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I have a problem with understanding how does the negative feedback of this circuit work.

In all books that I read it is said that "with the rise of temperature, collector current increases". Why is that? I know that \$\beta\$ increases as well and \$U_{BE}\$ decreases. Looking at the Thevenin's configuration of \$R_1\$ and \$R_2\$:

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We can write that base's potential (in Polish word for 'shunt' starts with b, so don't mind index at the image above)

\$V_B=E_s-I_BR_S=U_{CC}\cdot\frac{R_2}{R_1+R_2}-I_B\cdot\frac{R_1\cdot R_2}{R_1+R_2}\$

Hence, \$U_{BE}=V_B-V_E=U_{CC}\cdot\frac{R_2}{R_1+R_2}-I_B\cdot\frac{R_1\cdot R_2}{R_1+R_2}-I_ER_E\$

Now assume, that the temperature had increased. This means that \$U_{BE}\$ will decrease and \$\beta\$ will increase. What I don't understand is how the thought process goes next.

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  • \$\begingroup\$ Collector to emitter leakage current, Iceo, doubles for each 10 degree C rise in the junction temp. \$\endgroup\$
    – AlmostDone
    May 10 '18 at 18:56
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    \$\begingroup\$ Are you sure the temperature is related to the negative feedback? Because if someone showed me that circuit and asked me to explain the negative feedback, I'd probably point to the emitter degeneration resistor. \$\endgroup\$
    – Hearth
    May 10 '18 at 20:34
  • \$\begingroup\$ Yeah, whenever this topology was presented in my undergrad the emitter resistor was pointed to as a the source of negative feedback, in part to combat the temp/current positive feedback that can lead to thermal runaway. \$\endgroup\$
    – esilk
    May 10 '18 at 20:55
  • \$\begingroup\$ @Felthry yes I am sure, in the lecture slides it is written that negative feedback, which is possible thanks to the Re resistor, is to stabilize the quiescent point, i.e. to prevent various temps and betas from changing it \$\endgroup\$
    – SantaXL
    May 10 '18 at 22:23
  • \$\begingroup\$ SantaXL- No, Vbe will NOT decrease with temperature. The other way round: If you want to keep Ic constant (for rising temperatures) you must externally decrease Vbe in order to bring Ic back close to its desired value. \$\endgroup\$
    – LvW
    May 11 '18 at 10:09
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Assuming \$V_\text{TH}=V_\text{CC}\frac{R_2}{R_1+R_2}\$ and \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$, then:

$$I_\text{C}=\beta\cdot\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)R_\text{E}}=\frac{V_\text{TH}-V_\text{BE}}{\frac{R_\text{TH}}{\beta}+\frac{\beta+1}{\beta}R_\text{E}}\approx\frac{V_\text{TH}-V_\text{BE}}{R_\text{E}+\frac{R_\text{TH}}{\beta}} $$

There are two temperature-dependent variables in the above: \$V_\text{BE}\$ and \$\beta\$. As you point out, temperature tends to decrease \$V_\text{BE}\$ and increase \$\beta\$. Ignoring \$\beta\$, a decreasing \$V_\text{BE}\$ would tend to increase \$I_\text{C}\$. Ignoring \$V_\text{BE}\$ , an increasing \$\beta\$ would also tend to increase \$I_\text{C}\$. So the effects of temperature on \$V_\text{BE}\$ and \$\beta\$ tend to operate in the same direction on \$I_\text{C}\$.

As a side note, you can see that if \$R_\text{E}\gg \frac{R_\text{TH}}{\beta}\$ then this fact alone tends to stabilize the collector current against variations in \$\beta\$ (for temperature or for part variations.) \$R_\text{E}\$ also reduces variations due to \$V_\text{BE}\$, but it also simply reduces \$I_\text{C}\$, generally, too.


A more mathematical way of asking this question is to compare the impacts. For any given value of \$R_\text{E}\$, what percent variation in \$I_\text{C}\$ can we expect for a given percent variation in \$V_\text{BE}\$ or for a given percent variation in \$\beta\$. And knowing that, how do they compare with each other?

Here they are:

$$\begin{align*}\frac{I_\text{C}\: \text{% change}}{V_\text{BE}\: \text{% change}}&\left\{\begin{array}{l} \mu_\text{vbe}=\frac{\frac{\text{d}I_\text{C}}{I_\text{C}}}{\frac{\text{d}V_\text{BE}}{V_\text{BE}}}&=-\beta\cdot\frac{V_\text{BE}}{I_\text{C}\left(R_\text{TH}+\left(\beta+1\right)R_\text{E}\right)}\end{array}\right.\\\\\frac{I_\text{C}\: \text{% change}}{\beta\: \text{% change}}&\left\{\begin{array}{l} \mu_\beta=\frac{\frac{\text{d}I_\text{C}}{I_\text{C}}}{\frac{\text{d}\beta}{\beta}}&=\beta\cdot\frac{\left(V_\text{TH}-V_\text{BE}\right)\bigg[\frac{R_\text{TH}+R_\text{E}}{R_\text{TH}+\left(\beta+1\right)R_\text{E}}\bigg]}{I_\text{C}\left(R_\text{TH}+\left(\beta+1\right)R_\text{E}\right)}\end{array}\right.\end{align*}$$

To use them, just use them as in \$\% I_\text{C}=\mu_\text{vbe}\cdot \%V_\text{BE}\$ and \$\% I_\text{C}=\mu_\beta\cdot \%\beta\$.

For example, in one circuit I tested that was designed for \$I_\text{C}\approx 1\:\text{mA}\$ gave an actual \$I_\text{C}=1.04\:\text{mA}\$. After a \$25\:^\circ\text{C}\$ rise (using a controlled hot plate), I measured \$I_\text{C}=1.10\:\text{mA}\$. I also measured a \$-6\%\$ change in \$V_\text{BE}\$ and a \$+12\%\$ change in \$\beta\$. The above equations for the circuit I had gave me \$\mu_\text{vbe}\cdot \%V_\text{BE}=+5.8\%\$ and \$\mu_\beta\cdot \%\beta=+0.7\%\$. Combined, this suggests \$6.5\%\$ change on \$I_\text{C}\$. So:

$$I_\text{C}=1.065\cdot 1.04\:\text{mA}\approx1.11\:\text{mA}$$

Note that this is very close to what I actually got.

The ratio of the above two factors is:

$$\bigg\lvert{\frac{\mu_\text{vbe}}{\mu_\beta}}\bigg\rvert=\frac{1}{\frac{V_\text{TH}}{V_\text{BE}}-1}\cdot\frac{R_\text{TH}+\left(\beta+1\right)R_\text{E}}{R_\text{TH}+R_\text{E}}$$

Now, here you can see why \$\mu_\text{vbe}\$ dominates \$\mu_\beta\$. So long as \$V_\text{TH}\gt 2\:V_\text{BE}\$, the first factor will be somewhat less than 1. But the second factor is always greater than 1 and often a lot greater -- for example, 10 or so. So it's generally the case that with emitter degeneration in the well-designed CE circuit, the percent changes in \$V_\text{BE}\$ are more important than percent changes in \$\beta\$, even though the actual percent changes might be smaller. Their relative impacts are such that the base-emitter voltage changes are still the more important ones to worry about (if you need to worry at all.)

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  • \$\begingroup\$ Ok, now I see it. One more question, regarding transfer (I am not sure if this is a proper word, but I have \$I_C\left(U_{BE}\right)\$ on my mind) characteristic Doesn't decreasing \$U_{BE}\$ cause \$I_C\$ to decrease as well, so it is another thing that stabilizes the quiescent point? \$\endgroup\$
    – SantaXL
    May 10 '18 at 23:00
  • \$\begingroup\$ @SantaXL Yes, but that chart is with the temperature held solid. In the case of temperature change, the value of \$V_\text{BE}\$ decreases for the same \$I_\text{C}\$. Since the rest of the circuit is involved in setting \$I_\text{C}\$ (it's not just a BJT by itself, but a circuit of parts), the \$V_\text{BE}\$ declines for the same \$I_\text{C}\$. But then the circuit responds to that decline by increasing the base current a little bit. The base current change is now magnified still more by the slightly increasing \$\beta\$, as well. \$\endgroup\$
    – jonk
    May 10 '18 at 23:42
  • \$\begingroup\$ @SantaXL I'm going to add some interesting stuff at the bottom. You may, or may not, care. But it might be fun, anyway. \$\endgroup\$
    – jonk
    May 11 '18 at 0:12
  • \$\begingroup\$ A real world example is always appreciated :) Thanks again \$\endgroup\$
    – SantaXL
    May 11 '18 at 9:30
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Sometimes a picture can tell you more than many words (at least, it can support verbal explanations). The stabilization line clearly shows how a resistor RE (negative feedback) can reduce the variations in Ic (caused by temperature changes or other uncertainties like beta variations). Note that for RE=0 we have a vertical "stabilization line" (dotted) - and no stabilization at all!!

The second graph shows a similar effect caused by a large base resistor RB.

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