98-Phys-A5, May 2014, Q4 | Nivaar Exam Prep

Question 4 (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.

Symbol	Value
$V_i$	2 V
$R_1$	$5\,\text{k}\Omega$
$R_2$	$4\,\text{k}\Omega$
$R_3$	$1\,\text{k}\Omega$
$R_4$	$2\,\text{k}\Omega$
Floating source	8 V, in the $R_2$ feedback path between $V_C$ and $V_B$

Op-amp 1: inverting-type node $V_A$ (driven by $V_i$ through $R_1$), non-inverting input grounded, feedback through $R_2$ then the 8 V source to its own output $V_B$. Op-amp 2: non-inverting input driven directly by $V_B$; inverting input $V_D$ set by the $R_3$ (to ground)/$R_4$ (to $V_o$) divider.

Find. (a) $V_A$; (b) $V_o$.

Figure P4 — op-amp 1 (inverting node $V_A$, feedback through $R_2$ and an 8 V floating source to its output $V_B$) drives op-amp 2 (non-inverting amplifier, gain set by $R_3$/$R_4$).

Check

The battery symbol's polarity (long/short plate order): reading from $V_C$ toward $V_B$ the strokes are short–long–short–long, i.e. the positive terminal is at $V_B$ and the negative terminal is at $V_C$ (so $V_B=V_C+8\,\text{V}$).

Approach. Op-amp 1's grounded non-inverting input pins $V_A$ at virtual ground regardless of the feedback network; a KCL sweep from $V_A$ through $R_2$ and the 8 V source gives $V_B$; op-amp 2 is then a standard non-inverting amplifier with $V_B$ as its input and the $R_3/R_4$ divider as feedback.

Part (a) — $V_A$ by virtual ground. Op-amp 1's non-inverting input is tied directly to ground, and the ideal op-amp forces $V_-=V_+$, so $$\boxed{V_A=0\ \text{V}}$$ — this holds regardless of $R_1$, $R_2$ or the 8 V source; it is a direct consequence of the grounded non-inverting input (a "given but structurally irrelevant" datum for this particular sub-part).
Find $V_C$. No current enters the op-amp's inverting input, so the current pulled into $V_A$ from $V_i$ through $R_1$ must leave entirely through $R_2$ toward $V_C$ (KCL at node $V_A$): $$I_1=\frac{V_i-V_A}{R_1}=\frac{2-0}{5\,\text{k}\Omega}=0.4\,\text{mA}=I_2=\frac{V_A-V_C}{R_2}$$ $$\Rightarrow\ V_C=V_A-I_2R_2=0-(0.4\,\text{mA})(4\,\text{k}\Omega)=-1.6\,\text{V}$$
Find $V_B$ across the floating source. With the positive terminal at $V_B$ (per the polarity above), $$V_B=V_C+8\,\text{V}=-1.6+8=6.4\,\text{V}$$
Part (b) — $V_o$ from the non-inverting stage. $V_B$ drives op-amp 2's non-inverting input directly (no current, no drop). $R_3$ (to ground) and $R_4$ (to $V_o$) form the standard non-inverting feedback divider at $V_D$, and the ideal op-amp forces $V_D=V_B$: $$V_o=V_B\left(1+\frac{R_4}{R_3}\right)=6.4\left(1+\frac{2\,\text{k}\Omega}{1\,\text{k}\Omega}\right)=6.4(3)=\boxed{19.2\ \text{V}}$$

Quantity	Value
$V_A$	0 V
$V_C$	−1.6 V
$V_B$	6.4 V
$V_o$	19.2 V

Concept & Theory

Topic: Cascaded op-amp stages with a floating DC source in the feedback path.

Key relations: virtual ground/virtual short ($V_-=V_+$, no input current); KCL at a summing node; non-inverting gain $1+R_f/R_{in}$.

Why this works: an ideal battery in series with a feedback resistor does not block DC feedback current — it simply adds a fixed offset to the node voltage on the far side, so the same KCL method used for a plain resistive feedback network still applies; only the final voltage bookkeeping picks up the extra $\pm8\,\text{V}$ term.

Common pitfall: assuming $V_A$ depends on the feedback network (it does not — a grounded non-inverting input pins it at 0 V by itself); and mis-reading the floating source's polarity, which flips the sign of the $8\,\text{V}$ term and changes $V_o$ from $+19.2\,\text{V}$ to $-28.8\,\text{V}$.

Source: Sedra & Smith, Microelectronic Circuits, Ch. 2 (ideal op-amp analysis, inverting and non-inverting configurations).

Question 4 of 7

Question 4 (20 marks)