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A transistor, say $NPN$, can be used as an amplifier. It amplifies base current by a factor beta. Is there a minimum base current threshold below which it cannot amplify? Say, are there transistors which can amplify femtoampere?

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  • $\begingroup$ The beta coefficient is not the exact gain. $\endgroup$ Jun 2 '17 at 16:00
  • $\begingroup$ A femtoamp is only roughly 6000 electrons/second. For a DC current, electron drift speed is on the order of 0.0002 meters per second, so you're in a regime where transfer time across the p-n junctions matter. That suggests that "gain" becomes a not very useful concept. $\endgroup$ Jun 2 '17 at 16:05
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To answer the question, we can consider Ebers-Moll's ecuations: $$I_E=I_{ES}(e^{V_{EB}/V_T}-1) - \alpha_RI_{CS}(e^{V_{CB}/V_T}-1)$$ $$I_C=\alpha_FI_{ES}(e^{V_{EB}/V_T}-1) - I_{CS}(e^{V_{CB}/V_T}-1)$$ $$I_B=I_E-I_C$$ where $I_{ES}$ and $I_{CS}$ are the reverse saturation currents, $V_T=\frac{KT}{q}$ ($K$ Bolzmann constant) and $\alpha_F$ and $\alpha_R$ the forward and the reverse constants. As we can see, unless both PN junctions are in reverse direction (it is said, $V_{EB}=V_{CB}=0)$, which depends on the circuit and the input voltage, the bipolar junction will amplifiy the current of the emitter.

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  • $\begingroup$ it is not clear. will the base current of any positive amplitude however small amplified? $\endgroup$
    – jgyggiuig
    Jun 2 '17 at 9:01
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    $\begingroup$ I explain it in other way. the BJT has 4 operation states: cut-off, forward active, saturation and reverse active. The last one is the less commmon, so forget about it. When we connect a transistor and it is not in cut-off (amplification 0), it works in active (very high amplification) or in saturation, where it has less amplification but it is still considerable. Anyway, this depend essentially of the circuit, but the theory tell us BJT don't have a treshold voltage. $\endgroup$
    – Josemi
    Jun 2 '17 at 9:44
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In a normal NPN transistor we are dealing with three states (seen comment section on Josemi post). enter image description here

The red area, indicates the saturation the BC junction is forward polarised, the the green area is called linear, the BC junction is inverse polarised and the transistor works as an amplifier and when $I_B = 0$ we land in cut-off.

I prefere to work with the simplified version, so i take the threshold voltage of NP or PN junction as $0.7_v$. enter image description here

Let investigate all the three area a little bit more, starting with cut-off:

$$V_{BE} < 0.7_v \Rightarrow I_B = I_C = I_E = 0$$

Linear area:$$ \left\{ \begin{array}{ll} V_{BE} = 0.7_v & \\ V_{CE} > 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{ll} I_B>0 & \\ I_C = \beta I_B \end{array} \right. $$

saturation: $$ \left\{ \begin{array}{ll} V_{BE} = 0.7_v & \\ V_{CE} = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{ll} I_B>0 & \\ I_C < \beta I_B \end{array} \right. $$

As you can see there is no limitation on $I_B$. However, this just a theory, in reality, i don't think so they come in handy when the bias current is about some femtoamps, otherwise why, using CMOS then?

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The short answer is yes, there is a minimum, but it's device dependent and usually not specified. The GE Transistor Manual (no longer in print, but considered to be holy writ as recently as the 1970s)explains why ordinary leakage currents don't turn on SCRs. Referring to the 2 transistor SCR model (easily found online) it says that beta for transistors is dependent on collector current, and that SCR equivalent transistors have betas less than 1 for expected leakage currents.

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  • $\begingroup$ There being a minimum doesn't mean a transistor can't be used to amplify low currents - it just has to be connected for a much larger bias current that the signal current is added to. $\endgroup$
    – stretch
    Jul 6 '17 at 15:26

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