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Updated/Corrected question:

Realizing what confuses me the most, I will make this more straight-forward. Say that a transistor works in the forward active region and it receives an ac input signal at its base.

Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region? Is there any chance this might happen before first entering the saturation region?

Original post:

I'm soon taking an exam on transistors and I've just started my revision. After having a look at the conditions for each region of operation and some example problems, there is something I don't understand or that maybe I don't remember very well.

The condition for an npn bjt to be on is that the base emitter voltage is above 0.7 volts. From what I remember this is not an accurate value but let's say it is around there.

When we analyse the small signal model of a circuit , the values we calculate are superimposed on the dc values, right? So, when we apply an ac signal to the base of the bjt, the base emitter voltage changes.

So, my questions are : Is there a chance that the ac input signal will take the bjt in the cut off region?

If yes, how do we check it?

If yes, when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region? I mean, which comes first as the swing increases? Does it depend on the operation point?

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  • \$\begingroup\$ Normally we first do a DC analysis and then base on results we can see how far the BJT is from the saturation/cut-off. When we do small signal we do not care about the saturation or cut-off because the small-signal analysis is only valid if d(Vbe) is less than 10mV( we need 18mV of a Vbe change to double the collector current, 60mV to change Ic current 10 times). \$\endgroup\$ – G36 Jan 11 '20 at 15:37
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Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region?

Yes this is possible, and it would result in non-linear operation of your amplifier.

Is there any chance this might happen before first entering the saturation region?

Yes it's possible to (mis-)design an amplifier so this happens.

when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region?

You need to consider both when designing your amplifier.

Assuming we're talking about a common emitter amplifier, you need to be sure the amplifier can handle the complete range of input amplitudes desired without going into cut-off, and you need to be sure it can handle the complete range of output amplitudes without entering saturation.

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I guess this question is more about vocabulary than about BJTs.

"AC small signal analysis" or "small signal conditions" is a useful approximation to simplify the analysis of a circuit.

This approximation assumes the AC signal voltage wiggles around the DC operating point in a "small enough" interval that the operating point and all characteristics of circuit devices can be assumed to be constant. This allows to approximate all non-linear devices (like transistors) with linear equations.

For example, in small signal analysis, a diode is approximated by a resistor, and the value of this resistor depends on the DC current flowing in the diode.

So, my questions are : Is there a chance that the ac input signal will take the bjt in the cut off region?

No... and yes 😁

In a real circuit, yes of course.

In small signal analysis, no. If the "small signal" is large enough to cause a significant change in device operating parameters, like your example of turning the transistor on and off, then the small signal approximation is no longer valid... and this is no longer small signal analysis, but large signal analysis.

Small signal analysis replaces the system under study with a linear approximation. This is 100% correct if the AC signal is infinitely small, and will give correct results for gain, transfer function, impedance, etc. The accuracy of the approximation breaks down progressively as signal level increases, which manifest as distortion (for example the \$ g_m \$ of your transistor will vary a bit during the voltage swing). When the signal is strong enough to make devices switch on/off, the approximation breaks down completely.

If yes, how do we check it?

First do the small signal analysis. Then calculate the actual voltage swings and check by how much the operating point of your transistor varies or if will turn off or saturate.

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Yes the AC signal superimposed on the DC bias voltage can take the base-emitter voltage into the cutoff region

Some amplifiers are even designed to have the signal always do this, these are called class B or class C amplifiers. Amplifiers that have the singal always be in the linear region are called class A

To know whether your amplifier is operating as a class A, B or C amplifier, you need to calculate whether the input signal causes the base-emitter voltage to fall into the cutoff region, to do this you need to know the base-emitter bias current, the input impedance, and the source current (or voltage) and impedance.

You always design your amplifier so that it does not enter the saturation region, as long as the signal amplitude is within the operating range

Going into the saturation region means having a forward biased base-collector junction, and in this region the transistor stops amplifying the signal and just conducts large currents from base to emitter and collector.

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