Understanding the biasing of a transistor with negative feedback

I am trying to understand the following circuit.

However, I don't understand this step in the accompanying explanation:

"A transistor with a large gain will cause a large voltage drop over the collector resistor. Because of this, the collector voltage and the base current will decline."

I don't 'get' why this chain of reasoning is true. Why does a large gain causes a large voltage drop over the collector resistor? And then, why does this cause the collector voltage and base current to drop?

• Where did the quote come from - can you say? (it's true by the way). Might help Commented Jan 22, 2021 at 15:31
• @Andyaka The quote came from an old Elektor magazine. Commented Jan 22, 2021 at 15:32
• What is the question you want to have an answer? Commented Jan 22, 2021 at 15:33
• @Justme I added the question. Indeed, I wasn't clear. Commented Jan 22, 2021 at 15:39

The reasoning goes a bit like in circles. The more base current flows, the more collector current flows. The more collector current flows, it brings down collector voltage, and it reduces base current. So there is an equilibrium where voltages and currents are stable, so just the right amount of base current is in balance with just the right amount of collector current.

The collector current is roughly 100 to 400 times the base current, as shown in the image. Transistors have quite large manufacturing tolerances regarding the current gain.

Consider the case where transistor current gain $$\ \beta = 220 \$$...
Thus, collector current is 220 x base current, and current through the 1k collector resistor is dominated by collector current.
Since 220k base resistor is 220 x collector resistor, voltage drop across each of these resistors would be very nearly the same....because their ratio is the same as $$\\beta\$$. The sum of both voltage drops must equal approximately (5V - 0.6V base-emitter voltage).

simulate this circuit – Schematic created using CircuitLab

If transistor current gain is greater than 220, then base current tends to be greater than 220 x collector current, resulting in $$\V_{R_b} < V_{R_c}\$$. These voltage drops must still add to 4.4V.

This is negative feedback, since base current depends on collector voltage. In many cases, this circuit is designed to have collector voltage nearly half-way between DC supply and GND. Not much negative feedback in this case, but some.

"A transistor with a large gain will cause a large voltage drop over the collector resistor. Because of this, the collector voltage and the base current will decline."

Imagine the circuit running just fine with some collector and base currents and the collector voltage sitting at, let's say, $$\3\:\text{V}\$$. This means that the resistor to the $$\+5\:\text{V}\$$ rail ("collector resistor") must be dropping exactly $$\2\:\text{V}\$$. The reason it is dropping that voltage is almost entirely because of the current that the collector is pulling downward. (There is a tiny base current, but as you can see it's at least 100 times smaller, so you can discount it for now.)

We don't know what the $$\\beta\$$ is (because it's my example and I am only telling you that it is somewhere between 100 and 400.) But it is some value and it is the ratio of the collector current to the base current: $$\\beta=\frac{I_\text{C}}{I_\text{B}}\$$. Put another way: $$\I_\text{C}=\beta\cdot I_\text{B}\$$.

Suppose $$\\beta\$$ changes just slightly and rises up a tiny bit (perhaps due to heat?) What then happens? Well, assume for a moment that the base current does not change and see where that would take us, logically. If $$\I_\text{B}\$$ stays the same but $$\\beta\$$ rises a tiny bit then this means $$\I_\text{C}\$$ also rises, but magnified by this slightly larger $$\\beta\$$. That current will mean a larger voltage drop across the "collector resistor" and that will lower the voltage at the collector, itself. But lowering the collector voltage (assuming the base voltage doesn't change much by comparison) means that the resistor from collector to base ("base resistor") will have less current because there is a smaller voltage across it (the voltage difference taken between the collector and base.) So the base current must drop in response to an increase in $$\\beta\$$.

But that means that the collector current also drops, too.

We started this whole process by saying that the collector current increased, if the base current stayed the same. But now we've shown that the base current doesn't stay the same, but instead declines a little, and that this implies that the collector current declines -- which is the opposite of saying it increases.

In short, we've just demonstrated that an increase of the collector current, supposedly implied by a small increase in $$\\beta\$$, is then countered by a reduction in base current, so as to reduce the collector current. That's negative feedback and it helps stabilize the BJT's operating point over some useful range of $$\\beta\$$.

There's a way of quantifying all these details. It's called sensitivity analysis. Start with the simplified DC solution for the circuit:

\begin{align*} \begin{array}{r} {\text{collector node KCL:}}\vphantom{frac{V_\text{C}}{R_\text{C}}+\frac{V_\text{C}}{R_\text{B}}+\beta\cdot I_\text{B}}\\\\ {\text{base current:}}\vphantom{\frac{V_\text{C}-V_\text{B}}{R_\text{B}}} \end{array} && \begin{array}{r} \frac{V_\text{C}}{R_\text{C}}+\frac{V_\text{C}}{R_\text{B}}+\beta\cdot I_\text{B}\\\\ I_\text{B} \end{array} & \begin{array}{c} &\quad{=}\vphantom{frac{V_\text{C}}{R_\text{C}}+\frac{V_\text{C}}{R_\text{B}}+\beta\cdot I_\text{B}}\\\\ &\quad{=}\vphantom{\frac{V_\text{C}-V_\text{B}}{R_\text{B}}} \end{array} & \begin{array}{l} \frac{V_\text{CC}}{R_\text{C}}+\frac{V_\text{B}}{R_\text{B}}\\\\ \frac{V_\text{C}-V_\text{B}}{R_\text{B}} \end{array} \end{align*}

The solution for the collector voltage is:

$$V_\text{C}=\frac{V_\text{B}\,\left(\beta+1\right)R_\text{C}+V_\text{CC}\,R_\text{B}}{\left(\beta+1\right)R_\text{C}+R_\text{B}}$$

(This is the same equation you'd get if you had a resistor divider between two voltage sources. But save that thought for a later day.)

To find the sensitivity equation, you need to take the derivative of the above equation with respect to $$\\beta\$$ and then multiply it by another fraction. But you can get a value $$\S\$$ that you can simply use like this:

$$\%\:V_\text{C}=\%\:\beta\:\cdot\: S$$

In short, if you want to know the percent-change in the collector voltage for some given percent-change in $$\\beta\$$, then the above equation (if you know $$\S\$$) will help out a lot.

The value $$\S\$$ depends on the DC operating point, though. So it's not a constant for the entire range: $$\100\le\beta\le 400\$$. Perhaps the simplest way to use it would be to pick the mean: $$\\overline{\beta}=\frac{100+ 400}{2}=250\$$, figure out the DC operating point there, compute $$\S_{\beta=250}\$$, and see what that says.

(Note: using 250 means that $$\\beta\$$ can vary a balanced $$\\pm 60\%\$$ for your range of $$\100\le\beta\le 400\$$.)

For your circuit I get $$\S_{\beta=250}\approx -0.39361\$$. So, this means that an increase in $$\\beta\$$ will cause a decrease in the operating point for the collector voltage (the minus sign says so.) And that if we changed $$\\beta\$$ by 10%, we'd expect to see a -3.9361% change in the collector voltage.

Solving for the collector voltage with $$\\beta=250\$$ I get $$\V_\text{C}\approx 2.7085\:\text{V}\$$. Solving for the collector voltage with $$\\beta=275\$$ (a 10% increase) I get $$\V_\text{C}\approx 2.6073\:\text{V}\$$. The value predicted using $$\S_{\beta=250}\$$ would be $$\\left(1-0.039361\right)\cdot 2.7085\:\text{V}\approx 2.6019 \:\text{V}\$$. So the idea works reasonably well.

Roughly speaking, this says that the change in the collector voltage is about $$\\frac{2}{5}^\text{ths}\$$ the change in $$\\beta\$$, when speaking about percentages, and in the opposite direction of the change. (And when regarding your specific schematic values and using 250 as a medium value for $$\\beta\$$.)

To achieve the above, I just used sympy (free and easy to use):

var('vc vcc vb rb rc b ib')
ans = solve( [ Eq(vc/rc+vc/rb+b*ib, vcc/rc+vb/rb), Eq(ib, (vc-vb)/rb) ], [vc, ib])
for x in ans:x, ans[x].subs( {rc:1e3, rb:220e3, vb:.7, vcc:5, b:250} )
(ib, 9.12951167728238e-6)
(vc, 2.70849256900212)
for x in ans:x, ans[x].subs( {rc:1e3, rb:220e3, vb:.7, vcc:5, b:275} )
(ib, 8.66935483870968e-6)
(vc, 2.60725806451613)
sens = derivative( ans[vc], b ) * ( b / ans[vc] )
sens.subs( {rc:1e3, rb:220e3, vb:.7, vcc:5, b:250} )
-0.393605974955343
( (1+0.1*sens) * ans[vc] ).subs( {rc:1e3, rb:220e3, vb:.7, vcc:5, b:250} )
2.60188468317398


It's that simple.