How does the collector feedback bias circuit work?

Question

I'm a beginner in electronics, and I'm studying about BJT biasing.

I'm struggling with the "collector feedback biasing" method:

How does the collector feedback bias circuit work?

How does feedback in this circuit help stabilize the Q point?

Answer 1

There exists more than one way to draw the feedbacks. But this one comes easier to mind (for me):

^{simulate this circuit – Schematic created using CircuitLab}

The above illustrates certain important details. For example, the Shockley equation that, when the NPN BJT is in active mode, expresses the relationship between collector current and base-emitter voltage. Either can be used to control the other. Here, I chose to show the collector current as creating a base-emitter voltage. However, an entirely different (but equivalent) diagram could have made the choice to go the other way.

All of the Ebers-Moll DC Level 1 model-related part, temperature and time variations are identified and shown where they exist in the feedback loops.

Now, the above diagram has an element to it that violates the usual rules in laying out a linear feedback system: it includes a non-linear block called Shockley. This refers to the Shockley diode equation which is used in the simplified case for an active mode BJT.

Here's what the setup might look like in SymPy:

vcb, vc, ic, vbe, vrc, rc, rb, beta, vcc = symbols(
    "vcb, vc, ic, vbe, vrc, rc, rb, beta, vcc", real = True )   # vars
Shockley = Function( 'Shockley' )                               # unspecified
eq1 = Eq( vc, vcc - vrc )                                       # Vc
eq2 = Eq( vrc, vcb * (rc/rb) * (beta+1) )                       # V(Rc)
eq3 = Eq( vcb, vc - vbe )                                       # Vcb
eq4 = Eq( ic, (vcb/rb)*beta )                                   # Ic
eq5 = Eq( vbe, Shockley( ic ) )                                 # Shockley relation

# solve( [ eq1, eq2, eq3, eq4, eq5 ], [ vc, ic, vcb, vrc, vbe ] ) # won't solve

Before proceeding to solving the above, a note is needed.

Technically, a closed solution isn't possible without the application of the LambertW function. And many solvers will have trouble with this step. Usually, it's done manually by a human who can actually think for themselves. It could be done today by an AI if it were trained on the process. But so far as I know, the money hasn't been spent to do that.

If I tried to ask SymPy, using the above (commented out) solve, I would get this error: NotImplementedError: could not solve vbe.

But one can solve for the rest. (I'll ignore equation 5, above, and exclude \$V_{_\text{BE}}\$ from the solution list.)

Continuing now with SymPy and leaving, for now, \$V_{_\text{BE}}\$ as an input we get:

solve( [ eq1, eq2, eq3, eq4 ], [ vc, ic, vcb, vrc ] )
{vc: (rb*vcc + rc*vbe*(beta + 1))/(rb + rc*(beta + 1)),
 ic: beta*(-vbe + vcc)/(rb + rc*(beta + 1)),
 vcb: rb*(-vbe + vcc)/(rb + rc*(beta + 1)),
 vrc: rc*(beta + 1)*(-vbe + vcc)/(rb + rc*(beta + 1))}

Note that we get the exact same result here, for \$I_{_\text{C}}\$, as we'd get using KCL and KVL:

$$I_{_\text{C}}=\beta\cdot\frac{V_{_\text{CC}}-V_{_\text{BE}}}{R_{_\text{B}}+\left(\beta+1\right)R_{_\text{C}}}$$

(To see the SymPy code used to achieve the above equation using KCL and KVL, see this answer.)

So, rather than using KCL and KVL, we instead set up some standard closed-loop feedbacks and found the exact same result using this different approach.

That's probably a good thing.

Now, if we wanted, we could insert the Shockley equation (still unspecified) and see this:

$$I_{_\text{C}}=\beta\cdot\frac{V_{_\text{CC}}-\operatorname{Shockley}\left(I_{_\text{C}}\right)}{R_{_\text{B}}+\left(\beta+1\right)R_{_\text{C}}}$$

Unfortunately, we now have \$I_{_\text{C}}\$ on both sides of an equation and we already know that \$\operatorname{Shockley}\left(I_{_\text{C}}\right)\$ is non-linear and uses some kind of logarithm/exponential relationship.

Solving this equation, at least to get it in closed form, requires LambertW. I won't do that here. But if you are interested in seeing what steps may be similar, feel free to read here where I apply it in a different situation.

Some final notes about NFB.

I've already written enough and I'm not going to walk around each loop for you, now. Time enough has been spent getting to this point.

But you can readily see, as the single example I'll walk through now, that if there is a temperature effect that increases the saturation current of the BJT, which then acts to reduce the \$V_{_\text{BE}}\$ calculated from \$I_{_\text{C}}\$ by the Shockley equation, that this will have the effect of increasing \$V_{_\text{CB}}\$, which will then increase \$I_{_\text{C}}\$, which will increase \$V_{_\text{BE}}\$ and therefore counter the original effect due to temperature.

Feel free to consider all of the other ways that the above feedback diagram works to handle any number of various changes due to temperature, part variations, and drift over time (or even \$V_{_\text{CC}}\$.)

Answer 2

The resistor RB forms - together with the B-E path (DC resistance RBE) - a voltage divider driven by the voltage at the collector node. The current through RB is the base current producing the necessary B-E voltage of app. 0.65...0.7 volts.

Now - when the temperature rises the collector current Ic will rise causing an increased voltage drop across Rc. As a consequence, the driving voltage for the divider RB-RBE goes down and with it the base-emitter voltage VBE, which in turn again reduces Ic and (partly) compensates the temperature induced increase of Ic.

A verification of this sequence is based on the eqation

Ic=Is[exp(VBE/VT)-1].

In this equation, it is the current Is which is rather temperature sensitive. In order to avoid misunderstandings, it should be mentioned that any temperature change will also influence the temperature voltage VT - in addition to the current Is. However, this influence is rather small (in comparison to the Is sensitivity) and can be neglected.

Comment 1: A resistor Re between the emitter node and ground will have the same effect: Ic stabilization caused by VBE reduction for rising temperatures. However, in this case, it is more effective to use a separate resistive voltage divider for producing the required base voltage (as "stiff" as possible).

Comment 2: For both methods the stabilization effect works not only for temperature changes. The influence of other uncertainties on the desired Q-point is also reduced - parts tolerances and in particular the extremely large tolerances of B=Ic/Ib.

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EDIT 1:

Inspired by the block diagram as shown by jonk in his contribution, I have created an alternative block representation of the feedback effect.

(Quote jonk: "However, an entirely different (but equivalent) diagram could have made the choice to go the other way.")

Interestingly, the diagramm does not explicitely show the parameter B (resp. beta) - however, this quantity is "hidden" in the series connection of the two blocks "B-E path" and "Shockley".

Explantion: The block "B-E path" contains the function V_BE=f(I_B) which is identical to the inverse Shockley equation: V_BE=VT*ln[1+(I_B/I_Bo)]

^{simulate this circuit – Schematic created using CircuitLab}

EDIT 2:

It should be noted that both block diagrams (jonk`s post and my post) are valid for DC signals only. That means: They show how the DC collector current Ic is developed as a result of the supply voltage Vcc.

However, both diagrams can be linearized (and evaluated) for small signals values only. In this case, they show how the collector current changes (delta Ic=ic) as a function of a supply voltage variation (delta Vcc=vcc).

For the above diagram the result is

ic/vcc=1/[(1/gm)+(RB/beta)+(Rc/beta)+Rc]

In order to show the stabilization effect this expression is compared with the case when RB is connected to Vcc:

ic/vcc=1/[(1/gm)+(RB/beta)] .

Answer 3

As polarization is sometimes complicated, this tool may help you.
AppCAD from Avago Technologies.
It calculates the "stability" of this configuration.

Answer 4

The stabilization happens in terms of temperature dependency.

You have a \$\beta\$ that is temperature dependant. With the collector feedback bias you compensate for temperature changes in the way that if \$\beta\$ increases this directly tries to increase \$I_C\$ as well. With a higher \$I_C\$, the voltage drop across \$R_C\$ also tries to increase what in turn reduces the collector voltage and therefore the voltage across \$R_B\$. As a reduced voltage across \$R_B\$ also means a reduction in \$I_B\$ we have in combination with the increased \$\beta\$ from the start a more or less stable Q point.