Given. Fig.1: $V_i(t)=10\sin(2\pi t)$ V (peak 10V); $R_1=50\Omega$; clamp branches $V_1=3\text{V}+D_1$ (fires below $-3.7\text{V}$) and $V_2=4\text{V}+D_2$ (fires above $+4.7\text{V}$); $V_D=0.7\text{V}$. Fig.2: same $V_i(t)$; series $C_1$; shunt clamp $D_1$; series $D_2$ into shunt $R_1$ (output).
Given data
Quantity
Value
$V_i(t)$
$10\sin(2\pi t)$ V, both figures
$R_1$ (Fig.1)
$50\,\Omega$
$V_1,V_2$ (Fig.1 clamp batteries)
$3\text{V},4\text{V}$
$V_D$ (all diodes)
$0.7\text{V}$
Find. Fig.1: $V_i,V_o$ waveforms with peaks; a power rating for $D_1$; peak current in $R_1$. Fig.2: steady-state $V_o(t)$ with diode operating-region changes.
Fig. Q1(a) — Fig.1 double diode-battery limiter. Vo tracks Vi while -3.7V<Vo<4.7V (both diodes off, no drop across R1); D1 clamps the trough at −3.7V, D2 clamps the crest at +4.7V.
Approach. Fig.1: while both diodes are reverse biased no current flows in $R_1$, so $V_o=V_i$ exactly; once $V_i$ pushes the node past a clamp threshold the corresponding diode conducts and pins $V_o$ there, with the excess $(V_i-V_o)/R_1$ becoming the diode/battery-branch current. Fig.2: recognise the cascade as a negative clamper ($C_1+D_1$) feeding a series-diode envelope stage ($D_2+R_1$, no reservoir capacitor) — together the classic "clamp-then-rectify" voltage-doubler topology.
Part (a) — clamp thresholds and $V_o(t)$. $D_1$ fires when the node tries to go below $-(V_1+V_D)=-3.7\text{V}$; $D_2$ fires above $+(V_2+V_D)=+4.7\text{V}$ (KVL around each battery+diode branch once conducting, since no current flows elsewhere to drop volts anywhere else). Between these two levels both diodes are OFF and — because nothing else connects to the node — no current flows in $R_1$, so
$$V_o=V_i\quad\text{for }-3.7\text{V}\le V_i\le4.7\text{V}$$
Since $V_i$ swings a full $\pm10\text{V}$, both clamps are reached every cycle:
$$V_{o,\max}=\boxed{+4.7\text{ V}}\qquad V_{o,\min}=\boxed{-3.7\text{ V}}$$
Part (b) — rating $D_1$ for power. $D_1$ only conducts while $V_i<-3.7\text{V}$, i.e. while $10\sin(2\pi t)<-3.7$. The peak current through $D_1$ occurs at the $V_i=-10\text{V}$ trough:
$$I_{D_1,\text{pk}}=\frac{|V_{i,\min}|-3.7}{R_1}=\frac{10-3.7}{50}=\boxed{126\text{ mA}}$$
Averaging $i_{D_1}(t)=\max\!\big(0,\tfrac{-V_i(t)-3.7}{R_1}\big)$ numerically over the full period gives $I_{D_1,\text{avg}}\approx31.1\text{ mA}$, so the continuous (average) power dissipation in the diode junction itself is
$$P_{D_1,\text{avg}}=V_D\cdot I_{D_1,\text{avg}}=(0.7)(31.1\text{mA})\approx\boxed{21.8\text{ mW}}$$
while the instantaneous peak is $P_{D_1,\text{pk}}=V_D\cdot I_{D_1,\text{pk}}=(0.7)(126\text{mA})\approx88.2\text{ mW}$. $D_1$ should therefore be specified with a continuous (average) power rating comfortably above $\sim22\text{ mW}$ and a peak/surge forward-current rating above $126\text{ mA}$ — any small-signal switching diode (e.g. a 1N4148-class part, rated $\sim500\text{ mW}$, $I_F\le300\text{ mA}$) has ample margin. It should also block the peak reverse voltage it sees when off, $V_R=V_{i,\max}-V_1=10-3=7\text{V}$.
Part (c) — peak current in $R_1$. $R_1$ carries current whenever either diode conducts; by symmetry of construction the larger magnitude occurs on the $D_1$ (negative) side already computed. On the positive side, $I_{D_2,\text{pk}}=(10-4.7)/50=106\text{ mA}$. The overall peak current in $R_1$ is therefore
$$I_{R_1,\text{pk}}=\boxed{126\text{ mA}}\ \text{(at the }V_i=-10\text{V trough, flowing through }D_1\text{)}$$
Part (d) — Fig.2 clamp-then-rectify (voltage doubler). In steady state $D_1$ (shunt to ground) clamps node $V_1$'s minimum to $-V_D=-0.7\text{V}$ (same clamping action as Fig.1's branches, but against ground). Since $C_1$ blocks DC, $V_1(t)$ is just $V_i(t)$ shifted up by whatever DC level makes its trough sit at $-0.7\text{V}$:
$$V_1(t)=V_i(t)+\big(V_{i,\text{pk}}-V_D\big)=10\sin(2\pi t)+9.3$$
so $V_1(t)$ ranges from $-0.7\text{V}$ (trough, where $D_1$ conducts briefly to replenish $C_1$'s charge) up to $+19.3\text{V}$ (crest). $D_2+R_1$ then behave as a plain series-diode follower of $V_1$ (no reservoir capacitor at the output, so $V_o$ tracks $V_1-V_D$ instantaneously whenever $D_2$ is forward biased, and is pulled to $0$ by $R_1$ when it is not):
$$V_o(t)=\max\!\big(0,\ V_1(t)-V_D\big)=\max\!\big(0,\ 10\sin(2\pi t)+8.6\big)$$
$$V_{o,\text{pk}}=19.3-0.7=\boxed{18.6\text{ V}}\ \ (\approx2V_{i,\text{pk}}-2V_D,\text{ the classic doubler result})$$
$D_2$ cuts off only in the brief notch where $V_1(t)\lt V_D$, i.e. $10\sin(2\pi t)<-8.6\Rightarrow\sin(2\pi t)<-0.86$: this spans $t/T\approx0.665$ to $0.835$ (about $17\%$ of the cycle, centred on the trough), during which $V_o=0$ and $D_1$ is the only conducting device (briefly, right at the very bottom of that notch, to top the clamp back up to $-0.7\text{V}$).
Fig. Q1(d) — Fig.2 clamp-then-rectify (voltage-doubler) stage. C1+D1 shift Vi up so its trough sits at −0.7V (D1 clamps); D2+R1 then follow V1−0.7V, cutting off (Vo=0) only in the brief notch where V1<0.7V near the trough.
