Question 2 of 5: Op-Amp Transfer Function and Output Offset
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 07-Elec-A5 Electronics — December 2016.
Closed-book; any non-communicating calculator permitted; three hours. Five questions,
20 marks each — all five constitute a complete paper, so every question is solved
below. Per the paper’s notes, op-amps are ideal with supply rails of
±15 V unless a question states otherwise, and ground/chassis are common.
Reference texts (22-Elec-A5 Electronics): A. S. Sedra
& K. C. Smith, Microelectronic Circuits, 8th ed. (Oxford)
— diode circuits and rectifiers (Ch. 4), MOSFET biasing and amplifiers
(Ch. 5–7), op-amp circuits and offsets (Ch. 2/9), BJT amplifiers
(Ch. 6–7); R. C. Jaeger & T. N. Blalock,
Microelectronic Circuit Design (McGraw-Hill).
Question 2: Op-Amp Transfer Function and Output Offset (20 marks)
Given. (a) Equal resistors $R$ feed both op-amp inputs from $v_{in}$;
$C_2$ is the feedback capacitor (output to the inverting node) and $C_1$ shunts the
non-inverting node to ground. (b) A T-network feedback (two $R$ arms with a tap resistor
$R_1$ to ground) around an inverting stage; $R=R_{in}=100$ kΩ, $R_1=2$ kΩ,
$V_{os}=\pm5$ mV.
Find. (a) the phasor/time expression for $v_o$; (b) the dc output offset
voltage produced by $V_{os}$.
Figure 2a. Both inputs are driven from $v_{in}$
through equal resistors $R$; $C_1$ forms a low-pass at the $+$ input and $C_2$ closes the
feedback around the $-$ input.
Approach (a). Use the two ideal-op-amp rules: no input current, and
$v_+=v_-$. Get $v_+$ from the $R$–$C_1$ divider, force $v_-$ equal to it, then apply
KCL at the inverting node (through $R$ and $C_2$) to solve for $v_o$.
Non-inverting node (low-pass divider). With $v_{in}$ the phasor input
$V_I$, no current enters the $+$ terminal, so $R$ and $C_1$ form a divider:
$$v_+=V_I\,\frac{1/(j\omega C_1)}{R+1/(j\omega C_1)}=\frac{V_I}{1+j\omega RC_1}.$$
Inverting node KCL. Set $v_-=v_+$. Current in through $R$ equals current
out through $C_2$:
$$\frac{v_{in}-v_-}{R}+ (v_o-v_-)\,j\omega C_2=0
\;\Rightarrow\; v_o=v_- -\frac{v_{in}-v_-}{j\omega RC_2}.$$
Substitute and simplify. With $v_-=V_I/(1+j\omega RC_1)$ and
$v_{in}=V_I$, the numerator $v_{in}-v_-=V_I\,j\omega RC_1/(1+j\omega RC_1)$, giving
$$\boxed{\,v_o(j\omega)=V_I\,\frac{1-C_1/C_2}{1+j\omega RC_1}\,}.$$
As a check, if $C_1=C_2$ the output is identically zero — the two paths cancel.
Time-domain form. Writing the transfer function magnitude and phase,
$$v_o(t)=V_I\,\frac{|1-C_1/C_2|}{\sqrt{1+(\omega RC_1)^2}}\,\sin(\omega t+\phi),\qquad
\phi=\angle(1-C_1/C_2)-\tan^{-1}(\omega RC_1).$$
It is a first-order low-pass whose dc gain is $(1-C_1/C_2)$ and whose corner is at
$\omega=1/(RC_1)$.
Approach (b). Offset is a dc effect: ground the signal input, place $V_{os}$
between the op-amp inputs, and compute the output through the T-network’s noise gain.
Set up the offset model. Ground $v_{in}$; the ideal op amp forces its
inverting node (node A) to $v_-=V_{os}$ (with $v_+=0$). No current enters the input
terminals.
KCL at node A. A connects to ground through $R_{in}$ and to the T tap
node M through the first $R$. With $R_{in}=R$:
$$\frac{0-V_{os}}{R}+\frac{V_M-V_{os}}{R}=0\;\Rightarrow\;V_M=2V_{os}.$$
KCL at the tap node M. M connects to A ($R$), to ground ($R_1$), and to
the output ($R$):
$$\frac{V_{os}-V_M}{R}+\frac{0-V_M}{R_1}+\frac{v_o-V_M}{R}=0.$$
Substituting $V_M=2V_{os}$ and multiplying by $R=100$ kΩ:
$$-V_{os}-\frac{R}{R_1}V_M+ (v_o-2V_{os})=0
\;\Rightarrow\;v_o=\Big(3+2\frac{R}{R_1}\Big)V_{os}.$$
Noise gain and result. With $R/R_1=100/2=50$, the noise gain is
$3+2(50)=103$, so
$$\boxed{v_{o,\text{offset}}=103\,V_{os}=103(\pm5\ \text{mV})\approx\pm0.52\ \text{V}}.$$
The T-network multiplies the offset by 103 even though its signal gain uses the same
resistors — the reason a T-feedback network trades offset performance for large
effective feedback resistance.
