a/i4J N IFdl O -v . -J cr1 q_a C (r dd -_-io _b (sf ,*t l–,) l ” 1+q-x l= -? IIIItIIlIIII 2. al. In Figure 81, what are the peak positive and negative values of Vou, for V;n equal to +10 V? (Hint: Use Thevenin’s theorem.) (l) + 0.994,-0.994 (2) + 0.6,-7.519 (3) + 0.994,-7.519 (4) + 7.519-, 0.994 Y (orill Fig.81 – -. ,Y t r -In Fig. 81, if the diode (Dl) is reversedw, hat would be the peakp ositive and negative values of Vou, for Vin equal to +10 V peak? (1) + 0.994, -0.994 (2) + 0.6, -7.519 (Ilinr: Figs. 25 and 27) (3) + 0.994,-7.579 (4) + 7.519,-0.994 What is the cut-in angle for the circuit of Figure 82, forV6 = +10V peak? (L) 3.440″ (2) 26.104 (3) 34.056′ (4) 55.944″ (Ifinn V,)
-9!-87 Fig.82 4. Zener diodesh ave a reverse.biasedc onduction-pointv oltage of (1) around 0.6 V (2) between 0.6 V and 5 V (3) greater thanZ V but less than 18 V (4) from 1.8 V to 200 V. 5. The circuit of Figure 83 uses two 5.I-V zener diodes. If Vin is a sine wave having peak value of +50 V, the output will be a (l) good square-wavea pproximation. (2) sawtooth wave form (3) a sine wave. v(rl +!0 Fig.83 6. In Figure 83, reducing the peak amplitudeo f V;n will tend to decreaseth e rise time of the output waveform. (l) True (2) False (Hint Exp.2) 7. Although LEDs are in fact diodes, and have characteristics other types of diodes, their principle application is as similar to (1) fast rectifiers (3) clipper diodes. (2) audio-frequency detectors. (4) display indicators.V (cal
88 It IIIIIIIIIIIIIIII i 8 . Figure 84 is a biased series cripper. If vb is greater than V1n, the peak varueo f Vou1w, ill be (Hint:Exp. 3) (l) equal to Vb (2) equal to (Vb + V,n) (3) zero. (4) equal to (Vb -V6) while Dl is conductine. v0.l o–{ Y! Fig.84 Fig.85 9′ Figure8 5 is a double-biasepda railerc lipperw ith a back-to_baczke ner diodes huntw. hat wil be thep ositivea ndn egativpee ak (All diodes outputv ortage? havea n inrernabl anierp otentiaor r o.ov .) (r{inriNote gv) (l) +5.7,-5.7 (2)+ 8.6_. 8.6 (3)…