3 Biasing

Q 3.1

(a)

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

(b)

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

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

can be rewritten as

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

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

(e)

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

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 $\mathit{\alpha fe}$ 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 $I_C$.

Q 3.2

(a)

An answer:
Original situation: $$\begin{array}{rcll}& I_{C2}& =\mathit{\alpha fe}{I}_{B2}& \\ & & I_{B2}=I_{E1}& \\ & & I_{E1}=(\mathit{\alpha fe}+1)I_{B1}& \\ & & I_{B1}=\frac{V_{\mathit{CC}}-V_{B1}}{R_x}& \\ & & V_{B1}\approx 1.2V& \\ & I_{C2}& =\mathit{\alpha fe}(\mathit{\alpha fe}+1)\frac{V_{\mathit{CC}}-1.2V}{R_x}& \end{array}$$

Modified situation:

$$\begin{array}{rcll}& I_{C2}& =\mathit{\alpha fe}{I}_{B2}& \\ & & I_{B2}=I_{E1}& \\ & & I_{E1}=\frac{V_{\mathit{CC}}-V_{B1}}{R_x}& \\ & & V_{B1}\approx 1.2V& \\ & I_{C2}& =\mathit{\alpha fe}\frac{V_{\mathit{CC}}-1.2V}{R_x}& \end{array}$$

It follows from these relations that $I_{C2}$ is changed quite a lot when applying the described circuit modification.

Q 3.3

An answer:
$$\begin{array}{rcll}& V_{\mathit{CE}}& =V_{\mathit{CC}}-V_{R_C}& \\ & & =V_{\mathit{CC}}-I_C\cdot R_C& \\ & & I_C=\mathit{\alpha fe}{I}_B& \\ & & I_B=\frac{V_{\mathit{CC}}-V_B}{R_{B1}}& \\ & V_{\mathit{CE}}& =V_{\mathit{CC}}-\mathit{\alpha fe}\frac{V_{\mathit{CC}}-V_B}{R_{B1}}{R}_C& \end{array}$$

yields:

$$\begin{array}{rcll}& I_C& =0.84\mathit{mA}\dots 6.7\mathit{mA}& \\ & R_C& =1k\mathrm{\Omega }& \end{array}$$

and hence gives

$$\begin{array}{rcll}& V_{R_C}& =0.84\dots 6.7V& \\ & V_{\mathit{CE}}& =8.16\dots 2.3V& \\ & & & \end{array}$$

which is fine for biasing purposes.

Q 3.4

(a)

(b)

An answer:
Assuming the BJT satisfies $i_C=I_{C0}{e}^{\frac{q\cdot v_{\mathit{BE}}}{\mathit{kT}}}$ without any dependency on $V_{\mathit{CE}}$ : $$\begin{array}{rcll}& V_{C,T1}& =V_{\mathit{CC}}-I_{C,T1}R3& \\ & & I_{C,T1}=\mathit{\alpha fe}{I}_{B,T1}& \\ & & I_{B,T1}=\frac{V_{\mathit{CC}}-V_{B,T1}}{R2}& \\ & V_{C,T1}& =V_{\mathit{CC}}-\mathit{\alpha fe}\frac{V_{\mathit{CC}}-V_{B,T1}}{R2}R3& \end{array}$$

(c)

An answer:
Substituting the numerical values for the various components shows that for all $\mathit{\alpha fe}$ for the specified transistor type, the voltage drop across $R3$ would be larger than $V_{\mathit{CC}}$. This is in violation of the assumption on $V_{\mathit{CE}}$. In an actual circuit, this yields $V_{\mathit{CE}}\approx 0$ which does not correspond to proper biasing conditions.

(d)

An answer:
0A (zero)

Q 3.5

(a)

An answer:
A MOSFET works in saturation if $V_{\mathit{DS}}>V_{\mathit{GS}}-V_T$.

In the circuit, there is zero current through $R_g$, meaning $V_D=V_G$ from which it follows that $V_{\mathit{GS}}=V_{\mathit{DS}}$. Consequently, the saturation condition is always satisfied.

(b)

An answer:
$$\begin{array}{rcll}& R_D& =\frac{V_{R_D}}{I_D}& \\ & & =\frac{V_{\mathit{DD}}-V_{\mathit{DS}}}{I_D}& \\ & & =\frac{V_{\mathit{DD}}-V_{\mathit{GS}}}{I_D}& \\ & & V_{\mathit{GS}}=V_T+\sqrt{\frac{2I_D}{k}}& \\ & R_D& =\frac{V_{\mathit{DD}}-V_T+\sqrt{\frac{2I_D}{k}}}{I_D}& \end{array}$$

where the equation for $V_{\mathit{GS}}$ is a reworked version of the element equation (the square law relation).

Q 3.6

An answer:

$$\begin{array}{rcll}& I_{C1}& =\mathit{\alpha fe}\cdot \frac{V_{B1}-0.6}{R_1}& \\ & I_{C2}& =0& \end{array}$$

Q 3.7

(a)

An answer:

$$\begin{array}{rcll}& R_{G1}& =\frac{V_{\mathit{DD}}-V_G}{I_{\mathit{RG}1}}& \\ & & V_G=V_{\mathit{GS}}+V_S& \\ & & I_{\mathit{RG}1}=\frac{V_{\mathit{DD}}}{R_{G1}+R_{G2}}& \\ & & V_{\mathit{GS}}=V_T+\sqrt{\frac{2I_{\mathit{BIAS}}}{k}}& \\ & & V_S=I_{\mathit{bias}}\cdot R_S& \\ & R_{G1}& =\frac{V_{\mathit{DD}}-V_T-\sqrt{\frac{2I_{\mathit{BIAS}}}{k}}-I_{\mathit{BIAS}}\cdot R_S}{V_{\mathit{DD}}}\cdot (R_{G1}+R_{G2})& \\ & & =\frac{V_{\mathit{DD}}-V_T-\sqrt{\frac{2I_{\mathit{BIAS}}}{k}}-I_{\mathit{BIAS}}\cdot R_S}{V_T+\sqrt{\frac{2I_{\mathit{BIAS}}}{k}}+I_{\mathit{BIAS}}\cdot R_S}\cdot R_{G2}& \end{array}$$

This equation can of course be rewritten in many ways.

An alternative derivation can be:

Calculating the required $V_G$ from a biasing point-of-view and from the bias network (note that the gate current is zero) and equating the two:

$$\begin{array}{rcll}& V_G& =\frac{R_{G2}}{R_{G1}+R_{G2}}{V}_{\mathit{DD}}& \\ & V_G& =V_T+\sqrt{\frac{2I_{\mathit{BIAS}}}{k}}+I_{\mathit{BIAS}}\cdot R_S& \end{array}$$

Equating the two yields a proper expression for the required $R_{G2}$

(b)

An answer:
Filling in the previously derived equation yields $R_{G1}=10k\mathrm{\Omega }$

Without a) it can be done with iteratively calculating voltages and currents in the circuit, following a type of Sudoku-approach. Doing so: $V_S=1V$, need $V_{\mathit{GS}}=4V$ $\to$ $V_G=5V$, so $R_{G1}=R_{G2}$

(c)

An answer:
Clip: $V_D=V_{\mathit{DD}}=10V$

Saurationt: $V_D=V_{D{S}_{\mathit{min}}}+V_S=V_{\mathit{GS}}-V_T+V_S=4-2+1=3V$

Maximum swing for $V_D=6.5V$ $\to$ $V_{\mathit{RD}}=3.5V$, $R_D=3.5k\mathrm{\Omega }$

Q 3.8

(a)

An answer:

$$\begin{array}{rcll}& R_S& =\frac{V_S}{I_D}& \\ & & V_S=V_G-V_{\mathit{GS}}& \\ & & I_D=I_{\mathit{BIAS}}& \\ & & V_{\mathit{GS}}=V_T+\sqrt{\frac{2I_{\mathit{BIAS}}}{K}}& \\ & & V_G=V_{\mathit{DD}}\cdot \frac{R_{G2}}{R_{G1}+R_{G2}}& \\ & R_S& =\frac{V_{\mathit{DD}}\cdot \frac{R_{G2}}{R_{G1}+R_{G2}}-V_T-\sqrt{\frac{2I_{\mathit{BIAS}}}{K}}}{I_{\mathit{BIAS}}}& \end{array}$$

(b)

An answer:

$$\begin{array}{rcll}& V_G& =5V& \\ & V_{\mathit{GS}}& =3V& \\ & V_S& =2V& \\ & R_S& =\frac{2V}{4\mathit{mA}}=500\mathrm{\Omega }& \end{array}$$

(c)

An answer:
For saturation, $V_{\mathit{DS}}\ge V_{\mathit{GS}}-V_T$, and then $$\begin{array}{rcll}& V_D\ge V_S+V_{\mathit{GS}}-V_T& & \\ & V_D=V_{\mathit{DD}}-I_D\cdot R_D& & \end{array}$$

From these, it follows that

$$\begin{array}{rcll}& R_D<\frac{V_{\mathit{DD}}-V_S-V_{\mathit{GS}}+V_T}{I_D}=1.5k\mathrm{\Omega }& & \end{array}$$

(d)

An answer:
Parallel C: no more degeneration for signals $\to$ higher gain.

Parallel L to $R_S$: more degeneration for signals $\to$ lower gain.

Parallel L to $R_D$: higher impedance for the conversion of signal current (variation in the current) to output voltage change$\to$ higher gain.

Q 3.9

An answer:

$$\begin{array}{rcll}& I_C=& \frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}{I}_E& \\ & & I_E=\frac{V_E}{R_4}& \\ & & V_E=V_{\mathit{DD}}-I_B{R}_1-V_{\mathit{BE}}& \\ & & I_B=\frac{I_C}{\mathit{\alpha fe}}& \end{array}$$

back substitution and separation of variables yield

$$\begin{array}{rcll}& I_C=& \frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{\mathit{DD}}-\frac{I_C}{\mathit{\alpha fe}}{R}_1-V_{\mathit{BE}}}{R_4}& \\ & & =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{\mathit{DD}}-V_{\mathit{BE}}}{R_4+\frac{R_1}{\mathit{\alpha fe}+1}}& \end{array}$$

Q 3.10

(a)

An answer:
With the assumptions described in the hint (ignore base current of the proceeding stage) a derivation could be as shown below. Note that the supply voltage in this particular circuit is denoted as $V_{B1}$...

$$\begin{array}{rcll}& I_{C2}& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}{I}_{E2}& \\ & & I_{E2}=\frac{V_{E2}}{R_{E2}}& \\ & & V_{E2}=V_{B2}-V_{\mathit{BE}2}& \\ & & V_{B2}=V_{B1}\frac{R_{B21}}{R_{B21}+R_{B22}}-I_{B2}\cdot \frac{R_{B21}{R}_{B22}}{R_{B21}+R_{B22}}& \\ & & I_{B2}=\frac{I_{C2}}{\mathit{\alpha fe}}& \\ & & ⟹& \\ & I_{C2}& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{B1}\frac{R_{B21}}{R_{B21}+R_{B22}}-\frac{I_{C2}}{\mathit{\alpha fe}}\cdot \frac{R_{B21}{R}_{B22}}{R_{B21}+R_{B22}}-V_{\mathit{BE}2}}{R_{E2}}& \\ & & =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{B1}\frac{R_{B21}}{R_{B21}+R_{B22}}-V_{\mathit{BE}2}}{R_{E2}+\frac{1}{\mathit{\alpha fe}+1}\cdot \frac{R_{B21}{R}_{B22}}{R_{B21}+R_{B22}}}& \end{array}$$

Also $V_{E2}=\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}{I}_{C2}{R}_{E2}$

For the derivation of $I_{C3}$:

$$\begin{array}{rcll}& I_{C3}& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}{I}_{E3}& \\ & & I_{E3}=\frac{V_{E3}}{R_{E3}}& \\ & & V_{E3}=V_{E2}-V_{\mathit{BE}3}& \\ & & ⟹& \\ & I_{C3}& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{E2}-V_{\mathit{BE}3}}{R_{E3}}& \end{array}$$

Where you may substitute the earlier found relation for $V_{E2}$.

(b)

An answer:

$$\begin{array}{rcll}{I}_{C2}\approx 2.05\mathit{mA}& & & \\ I_{C3}\approx 1.6\mathit{mA}& & & \end{array}$$

(c)

An answer:
This relates to small signal properties, in this case to voltage gain of the stages. See therefore chapters 4 and 5.

$R_9$ is decoupled because any (signal induced) change in the base voltage appears in full across $V_{\mathit{BE}}$ of $Q_3$ which results in maximum sensitivity. For biasing purposes, the circuit should be insensitive to changes: at DC we want to have a relatively high ohmic emitter series resistance.

If $R_6$ would be similarly decoupled there would be no signal present at the base of $Q_3$. Then the voltage gain of the stage with $Q3$ would be zero.

Q 3.11

An answer:
$$\begin{array}{rcll}& I_C& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}{I}_E& \\ & & I_E=\frac{V_E}{R_6}& \\ & & V_E=V_B-V_{\mathit{BE}}& \\ & & V_B=V_{\mathit{CC}}\cdot \frac{R_4+R_5}{R_3+R_4+R_5}-I_B\cdot \frac{(R_4+R_5)\cdot R_3}{R_3+R_4+R_5}& \\ & & ⟹& \\ & I_C& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{\mathit{CC}}\cdot \frac{R_4+R_5}{R_3+R_4+R_5}-0.6V}{R_6+\frac{1}{\mathit{\alpha fe}+1}\cdot \frac{(R_4+R_5)\cdot R_3}{R_3+R_4+R_5}}& \end{array}$$

In this question, letting $\mathit{\alpha fe}\to \infty$ from the start:

$$\begin{array}{rcll}& I_C& =I_E& \\ & & I_E=\frac{V_E}{R_6}& \\ & & V_E=V_B-V_{\mathit{BE}}& \\ & & V_B=V_{\mathit{CC}}\cdot \frac{R_4+R_5}{R_3+R_4+R_5}& \\ & & ⟹& \\ & I_C& =\frac{V_{\mathit{CC}}\cdot \frac{R_4+R_5}{R_3+R_4+R_5}-0.6V}{R_6}& \end{array}$$

Q 3.12

An answer:
$$\begin{array}{rcll}& I_C& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_E}{R_4}& \\ & & V_E=V_B-V_{\mathit{BE}}& \\ & & V_B=V_{\mathit{CC}}\frac{R_2}{R_1+R_2}-(R_{!}∕∕R_2)\cdot I_B& \\ & & I_B=\frac{I_C}{\mathit{\alpha fe}}& \\ & & ⟹& \\ & I_C& =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{\mathit{CC}}\frac{R_2}{R_1+R_2}-R_{!}∕∕R_2\frac{I_C}{\mathit{\alpha fe}}-V_{\mathit{BE}}}{R_4}& \\ & & =\frac{\mathit{\alpha fe}}{\mathit{\alpha fe}+1}\frac{V_{\mathit{CC}}\frac{R_2}{R_1+R_2}-V_{\mathit{BE}}}{R_4+\frac{1}{\mathit{\alpha fe}+1}\cdot R_{!}∕∕R_2}& \end{array}$$

Substituting the numerical values yields $I_C=0$ because $V_B\approx 0$. Most likely $R2$ should be 330k$\mathrm{\Omega }$.

Q 3.13

An answer:
$$\begin{array}{rcll}& I_C& =\mathit{\alpha fe}\cdot I_B& \\ & & I_B=\frac{V_C-0.6}{R_2}& \\ & & V_C=V_{\mathit{CC}}-R_3\cdot (I_C+I_B)& \\ & & ⟹& \\ & I_C& =\mathit{\alpha fe}\cdot \frac{V_{\mathit{CC}}-R_3\cdot I_C\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}-0.6}{R_2}& \\ & & =\mathit{\alpha fe}\cdot \frac{V_{\mathit{CC}}-0.6}{R_2+(\mathit{\alpha fe}+1)R_3}& \\ & & \approx 3\mathit{mA}& \end{array}$$