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Let us say I bias my base emitter region so that the depletion region is reduced (Vbe=0.7 volts) so the the base emitter region conducts. Now if I apply a small base current why does it result in a large collector current since I already have the base emitter region forward biased . Does the base current inn some way cause a change in the collector base depletion region? In what way do I have to look at this to be able to understand?

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In what way do I have to look at this to be able to understand?

The base current increase (decrease) is due to an increase (decrease) in \$v_{BE}\$.

The increase (decrease) in \$v_{BE}\$ increases (decreases) the injection of carriers from the heavily doped emitter.

Most of these carriers cross the thin base region without recombining and are then swept across the base-collector junction into the collector region. A small percentage don't and these form the base current.

Update to answer a comment:

Once it has been biased isnt Vbe=0.7V. Now when we apply a small ac signal of the order of milli volt isnt the change negligible?

No, the collector current is exponential in the base-emitter voltage:

\$i_C = I_Se^{(v_{BE}/V_T)}\$

To get a feel for this, consider this question: to double the collector current, how much would \$v_{BE}\$ need to increase?

For example, assuming the bias value is \$V_{BE} = 0.7V \$, increasing this voltage by a mere \$17 mV \$ (an increase of just under 2.5%) will double the collector current.

Another approach:

According to the collector current equation, if we change the base-emitter voltage from its quiescent value by some small amount, the change in collector current is approximately:

\$\Delta i_C = \dfrac{I_C}{V_T} \Delta v_{BE}\$

As a typical example, let the quiescent collector current \$I_C = 1mA \$. At room temperature, \$V_T = 25mV\$.

Then, for these numbers, the collector current changes by 4% when the base-emitter voltage changes by 1mV.

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  • \$\begingroup\$ Once it has been biased isnt Vbe=0.7V. Now when we apply a small ac signal of the order of milli volt isnt the change negligible? \$\endgroup\$ – Developer Android Apr 8 '13 at 12:11
  • \$\begingroup\$ @DeveloperAndroid, see update. \$\endgroup\$ – Alfred Centauri Apr 8 '13 at 13:04
  • \$\begingroup\$ I agree and completely accept what you are saying. But the thing I am not able to understand is when we apply the 0.7V across Vbe then the base emitter junction is forward biased completely. In fact it acts very much like a conductor(A short circuit from the diode graphs since current increases exponentially). Now when base current is supplied to the transistor, how does it change the collector current. How does it increase or decrease the collector current. What is PHYSICALLY happening inside the transistor once it has been biased and then a small ac signal is given(not in terms equations)? \$\endgroup\$ – Developer Android Apr 8 '13 at 13:09
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    \$\begingroup\$ @DeveloperAndroid, I think your difficulty is rooted in your notion of "forward biased completely". When we say that the junction is "forward biased" , or "on", we simply mean that forward voltage is sufficient to produce "significant" current, e.g., micro or milli-amps versus, say, pico or nano-amps. But, if you do understand the equation I give in my answer, you see that it is continuous and increasing; there's no "turning on completely" anywhere in that equation. Examine you ideas closely here. I think you'll find one or more of them is wrong. \$\endgroup\$ – Alfred Centauri Apr 8 '13 at 14:08
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Yes, the base current causes things to happen in the transistor: a proportional amount of collector current flows. So changes in base current are amplified into larger changes in collector current.

(This is just an simplified model of the transistor which happens to work in many kinds of circuits. In fact the BJT is a voltage controlled device: the amount of collector current is a function of \$V_{BE}\$, namely the Ebers-Moll equation. That also factors in temperature, which the simplified model doesn't.)

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  • \$\begingroup\$ I wanted to know how a change in base current after the transistor has been forward biased causes such a large change in collector current. \$\endgroup\$ – Developer Android Apr 6 '13 at 2:26
  • \$\begingroup\$ The simplified explanation of it is that the base region is very thin. As the electrons traverse the base region toward the base terminal, they easily "fall" into the collector. Most of them do not make it. So to maintain the small base current in the face of this attrition of electrons into the collector, extra current has to come into the emitter. \$\endgroup\$ – Kaz Apr 6 '13 at 2:30

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