For the circuit with an ideal OPAMP shown in the figure. $V_{\text {REF }}$ is fixed.

If $V_{\text {OUT }}=1$ volt for $V_{\text {IN }}=0.1$ volt and $V_{\text {OUT }}=6$ volt, for $V_{\text {IN }}=1 \mathrm{~V}$, where $V_{\text {OUT }}$ is measured across $R_{\mathrm{L}}$ connected at the output of this OPAMP. The value of $R_{\mathrm{F}} / R_{\text {IN }}$ is

For the transistor $M_1$ in the circuit shown in the figure, $\mu_n C_{o x}=100 \mu \mathrm{~A} / V^2$ and $\frac{W}{L}=10$, where $\mu_n$ is the mobility of electron, $C_{o x}$ is the oxide capacitance per unit area. $W$ is the width and $L$ is the length.

The channel length modulation coefficient is ignored. If the gate-to-source voltage $V_{G S}$ is 1 V to keep the transistor at the edge of saturation. Then the threshold voltage of the transistor (rounded off to one decimal place) is $\_\_\_\_$ V.

Consider the circuit with an ideal OPAMP shown in the figure.

Assuming $\left|V_{\mathrm{IN}}\right| \ll\left|V_{\mathrm{CC}}\right|$ and $\left|V_{\mathrm{REF}}\right| \ll\left|V_{\mathrm{CC}}\right|$. The condition at which $V_{\text {OUT }}$ equals to zero is

An asymmetrical periodic pulse train $v_{\text {in }}$ of 10 V amplitude with on-time $T_{\mathrm{ON}}=1 \mu \mathrm{~s}$ is applied to the circuit shown in the figure. The diode $D_1$ is ideal.

The difference between the maximum voltage and minimum voltage of the output waveform $v_0$ (in integer) is $\_\_\_\_$ V.