Question 2 of 5: Op-Amp with a Grounded-Gate MOSFET in the Feedback Path — Transfer Law and Function

Question 2: Op-Amp with a Grounded-Gate MOSFET in the Feedback Path — Transfer Law and Function (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An ideal op-amp $A_1$ on the paper's default $\pm 15$ V rails, non-inverting input grounded; $v_{IN}$ drives the inverting node $v_X$ through $R_1$; an n-channel MOSFET $M_1$ with parameters $K$ and $V_{TH}$ closes the loop, its gate grounded and its body tied to its source. The triode and saturation laws quoted above are supplied with the question; channel-length modulation is neglected.

Find. (a) a closed-form expression for $v_{OUT}$ in terms of $v_{IN}$, $K$, $V_{TH}$ and $R_1$, and hence the ratio $v_{OUT}/v_{IN}$; (b) the signal-processing function the block performs.

Approach. Use the virtual short to pin $v_X$, force the whole input current through the channel of $M_1$, show from the terminal voltages that $M_1$ is diode-connected in effect and therefore saturated, then invert the square law to get $v_{OUT}$.

1. Read the terminals off the schematic before writing a single equation. The gate of $M_1$ goes to ground; the terminal joined to the summing node carries the higher potential and is therefore the drain; the terminal on the op-amp output carries the body tie and is the source. That assignment is what makes the loop negative: if $v_X$ drifts up, the op-amp output drives further negative, which increases $V_{GS} = -v_{OUT}$, which pulls more current out of the summing node and brings $v_X$ back down.
2. Apply the virtual short at the op-amp input. The non-inverting terminal is grounded and the op-amp is ideal with negative feedback closed, so $$v_X = 0 \quad\Longrightarrow\quad V_G = 0,\ \ V_D = 0 .$$
3. Force the entire input current through the channel. No current enters the op-amp input terminal, so every ampere delivered by the source through $R_1$ must leave through $M_1$: $$i_{DS} = \frac{v_{IN} - v_X}{R_1} = \frac{v_{IN}}{R_1}.$$ This is the step that makes the block a current-in device: $R_1$ converts the input voltage to a current, and $M_1$ converts that current back to a voltage by its own law.
4. Establish the region of operation of $M_1$. With the gate and drain both held at 0 V and the source at $v_{OUT}$, $$V_{GS} = 0 - v_{OUT} = -v_{OUT}, \qquad V_{DS} = 0 - v_{OUT} = -v_{OUT},$$ so $V_{DS} = V_{GS}$ exactly. Since $V_{GS} - V_{TH}$ is smaller than $V_{GS}$ for a positive threshold, the saturation condition $V_{DS} \ge V_{GS} - V_{TH}$ always holds: $M_1$ is effectively diode-connected and stays saturated at every operating point, however large the signal. Conduction needs $V_{GS} = -v_{OUT}$ to exceed $V_{TH}$, so $v_{OUT}$ is negative and $v_{IN}$ must be positive.
5. Write the saturation law with the channel-length term dropped. $$i_{DS} = \tfrac{1}{2}K\,(V_{GS} - V_{TH})^{2} = \tfrac{1}{2}K\,(-v_{OUT} - V_{TH})^{2}.$$
6. Equate the two expressions for the same current and invert the square law. Setting the current from Step 3 equal to the current from Step 5, $$\tfrac{1}{2}K\,(-v_{OUT} - V_{TH})^{2} = \frac{v_{IN}}{R_1} \quad\Longrightarrow\quad (-v_{OUT} - V_{TH})^{2} = \frac{2\,v_{IN}}{K R_1},$$ and taking the physical (positive-overdrive) root gives the closed form asked for: $$\boxed{\;v_{OUT} = -\left[\,V_{TH} + \sqrt{\dfrac{2\,v_{IN}}{K R_1}}\,\right], \qquad \frac{v_{OUT}}{v_{IN}} = -\,\frac{V_{TH} + \sqrt{2 v_{IN}/(K R_1)}}{v_{IN}}\;}$$
7. Interpret the ratio: there is no constant gain here. The ratio depends on the signal itself, so the block is deliberately non-linear. Differentiating gives the incremental slope $$\frac{dv_{OUT}}{dv_{IN}} = -\frac{1}{g_m R_1}, \qquad g_m = \sqrt{2K i_{DS}} = \sqrt{2K v_{IN}/R_1},$$ which shrinks as the input grows, so the transfer characteristic compresses large signals. With the illustrative set $K = 1\ \text{mA/V}^2$, $V_{TH} = 1$ V and $R_1 = 10\ \text{k}\Omega$, an input of 5 V gives $i_{DS} = 0.5$ mA and $v_{OUT} = -2.00$ V, while 20 V gives 2 mA and $-3.00$ V. Quadrupling the input has exactly doubled the overdrive term, which is the square-root law made visible.
8. Part (b): name the function. The output is the square root of the input, offset by the threshold and inverted: $$\bigl|v_{OUT}\bigr| - V_{TH} = \sqrt{\frac{2}{K R_1}}\;\sqrt{v_{IN}} \;\propto\;\sqrt{v_{IN}} .$$ The circuit is therefore an analogue square-root extractor — the inverse-function block obtained by putting a square-law device inside the feedback loop of an inverting op-amp, exactly as a diode in that position gives a logarithmic amplifier. Its classical use is in flow measurement, where a differential-pressure cell across an orifice plate produces a signal proportional to the square of flow rate and the transmitter must linearise it; it is equally used as a compressor ahead of an ADC to extend dynamic range, and as the square-root element of an analogue RMS-to-DC converter.