6 Feedback

Q 6.1

(a)

An answer:
$$\begin{array}{rcll}& H(j\omega )& =\frac{v_{out}}{v_{in}}& \\ & & v_{out}=Z_{C1}\cdot i_{C_1}& \\ & & i_{C_1}=\frac{0-v_{in}}{R_2}& \\ & & v_{out}=\frac{1}{j\omega C_1}\cdot -\frac{v_{in}}{R_2}& \\ & H(j\omega )& =-\frac{1}{j\omega C_1{R}_2}& \end{array}$$

(b)

An answer:
$$\begin{array}{rcll}& v_{out}& =v_{-}-v_{C1}(t)& \\ & & v_{-}=0& \\ & & v_{C1}(t)=v_{C1}(0)+\frac{1}{C_1}{\int \; <!--nolimits-->}_0^t{i}_{C}dt& \\ & & i_C=\frac{v_{in}(t)}{R_2}& \\ & v_{out}& =v_{out}(0)-\frac{1}{R_2{C}_1}{\int \; <!--nolimits-->}_0^t{v}_{in}(t)dt& \end{array}$$

(c)

An answer:
$$\begin{array}{rcll}& H(j\omega )& =\frac{v_{out}}{v_{in}}& \\ & & v_{out}=R_1\cdot i_{R_1}& \\ & & i_{R_1}=\frac{0-v_{in}}{j\omega L_2}& \\ & & v_{out}=R_1\cdot -\frac{v_{in}}{j\omega L_2}& \\ & H(j\omega )& =-\frac{1}{j\omega \frac{L_2}{R_1}}& \end{array}$$

(d)

An answer:
$$\begin{array}{rcll}& v_{out}& =v_{-}-v_{R1}(t)& \\ & & v_{-}=0& \\ & & v_{R1}(t)=R_1\cdot i_{R1}& \\ & & i_{R1}=i_{L2}(0)+\frac{1}{L_1}{\int \; <!--nolimits-->}_0^t{v}_{in}dt& \\ & v_{out}& =v_{out}(0)-\frac{R_1}{L_2}{\int \; <!--nolimits-->}_0^t{v}_{in}(t)dt& \end{array}$$

Q 6.2

An answer:

For an ideal opamp with a finite output voltage, a sudoku-approach usually is the easiest. Then

$$\begin{array}{rcll}& & v_{+}=v_{-}=V_{REF}& \\ & & V_{REF}=v_{OUT}\cdot \frac{R_1}{R_1+R_2}& \\ & v_{OUT}& =V_{REF}\cdot \frac{R_1+R_2}{R_1}& \end{array}$$

An alternative derivation could be:

$$\begin{array}{rcll}& v_{OUT}& =I_{M1}\cdot (R_1+R_2)& \\ & & I_{M1}=\frac{1}{2}K{(V_{DD}-v_G-V_T)}^2& \\ & & v_G=\infty \cdot (v_{OUT}\cdot \frac{R_1}{R_1+R_2}-V_{REF})=finite& \end{array}$$

The last line yields the correct expression for $v_{OUT}$. Note that with a finite $a_v$, the derivation would be quite cumbersome. That is the reason that $a_v\to \infty$ was stated in the question.

A derivation using a small signal equivalent cannot be done as this circuit needs to make a DC output voltage. small signal analyses can be used to get output impedance, bandwidth, ... for this circuit.

Q 6.3

(a)

An answer:

$$\begin{array}{rcll}& z_{in}& =\frac{v_{in}}{i_{in}}& \\ & & i_{in}=\frac{v_{in}-v_{out}}{R_2}& \\ & & v_{out}=v_i\cdot (1+\frac{Z}{R_1})& \\ & & i_{in}=\frac{v_{in}(1-(1+\frac{Z}{R_1}))}{R_2}& \\ & & \Rightarrow & \\ & z_{in}& =\frac{v_{in}}{i_{in}}=-\frac{R_1{R}_2}{Z}& \end{array}$$

(b)

An answer:
$$\begin{array}{rcll}{z}_{in}=-\frac{R_1{R}_2}{\frac{1}{jwC}}=-jw{R}_1{R}_{2}C& & & \end{array}$$

(c)

An answer:

In general, the Nyquist plot should not right-handedly encircle the -1 point. For a frequency independent loop the Nyquist point should stay at the right hand side of the -1 point.

$$\begin{array}{rcll}& & A\cdot \beta >-1& \end{array}$$

which translates into $\beta >0$ for $A\to \infty$ and $beta$ is the net feedback as defined for negative feedback systems. In this system, both positive feedback ${\beta }_{pos}$ and negative feedback ${\beta }_{neg}$ is applied:

$$\begin{array}{rcll}& {\beta }_{neg}& >{\beta }_{pos}& \\ & & {\beta }_{neg}=\frac{R_1}{R_1+R_{Z_R}}& \\ & & {\beta }_{pos}=\frac{R_6}{R_G+R_2}& \\ & \Rightarrow & & \\ & R_G& <\frac{R_1{R}_2}{R_{Z_R}}& \end{array}$$

This is the same as $R_G∕∕z_{in}>0$

Q 6.4

(a)

Derive an equation for (the large signal) $v_{OUT}(v_{IN})$

(b)

An answer:
The circuit includes a PMOS transistor. Its SSEC is the same as that of an NMOS transistor.

The combination of the transistor and the resistor forms a CSC that has negative gain. With the feedback to the non-inverting input of the opamp, this gives negative feedback.

(c)

An answer:

$$\begin{array}{rcll}& v_X& =V_{DD}+v_{GS}& \\ & & v_{GS}=V_T-\sqrt{\frac{2i_D}{k}}\mathrm{\text{(compared to an NMOS some signs are reversed)}}& \\ & & i_D=\frac{v_{OUT}}{R_1}& \\ & v_X& =V_{DD}+V_T-\sqrt{\frac{2v_{IN}}{k\cdot R_1}}& \end{array}$$

(d)

It was assumed that the transistor works in saturation, for which the square law relation is valid. Identify the range of $v_{IN}$ where this assumption is satisfied.

(e)

An answer:
$v_X$ will not be in saturation, but in its linear region. The assumed square law then is no longer satisfies resulting in different behavior. For the $i-v$-relation in that region, with $v_{DS}\to 0$, $v_X$ goes to minus infinity.

Q 6.5

An answer:

$$\begin{array}{rcll}& & v_{OUT}=a_V\cdot A_{sig}\cdot sin(\omega t)& \\ & & \frac{\partial v_{OUT}}{\partial t}=\omega \cdot a_V\cdot A_{sig}\cdot cos(\omega t)& \\ & & \text{max}(\frac{\partial v_{OUT}}{\partial t})=\omega \cdot a_V\cdot A_{sig}\cdot cos(0)=\omega \cdot a_V\cdot A_{sig}=2\pi f\cdot a_V\cdot A_{sig}& \end{array}$$

Hence the slew rate must be at least $SR=50\pi f$ [V/s].

Q 6.6

(a)

An answer:

(b)

(c)

An answer:
Then the specified equation is not valid anymore: the term that includes the $v_{CE}$-dependency into account must be included. That shows that $i_C$ is (almost) zero for $v_{CE}$ that is (almost) zero. The opamp tries to compensate this by letting $v_X$ as high as possible (usually limited by the $i_B$ that grows exponentially with $v_X$).

Q 6.7

(a)

(b)

An answer:
The feedback loop gain contains only the opamp and a resistive divider; resistive dividers do not add any gain to the loop. Unity gain stable opamps are defined to be stable for any passive (attenuating) feedback network that provides zero phase shift, the limit being unity feedback.

So this circuit should be stable.

(c)

An answer:

$$\begin{array}{rcll}& H(j\omega )& =\frac{v_{out}}{v_{in}}& \\ & & v_{out}=v_{-}+R_3\cdot i_{R_3}& \\ & & i_{R_3}=\frac{v_{-}-v_{in}}{R_2}& \\ & & A\to \infty \Rightarrow v_{-}=v_{+}& \\ & & v_{-}=v_{+}=v_{in}\cdot \frac{1}{1+j\omega R_{1}C}& \\ & & v_{out}=v_{in}\cdot \frac{1}{1+j\omega R_{1}C}\cdot (1+\frac{R_3}{R_2})-v_{in}\cdot \frac{R_3}{R_2}& \\ & H(j\omega )& =(1+\frac{R_3}{R_2})(\frac{1}{1+j\omega R_{1}C})-\frac{R_3}{R_2}& \\ & & =\frac{2}{1+j\omega R_{1}C}-1\text{ ; when }{R}_2=R_3& \end{array}$$

(d)

An answer:
Towards a standard form:

$$\begin{array}{rcll}& H(j\omega )& =\frac{v_o}{v_i}=\frac{2}{1+j\omega R_{1}C}-1& \\ & & =\frac{2}{1+j\omega R_{1}C}-\frac{1+j\omega R_{1}C}{1+j\omega R_{1}C}& \\ & & =\frac{1-j\omega R_{1}C}{1+j\omega R_{1}C}& \end{array}$$

Or if you do not like the $-j\omega$ term, you could rewrite into

$$\begin{array}{rcll}& H(j\omega )& =\frac{1+{(\omega R_{1}C)}^2}{{(1+j\omega R_{1}C)}^2}& \end{array}$$

The Bode plot is shown below. From the relations above you can directly derive that the modulus of the gain is unity, and that the phase is between $0°$ and $-180°$.

Q 6.8

An answer:
$$\begin{array}{rcll}& H(j\omega )& =\frac{v_{out}}{i_{in}}& \\ & & v_{out}=A(v_{+}-v_{-})=-A{v}_{-}& \\ & & v_{-}=(\frac{v_{out}-v_{-}}{R_2}-i_{in})\frac{1}{j\omega C_1}& \\ & & =\frac{v_{out}-i_{in}{R}_2}{1+j\omega R_2{C}_1}& \\ & & v_{out}=-A\frac{v_{out}-i_{in}{R}_2}{1+j\omega R_2{C}_1}& \\ & & v_{out}=\frac{\frac{A{i}_{in}{R}_2}{1+j\omega R_2{C}_1}}{1+\frac{A}{1+j\omega R_2{C}_1}}& \\ & H(j\omega )& =\frac{A{R}_2}{1+j\omega R_2{C}_1+A}& \\ & & =\frac{A}{1+A}\cdot \frac{R_2}{1+j\omega (\frac{R_2{C}_1}{1+A})}& \end{array}$$

From which is follows that

$$\begin{array}{rcll}& & H(0)=\frac{A}{1+A}\cdot R_2& \\ & & {\omega }_0=\frac{1+A}{R_2{C}_1}& \end{array}$$

The magnitude part of a Bode plot is:

Q 6.9

(a)

An answer:

$$\begin{array}{rcll}& z_o& =\frac{v_o}{i_o}& \\ & & i_o=\frac{v_o-A(v_{+}-v_{-})}{z_x}& \\ & & =\frac{v_{-}(1+A)-A{v}_{+}}{z_x}& \\ & & v_{+}=v_i\equiv 0& \\ & & i_o=\frac{v_o(1+A)}{z_x}& \\ & z_o=& \frac{R_2}{1+A}\cdot \frac{1}{1+j\omega R_2{C}_1}& \end{array}$$

(b)

An answer:

Q 6.10

(a)

An answer:

$$\begin{array}{rcll}& r_{out,circuit}& =\frac{v_o}{i_o}& \\ & & i_o=\frac{(v_o-\frac{A_o{v}_x}{1+j\omega {\tau }_1})}{r_{out}}& \\ & & v_x=-v_o& \\ & & i_o=\frac{v_o}{r_{out}}+\frac{A_o{v}_o}{r_{out}(1+j\omega {\tau }_1)}& \\ & & =\frac{v_o}{r_{out}}(1+\frac{A_o}{1+j\omega {\tau }_1})& \\ & r_{out,circuit}& =\frac{r_{out}}{1+\frac{A_o}{1+j\omega {\tau }_1}}& \\ & & =\frac{r_{out}(1+j\omega {\tau }_1)}{1+j\omega {\tau }_1+A_o}& \\ & & =\frac{r_{out}}{1+A_o}\cdot \frac{1+j\omega {\tau }_1}{1+\frac{j\omega {\tau }_1}{1+A_o}}& \end{array}$$

Assuming for simplicity reasons that $1+A_0=100$, the Bode plot would be similar to:

Q 6.11

(a)

An answer:

$$\begin{array}{rcll}& v_{out}& =v_{-}-v_{BE}& \\ & & v_{+}=v_{-}=0& \\ & & v_{BE}=\frac{kT}{q}\ln <!--FUNCTION\; APPLICATION-->\frac{i_c}{I_{C0}}& \\ & v_{out}& =-\frac{kT}{q}\ln <!--FUNCTION\; APPLICATION-->\frac{v_{in}}{I_{C0}R}& \end{array}$$

(b)

An answer:

$$\begin{array}{rcl}& g_{BJT}& =\frac{\partial i_C}{\partial v_{BE}}=\frac{q}{kT}{I}_{C0}{e}^{}& q{V}_{BE}kT=\frac{q}{kT}{I}_C\end{array}$$

$$$$

$$\begin{array}{rcll}& r_{out,circuit}& =\frac{v_o}{i_o}& \\ & & i_o=\frac{v_o}{R_1+r_{bjt}}+\frac{v_o+A\cdot v_o\frac{R_1}{R_1+r_{bjt}}}{r_{out}}& \\ & & =v_o(\frac{1}{R_1+r_{bjt}}+\frac{1+A\frac{R_1}{R_1+r_{bjt}}}{r_{out}})& \\ & r_{out,circuit}& =(R_1+r_{bjt})∕∕\frac{r_{out}}{1+A\frac{R_1}{R_1+r_{bjt}}}& \\ & \mathrm{\text{with }}& r_{bjt}=\frac{kT}{q}\frac{1}{I_C}=\frac{kT}{q}\frac{R}{V_{IN}}& \end{array}$$

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