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I'm covering BJT's in my semiconductor class and in lecture, the professor said that BJTs are used as amplifiers in the linear region. I'm confused because I thought the gain was proportional to the base current.

By that logic wouldn't the highest gain be in the saturation region?

Can someone clear up this confusion?

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    \$\begingroup\$ What can be confusing is that the saturation region of a MOSFET is sometimes called the linear region in a BJT whereas the sat region in a BJT is sometimes referred to as the linear region where Vgs can control drain-source resistance! \$\endgroup\$ – Andy aka Jun 10 '15 at 20:43
  • \$\begingroup\$ Indeed, the names are swapped. VERY confusing, whoever came up with that ??? \$\endgroup\$ – Bimpelrekkie Jun 10 '15 at 21:25
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    \$\begingroup\$ @Rimpelbekkie In the BJT it refers to current saturation, in the MOSFET it refers to voltage saturation and the names were chosen since the BJT is a current-controlled device and the MOSFET is a voltage-controlled device (to first order). It makes sense from a device physics standpoint. From a circuit design standpoint not so much. \$\endgroup\$ – crgrace Jun 10 '15 at 21:37
  • \$\begingroup\$ Why was this question downvoted? I thought it was valid question that met the guidelines of this site. Can someone provide feedback so I can make better questions in the future? \$\endgroup\$ – Adam Jun 11 '15 at 13:32
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If you´re just going to switch a large current on/off with a small controlling current (at the base) then you want the transistor to go into saturation.

But if you want the collector current to follow, for instance an oscillating signal at the base, then you want the transistor to stay in the linear region.

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  • \$\begingroup\$ I think I understand, so the current coming from the supply would follow and amplify the input signal while in the linear region? \$\endgroup\$ – Adam Jun 10 '15 at 19:08
  • \$\begingroup\$ That is correct Adam. \$\endgroup\$ – Dejvid_no1 Jun 10 '15 at 19:13
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Consider the characteristic curves of a typical small-signal NPN transistor:

enter image description here

Referring to the right-hand graph, we can see that the DC Current Gain hFE is almost constant with collector current provided Vce is maintained constant at 1V. It starts to drop as you approach Ic of 100mA and (not shown) it will also drop at very low collector current.

If you now look at the left-hand graph, the intersection of Vce = 1.0V corresponds to a point on the right-hand graph for each of the base currents from 0.1mA to 1.0mA. As the collector current increases the gain is less, so the space between the evenly spaced base currents becomes less. Now, imagine the transistor is entering saturation, so Vce is less than 1V, approaching zero. For example at 1mA base current, and 0.1V Vce you can see that the collector current is only about 25mA- that means the current gain has dropped to 25 from more than 200. At 0.1V and 0.1mA base current, the gain is more like 50, still a far cry from the 300 at 1V Vce.

As Vce approaches zero, the gain current drops to zero. This makes sense because the transistor needs some voltage to work. At some point very near zero it will even turn slightly negative as increasing base current causes current to flow out of the collector.

The other thing you might find interesting is that at low base current the curves are fairly flat- the gain does not change much with Vce provided it sees "enough".

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Please note that the "saturation region" for a BJT is the region where Vce < Vce_sat. In this region of operation, Ic is not only determined by Ib and Vbe but also by Vce. If you would determine a small signal model of the BJT in the saturation region you would find an extra component "eating up" part of the collector current resulting in less gm (what you call gain).

The linear region is where Vce > Vce_sat. In this region of operation Ic is mainly determined by Ib and Vbe but not so much by Vce (only through the Early effect). This results in a higher output impedance (at the collector) if compared with the saturated region, resulting in more usable gm.

Although a signal can be amplified by a BJT in either region, the linear region is more convenient to use as the Vce dependancy is (in 1st order) removed.

The (small signal) transconductance gm = dIc/dVbe of a BJT is mainly dependent on Ic but since beta = Ic/Ib (assuming we're in the linear region !) you could say that the gm (I would not call that gain as gm has unity [A/V]) is dependent on Ib. But it is more convenient to say the gm is dependant on Ic since gm = Ic/Vt.

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  • \$\begingroup\$ Just a small correction: The Early effect causes a LOWERING of the output resistance. \$\endgroup\$ – LvW Jun 10 '15 at 20:19
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    \$\begingroup\$ A BJT is saturated when both junctions are forward-biased. The \$V_{\text CE}{\text (SAT)}\$ is a nominal value that depends on the collector current and such. \$V_{\text CE}\$ cannot be lower than \$V_{\text CE}{\text (SAT)}\$, by definition. Whatever \$V_{\text CE}\$ value is measured at saturation is \$V_{\text CE}{\text (SAT)}\$ (for the given additional conditions that are present). \$\endgroup\$ – Kaz Jun 10 '15 at 20:49
  • \$\begingroup\$ @ LvW: Please read carefully, the point I wanted to make is that the in the linear region, the Early effect is the main culprit affecting Ic(Vce) and that this results in a higher output impedance COMPARED to saturation region. I never suggested the Early effect increases Rout because it doesn't. \$\endgroup\$ – Bimpelrekkie Jun 10 '15 at 21:19
  • \$\begingroup\$ @ Kax: "VCE cannot be lower than VCE(SAT), by definition". Then why do all transistor curves show VCE from 0 to whatever while for a certaing VBE the VCE_sat would be 200 mV for example ? What you mean is that the linear region starts where VCE > VCE_sat, exactly as I mentioned above. \$\endgroup\$ – Bimpelrekkie Jun 10 '15 at 21:23
  • \$\begingroup\$ @Rimpelbekkie,OK - I see. To avoid such misinterpretations it would help to say always "higher ...if compared with..". \$\endgroup\$ – LvW Jun 11 '15 at 8:55

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