A voltage source circuit including an opamp

Exercise 6.2 A voltage source circuit including an opamp

An answer:

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

\begin{array}{rcl} 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}{rcl} v_{OUT} & = I_{M 1} \cdot (R_{1} + R_{2}) \\ I_{M 1} = \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.

[next] [prev] [prev-tail] [front] [up]