage. Where does this break occur? It's based on \$V_T\$ which is the thermal voltage (a constant) and \$n\$, which is an "ideality" factor (typically between 1 and 2). Needless to say, the mere fact that there is such an "ideality" fudge factor in the equation shows why this concept of a forward bias is more empirical than anything else.

So really, if you think about it, the concept of forward bias divides the behavior of the transistor into three regions, including some \$\epsilon\$ bound around \$V_{th}\$ to account for the murky transition:

  • Forward biased (\$V_{BE} > V_{th} + \epsilon\$): equations may be simplified by ignoring most of the low voltage effects.
  • Reverse biased (\$V_{BE} < V_{th} - \epsilon\$): equations may assume the transistor is "off," where the (exponential-1) is very close to 0.
  • Transition (\$V_{BE} \approx V_{th}\$): neither simplified equation will really accurately model the behaviors in this region. If you're really really close to \$V_{th}\$, you may see strange artifacts which are only explained by a more complete transistor model which is not dependent on \$V_{th}\$ as a fundamental parameter.

Note, there is an effect called avalanche breakdown, where the diode ceases to prevent current flow (due to quantum effects) if you put a high enough negative voltage on it. Its behavior is also well modeled by an exponential, so you have another fuzzy region where you need a more complicated model, after which the equations simplify again. This suggests five total regions (from most negative \$V_{BE}\$ to most positive): breakdown, transition 1, reverse biased, transition 2, forward biased. Fortunately for most cases, you can ignore the transitions because they are small enough, but if you want to understand precisely what is happening, it helps to remember that they are there.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Is understanding Ebers-Moll model enough to understand precisely what is happening? Or do I need even more information? \$\endgroup\$ – Utku Jan 23 '16 at 19:40
  • \$\begingroup\$ @Utku It looks like that is sufficient for DC modeling, if you're using the full unapproximated equations (which get the diode behaviors right). If it helps, one of the particular things I looked for was that the equations had no "threshold voltage" parameter at all, but instead had things like the diode reverse saturation current. (And, while this model should be enough to understand the question at hand, know that they make far more advanced models for handling other situations, like AC or even transient behaviors. Don't be surprised when you have to switch models again later!) \$\endgroup\$ – Cort Ammon Jan 23 '16 at 20:11
  • \$\begingroup\$ When I was doing a SPICE model for a project in school, I used a model that captured the first 42 derivatives of the voltage/current equations! (Of course this is school, not real life: i didn't actually gain anything from using all of those terms, but the datasheet had them, and it made my report look really spiffy when I had a half page block of coefficients to terrorize the TA with!) \$\endgroup\$ – Cort Ammon Jan 23 '16 at 20:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.