Question 5 of 5: Ideal-diode output-waveform sketches

Question 5: Ideal-diode output-waveform sketches

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. A $\pm10\text{ V}$ sinusoidal input to five diode–resistor (and battery) networks; the diode is ideal with $V_f=0$.

Find. For each network, the conducting condition, the piecewise output law, and the one-cycle output sketch.

Approach. In each part decide the diode state per half-cycle (an ideal diode is a closed switch when forward-biased, open when reverse), reduce the resulting resistor network, and read off the output law and its peaks.

Part (a)

Part (a) has a $+3\text{ V}$ battery ($+$ toward the source) feeds node $A$; the series diode has its cathode at $A$ (it can only pass current $B\!\to\!A$), and two $1\text{ k}\Omega$ resistors shunt nodes $A$ and $v_{OUT}$.

Node $A$ is a stiff $v_A=v_{in}-3$ (ideal source through the ideal battery), so its $1\text{ k}\Omega$ only loads it — it does not set $v_{OUT}$. The diode conducts ($B\!\to\!A$) only while $v_A\lt0$, i.e. $v_{in}\lt3\text{ V}$, clamping $v_{OUT}=v_A$; for $v_{in}\ge3\text{ V}$ it blocks and the output resistor pulls $v_{OUT}=0$. Hence $v_{OUT}=v_{in}-3\ (v_{in}\lt3);\quad v_{OUT}=0\ (v_{in}\ge3)$. The output rides at $0$ over the positive excursion and swings down to $\boxed{-13\text{ V}}$ at $v_{in}=-10\text{ V}$.

Part (b)

Part (b) has a $3\text{ V}$ battery ($-$ toward the source) sets $v_A=v_{in}+3$; the diode cathode is at $A$; a single $1\text{ k}\Omega$ shunts $v_{OUT}$.

The diode conducts only while $v_A=v_{in}+3\lt0$, i.e. $v_{in}\lt-3\text{ V}$, passing $v_{OUT}=v_{in}+3$; otherwise it blocks and $v_{OUT}=0$. Hence $v_{OUT}=v_{in}+3\ (v_{in}\lt-3);\quad v_{OUT}=0\ (v_{in}\ge-3)$. The output is $0$ for all of the positive half and most of the negative half, dipping only to $\boxed{-7\text{ V}}$ near $v_{in}=-10\text{ V}$.

Part (c)

Part (c) has a $1\text{ k}\Omega$ series resistor into $v_{OUT}$, which is shunted by a $1\text{ k}\Omega$ to ground and by a downward diode in series with a second $1\text{ k}\Omega$ to ground.

For $v_{in}\gt0$ the diode conducts, placing the two $1\text{ k}\Omega$ legs in parallel ($500\ \Omega$) so $v_{OUT}=v_{in}\,\tfrac{500}{1500}=v_{in}/3$. For $v_{in}\lt0$ the diode is off and the plain $1\text{ k}\Omega$ divider gives $v_{OUT}=v_{in}/2$. Hence $v_{OUT}=v_{in}/3\ (v_{in}\gt0);\quad v_{OUT}=v_{in}/2\ (v_{in}\lt0)$. A dual-slope clipper: the positive lobe peaks at $\boxed{+3.33\text{ V}}$ and the negative lobe at $\boxed{-5\text{ V}}$ — both halves pass but the positive half is attenuated more.

Part (d)

Part (d) has a $3\text{ V}$ battery ($-$ toward the source) sets $v_A=v_{in}+3$; the diode anode is at $A$ (passes $A\!\to\!B$); two $1\text{ k}\Omega$ resistors shunt $A$ and $v_{OUT}$.

The diode conducts ($A\!\to\!B$) while $v_A=v_{in}+3\gt0$, i.e. $v_{in}\gt-3\text{ V}$, giving $v_{OUT}=v_{in}+3$; for $v_{in}\le-3\text{ V}$ it blocks and $v_{OUT}=0$. This is the mirror image of part (a). Hence $v_{OUT}=v_{in}+3\ (v_{in}\gt-3);\quad v_{OUT}=0\ (v_{in}\le-3)$. The output tracks $v_{in}+3$ up to $\boxed{+13\text{ V}}$ and rests at $0$ through the deep negative excursion.

Part (e)

Part (e) has a $100\ \Omega$ series resistor into $v_{OUT}$, shunted by a downward diode straight to ground and by a $1\text{ M}\Omega$ resistor.

For $v_{in}\gt0$ the diode conducts and clamps $v_{OUT}=0$. For $v_{in}\lt0$ the diode is off and, since $1\text{ M}\Omega\gg100\ \Omega$, $v_{OUT}=v_{in}\tfrac{1\text{M}}{1\text{M}+100}\approx v_{in}$. Hence $v_{OUT}\approx0\ (v_{in}\gt0);\quad v_{OUT}\approx v_{in}\ (v_{in}\lt0)$. A negative-passing half-wave clipper: the positive half is removed and the negative half passes almost unattenuated to $\boxed{-10\text{ V}}$.