Why base current is affected most on changing beta in voltage divider bias configuration of BJT?

Question

In the given voltage divider bias configuration when the value of \$ \beta \$ is changed from 80 to 140 the base current changes from 21.26\$\mu\text{A}\$ to 12.9\$\mu\text{A}\$ which is a change of 39 % where as the change in quantities like collector current,collector to emitter voltage is merely 5.4 % and 5.6 % respectively.

^{simulate this circuit – Schematic created using CircuitLab}

So,my question is why has the base current changed the most?

My guess to this question is that base current is in micro amperes so a change of about 10 micro ampere ,wouldn't bring much change to the circuit,although the operating point will be shifted.

Answer 1

What if beta were infinity? The base voltage would be exactly defined by the potential divider formed by R1 and R2 and the 16 volt supply: -

i.e. Vb would be 16V \$\times \dfrac{9.1}{9.1+62}\$ = 2.048 volts.

If beta dropped from infinity, the current taken by the base (although small) would lower that 2.048 volts a little bit. With a base current of 20 uA, Vb drops to about 1.89 V (a drop of about 7.8%).

I converted the 16 volt source to a current source and paralleled R1 and R2 then subtracted 20 uA to get the new base voltage by the way

I'm just trying to demonstrate that a fundamentally massive swing in beta from around 90 to infinity results in a change in Vb of about 8%. This change in bias voltage has a knock-on effect on Ic and Ib of approximately the same amount.

If R1 and R2 were lowered in value by 10 the influence of the 20 uA base bias current would take Vb from 2.048 volts to 2.03 volts i.e. a drop of about 1%.

This is why "strong" bias resistor values are preferred because (in conjunction with an emitter resistor) the change in beta from one transistor to another (or from one temperature to another) results in a far smaller change in quiescent operating conditions.

Answer 2

Compute the Thevenin equivalent for the biasing point set by \$R_1\$ and \$R_2\$. In your case, this is \$V_{TH}=V\cdot\frac{R_2}{R_1+R_2}\approx 2.048\:\textrm{V}\$ and \$R_{TH}\approx 7.935\:\textrm{k}\Omega\$. The base current in the BJT then follows this calculation (which takes into account the thermal voltage's equivalent resistance):

$$\begin{align*} I_B&=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)\cdot R_E}\end{align*}$$

If I start out by estimating \$V_{BE}\approx 680\:\textrm{mV}\$, then I get the following results:

$$\begin{array}{l@{}r@{}c} & & I_{B} \\ \begin{array}{l} \beta \end{array} & \begin{array}{r} 80 \\ 140 \end{array} & \begin{array}{l} { \begin{array}{cc} 21.95\:\mu\textrm{A} \\ 13.26\:\mu\textrm{A} \end{array} } \end{array}\end{array}$$

Which is close to your results. The above equation tells you all you need to know about why \$\beta\$ multiplies the effect of the emitter resistor. (The Early effect also affects this and would work to slightly reduce the base current from the above figures.)

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You can see a related discussion I had a while back here: BJT amplifier (Vce) voltage!

Answer 3

Your confusion seems to come from an understanding of the BJT function which is not correct (better: not in accordance with the physical behaviour of the BJT). Please note that - in contrast to the explanations which can be found, unfortunately, in some textbooks - the BJT is a voltage-controlled device and NOT a current-controlled part.

That means: It is the voltage VBE which detrmines the current Ic=f(Vbe). This function is the well-known transfer function as given by the famous W. Shockley.

Unfortunately, the BJT has a base current which cannot be avoided (but it has no controlling function!) and, therefore, it is also to be considered during calculation of the base resistors. However, if the resistor niveau is chosen relatively low the actual value of the base current (and the corresponding uncertainties) play a minor role only.

And exactly this was the effect you have observed.

For this purpose, it is common practice to select these resistors so that the current through the divider is at least 10 times the base current. In your example, the divider current is app. 230µA.

(Remark: Your question is a good example which shows why a false understanding of the BJT`s working principle can cause misunderstandings and confusion).

Answer 4

We want the bipolar to operate as a constant-current device, so the operating point, the Vce, is constant. Picking resistors to accomplish that ---the constant-current --- is an acquired design skill.