Question 1 of 5: Zener + MOSFET Series Voltage Regulator
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 1: Zener + MOSFET Series Voltage Regulator (20 marks)
Given. A zener reference feeds the gate of an n-channel source follower
that delivers the load current to $R_L$.
Given data (Question 1)
Parameter
Value
Parameter
Value
Zener $V_Z@I_Z$
6.8 V @ 5 mA
$r_Z$
20 Ω
$I_{ZK}$ (knee)
0.1 mA
$V_{TH}$
1 V
$K$
2 mA/V$^2$
$\lambda$
0
$V_{DD}$ (avg)
20 V
$v_{ripple}$
±1 V
$R_1$ (from figure)
5 kΩ
$R_L$ (from figure)
1 kΩ
Find. (a) the average dc output voltage $V_{OUT}$; (b) the amplitude
of the residual ripple that appears at $V_{OUT}$ when $V_{DD}$ carries ±1 V.
Figure 1. Series regulator: $R_1$ biases the zener
$D_{Z1}$, whose voltage $V_Z$ drives the gate of source follower $M_1$; the source delivers
$V_{OUT}$ to $R_L$. The gate draws no dc current.
Approach. First find the zener operating voltage from its small-signal
model (since the gate draws no current, all of $R_1$’s current is zener current); then
solve the source-follower square-law equation $V_{OUT}/R_L=\tfrac12K(V_Z-V_{OUT}-V_{TH})^2$
for the dc output; finally propagate the $V_{DD}$ ripple through the $R_1$–$r_Z$
divider and the follower’s gain.
Zener knee (extrapolate the model to $I=0$). The stated point plus the
slope resistance give the projected knee voltage
$$V_{Z0}=V_Z-I_Zr_Z=6.8-(5)(0.02)=\boxed{6.7\ \text{V}}.$$
Zener operating point. No gate current, so the whole $R_1$ current is
zener current: $I_Z=(V_{DD}-V_Z)/R_1$ with $V_Z=V_{Z0}+I_Zr_Z$. Solving,
$$V_Z=\frac{V_{Z0}+r_Z V_{DD}/R_1}{1+r_Z/R_1}
=\frac{6.7+0.02(20)/5}{1+0.02/5}=6.75\ \text{V},\qquad
I_Z=\frac{20-6.75}{5}=2.65\ \text{mA}.$$
This is well above $I_{ZK}=0.1$ mA, so the zener is in hard breakdown — the
reference is valid.
Source-follower output (square law). $M_1$ carries the load current,
$I_D=V_{OUT}/R_L$, and $V_{GS}=V_Z-V_{OUT}$. With $\lambda=0$,
$$\frac{V_{OUT}}{R_L}=\tfrac12K\,(V_Z-V_{OUT}-V_{TH})^2
\;\Rightarrow\; V_{OUT}=(5.75-V_{OUT})^2\quad(R_L=1\text{ k}\Omega,\ K=2).$$
This quadratic $V_{OUT}^2-12.51V_{OUT}+33.1=0$ has roots $8.70$ and $3.80$ V; only
$V_{OUT}=3.80$ V keeps $V_{GS}=V_Z-V_{OUT}=2.95\gt V_{TH}$, so
$$\boxed{V_{OUT}\approx 3.80\ \text{V}},\qquad I_D=V_{OUT}/R_L=3.80\ \text{mA}.$$
Confirm saturation of $M_1$. $V_{DS}=V_{DD}-V_{OUT}=16.2$ V, while
$V_{GS}-V_{TH}=1.95$ V. Since $V_{DS}\ge V_{GS}-V_{TH}$, $M_1$ is saturated as assumed,
and its transconductance is $g_m=K(V_{GS}-V_{TH})=2(1.95)=3.90$ mA/V.
(b) Ripple at the gate. For small signals the zener is $r_Z=20$ Ω
to ac ground, so the $V_{DD}$ ripple divides down through $R_1$:
$$v_{z,\text{rip}}=v_{ripple}\,\frac{r_Z}{R_1+r_Z}
=1\cdot\frac{20}{5020}=3.98\ \text{mV (amplitude)}.$$
Ripple at the output (follower gain). With $\lambda=0$ the drain ripple
has no effect ($r_o\to\infty$), so only the gate ripple couples through, attenuated by the
source-follower gain $A_f=g_mR_L/(1+g_mR_L)=3.9/4.9=0.796$:
$$v_{out,\text{rip}}=A_f\,v_{z,\text{rip}}=0.796(3.98\ \text{mV})
=\boxed{\approx 3.2\ \text{mV amplitude}}\;(\approx 6.3\ \text{mV peak-to-peak}).$$
The regulator suppresses the ±1 V supply ripple by a factor of about 300.
