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The bipolar transistor

Also called BJT transistor, we can see it as a switch and a current amplifier. There are 2 types: NPN and PNP.

Symbols :

bipolar transistor

3 branches: the base (B), the collector (C) and the emitter (E).

Feature :

Let’s study the characteristic of an NPN:

characteristic of an NPN

The principle is simple: always for an NPN:

  • if a current Ib is applied to the base and the voltage Vce is positive, then the transistor is conducting => presence of Ic and Ie.
  • otherwise the transistor is blocked => no Ic or Ie.

Let’s study the different operating modes of an NPN:

For the explanations, let’s use the small assembly (called polarization) below:

NPN polarization

   

According to the mesh law, we have: E = Rc.Ic + Vce => Ic = (E – Vce) / Rc, this equation is called “load line”.
Let us draw it (in blue) on the characteristic of the npn:

npn transistor charge right

The different operating modes:

  • if Ib = 0 or Vbb <0 or Vbe = 0 then the transistor is blocked => Ic = 0 => Vce = E (according to the load line). The transistor can be seen as an open switch. On the above characteristic we are therefore at point B.
  • if Ib> 0 the transistor is on. Semiconductor physics then automatically dictates Vbe = 0.6 V (approximately). We say that the BE junction is forward biased. But 2 cases can then arise:
  • if Vce> Vce_sat (= 0.2 V approximately), for example: Vce = 2 V. Then Vbc = Vbe – Vce = 0.6 – 2 = -1.4 V <0 therefore the BC junction is reverse biased. In this case Ic = ß.Ib (ß: gain in current). The transistor operates in a regime called linear, and can be seen as a current amplifier. In addition, according to the law of nodes: Ie = Ib + Ic => Ie = (ß + 1) .Ib (usually ß is greater than 100, so we can make the approximation Ie = ß.Ib = Ic) . This corresponds to point A of the characteristic.
  • if Vce <Vce_sat then the BC junction is also forward polarized, and we no longer have the previous relations. Indeed, in this case Ic <ß.Ib. The transistor is then in saturation regime. Vce_sat being very low (approximately 0.2 V), we will automatically say that when the transistor is saturated then Vce = Vce_sat. This corresponds to point S of the characteristic.

Small recap:

Junction BEBC JunctionNPN transistor state
BlockedBlockedBlocked <=> switch open => Ic = 0
PasserbyBlockedLinear regime: Vbe = 0.6V, Ic = ß.Ib, Vce> Vce_sat
PasserbyPasserbySaturated regime: Vbe = 0.6V, Ic <ß.Ib, Vce = Vce_sat = 0.2V

Equivalent diagram of an NPN transistor in AC LF (in alternating small signals, for low frequencies):

NPN transistor in AC LF

rb = ß.Vt / Ic (rb of the order of 1kOhms in general) with Vt = kT / q (k: 1.3806 × 10-23 J / K, T: temperature in K, q: elementary charge: 1 , 60217646 × 10-19 C), Vt = 25mV approximately at 25 ° C.

To note :

  • there is also an equivalent diagram in AC HF (high frequencies) but very very complicated: other resistors and capacitors are involved.
  • we have studied the transistors from the example of the NPN, but for the PNP it is the same thing with reversed voltages. For example when a PNP is driving, then Vbe = -0.6 V => Veb = 0.6 V.

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