Exercise 3.1 Biasing a circuit with an NPN

a)

An answer:
$$\begin{array}{rcll}& R_B& =\frac{V_{\mathit{RB}}}{I_{\mathit{RB}}}& \\ & & V_{\mathit{RB}}=V_{\mathit{CC}}-V_{\mathit{BE}}& \\ & & I_{\mathit{RB}}=I_B& \\ & & V_{\mathit{BE}}=0.6V(\mathit{assumption})& \\ & & I_B=\frac{I_C}{\mathit{\beta fe}}& \\ & R_B& =\frac{\mathit{\beta fe}(V_{\mathit{CC}}-0.6)}{I_C}& \end{array}$$

b)

An answer:
$$\begin{array}{rcll}& R_B& =\frac{V_{\mathit{RB}}}{I_{\mathit{RB}}}& \\ & & V_{\mathit{RB}}=V_{\mathit{CC}}-V_{\mathit{BE}}& \\ & & I_{\mathit{RB}}=I_B& \\ & & I_B=\frac{I_C}{\mathit{\beta fe}}& \\ & & V_{\mathit{BE}}=\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->(\frac{I_C}{I_{C0}}+1)& \\ & R_B& =\frac{\mathit{\beta fe}\cdot (V_{\mathit{CC}}-\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->(\frac{I_C}{I_{C0}}+1))}{I_C}& \end{array}$$

Note that typically $\frac{I_C}{I_{C0}}>>>1$ and hence the ”+1” can be ignored.

c)

An answer:
a: $R_B=\frac{200\cdot 9.4}{10^{-3}}=1.88M\mathrm{\Omega }$ b: Calculating $V_{\mathit{BE}}$ from the $I_C$ and $I_{C0}$ yields a little different $R_B=\frac{200\cdot 9.43}{10^{-3}}=1.89M\mathrm{\Omega }$

d)

An answer:
$$\begin{array}{rcll}& R_B& =\frac{\mathit{\beta fe}\cdot (V_{\mathit{CC}}-\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->(\frac{I_C}{I_{C0}}+1))}{I_C}& \end{array}$$

can be rewritten as

$$\begin{array}{rcll}& \mathit{\beta fe}& =\frac{R_B{I}_C}{(V_{\mathit{CC}}-\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->(\frac{I_C}{I_{C0}}+1))}& \end{array}$$

which leads, with unchanged $I_{C0}$, $R_B$, $V_{\mathit{CC}}$, $T$ to almost a doubled $\mathit{\beta fe}$: $\mathit{\beta fe}\approx 400$

e)

An answer:
The equation for $R_B$ $$\begin{array}{rcll}& R_B& =\frac{\mathit{\beta fe}\cdot (V_{\mathit{CC}}-\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->(\frac{I_C}{I_{C0}}+1))}{I_C}& \end{array}$$

can be recycled to get an answer. Note that rewriting this into $I_C=\dots$ is hard because $I_C$ appears in the denominator and in the $\mathit{ln}(\dots )$ term in the numerator. In this specific question, no exact number is required: only a larger/smaller/equal answer is requested. This can be done by e.g. assuming an answer and then verifying it or falsifying it. The impact of doubling a transistor is that:

• the $I_{C0}$effectively doubles
• the $\mathit{\beta fe}$ is the same

We now can verify/falsify the 3 conditions:

• assume that $I_C$ is doubled
  In this case, $I_C∕I_{C0}$ is unchanged, and according to the equation for $R_B$, the value for $R_B$ should be halved. This is in contradiction with having an unchanged circuit.
• assume that $I_C$ more than doubled
  In this case, $I_C∕I_{C0}$ increases, and according to the equation for $R_B$, the value for $R_B$ should be more-than-halved. This is in contradiction with having an unchanged circuit.
• assumed that $I_C$ is less than doubled
  obviously, this does not have to be worked out: if 2 out of 3 possible options are false, the remaining option must be true.

Other reasoning:

• driving the transistor(s) from a voltage source, the $V_{\mathit{BE}}$ does not change when adding a second BJT. Then the summed $I_C$ is twice the current of one BJT. This however assumes a voltage source and hence $R_B=0$.
• driving the transistor(s) from a current source, the summed $I_B$ does not change and then the summed $I_C$ does not change.
• driving the transistor(s) from a source with (positive) output resistance yields a source that is somewhere between the previous 2 ideal sources. The resulting behavior is then also between the two previously derived $I_C$ results: the total $I_C$ will be somewhat increased.

Yet another alternative reasoning is:

a)

Assuming as starting point that that the total $I_C$ and hence the total $I_B$ do not change.

b)

with the doubled $I_{C0}$ and the same total $I_C$, the $V_{\mathit{BE}}$ must be lowered by $\mathrm{\Delta }{V}_{\mathit{BE}}=\frac{\mathit{kT}}{q}\mathit{ln}(2)$: numerically this is a decrease by 18 mV.

c)

the slightly decreased $V_{\mathit{BE}}$ results in a small increase in $I_B$ and hence results in a small increase in $I_C$.