- 2073 Magh · 1+4 marks
Pulse shaping circuit and output waveform
Short answer
Pulse shaping circuits alter input waveform shapes to produce desired outputs, using clipping, clamping, or RC networks. For this circuit, positive peaks are clipped flat at +3 V when input exceeds 9 V; negative peaks pass through a 1/3 voltage divider, reaching −5 V.
Define pulse shaping circuits. Find the output wave form of the given circuit. [Figure: Vin is a sine wave of amplitude ±15 V; series resistor 2 kΩ, diode in parallel branch with 3 V battery, load resistor 1 kΩ; some values hard to read.]
Pulse shaping circuits are circuits that change the shape of an input waveform (for example clipping, clamping, differentiating, integrating) to produce a desired output waveform such as a pulse or square wave. Diode clippers and clampers and RC circuits are common examples.
Output of the given circuit
Assumptions (values were hard to read): Vin = 15 sin ωt, series resistor Rs = 2 kΩ, load RL = 1 kΩ, ideal diode in series with a 3 V battery in a shunt branch. The diode anode is on the output side, so it conducts when Vo exceeds +3 V.
Vin o--[2k]--+----------+--o Vo
| |
D RL 1k
| |
3 V |
GND o--------+----------+
Diode OFF (Vo < 3 V): the circuit is a voltage divider: Vo = Vin × RL/(Rs + RL) = Vin × 1/3 = Vin/3
The diode turns ON when Vin/3 = 3 V, that is Vin = 9 V.
Diode ON (Vin > 9 V): Vo is held at 3 V.
Negative half cycle: the diode stays OFF, Vo = Vin/3, so the negative peak is −15/3 = −5 V.
| Vin | Diode | Vo |
|---|---|---|
| < 9 V (including negative) | OFF | Vin/3 |
| > 9 V | ON | 3 V |
Vo
+3 V | ______ ______ (clipped)
| / \ / \
0 +--/--------\-------/--------\--
| \_____/
-5 V |
Result: the positive peak is clipped flat at +3 V; the negative peak reaches −5 V (reduced by the divider).
✓ Answer checked by Claude Opus 5.5 (high effort)
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