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A A

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. . :

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/

, .

, .

.

A

A

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Analysis at:

vo = vi -5+Vvi < 5 - V ------Diode on

vi > 5 - V ------Diode offvo (max)= -Vs-5+V

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Example:

The transition level will beat vi= 4+V since iD=0

When vi < 4+V , vo will follow viWhen vi > 4+V ,vo will stay at 4+V

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A

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. A 0.7.

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A

. .

1. During T/4 thecapacitor is charged to

max vI.2. When v

I

decreases,diode is reverse-biased,capacitor cannotdischarge.

3. After T/4, vC remains at

VM

. Then vO

=vI

- vC=vI - VM

4. Diode remains reversed-biased when vI becomesnegative.

5. vO

= vI

- VM= VM sin t -VM

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v-O=VBvC is charged to VS

vO=-vC+vSv+O=VB-VS+VS sin tvO(min)=VB-2VS

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:

1. A .

2. ,

.3. /4 .

. ,

: =

+ 5.

(4).6. ,

, .

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:

2

V1 (V) V2 (V) VO (V)

0 0 0

.

0 5 4.3

5 5 4.3

= 0.7

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:

2 A

V 1( V ) V 2

( V ) V O

( V )

0 0 0

0 5 0

5 5 4.3

= 0.7