- 2081 Chaitra (new course) · 1+4 marks
Clamper circuit output waveform
Short answer
The circuit with the series resistor acts as a biased clipper: during positive half-cycle the diode conducts and vo is clamped to 3.7 V (3 V battery + 0.7 V diode drop); during negative half-cycle diode is off, vo = −6 V.
What is a clamper circuit? Draw the output waveform of the circuit by applying the given input waveform. [Figure: input vin is a square wave switching between +6 V and −6 V. Circuit: source vin in series with resistor R to the output node; from the output node to ground a Si diode (anode at the output node, cathode downward) in series with a 3 V battery (positive terminal toward the diode cathode); vo is taken across the diode–battery branch.]
Clamper
A clamper (DC restorer) is a circuit of a capacitor, a diode and a resistor that shifts the whole input waveform up or down by a DC level without changing its shape or peak-to-peak value. A battery in series with the diode sets the level at which the waveform is clamped. The time constant RC must be much larger than the half-period of the input.
Output waveform
Assumption: a clamper needs a series capacitor, so the series element is taken as a capacitor C (large RC). Si diode: V_D = 0.7 V. The diode with the 3 V battery conducts when vo ≥ 3 + 0.7 = 3.7 V.
Positive half (vin = +6 V): the diode is forward biased. The output is held at
vo = 3 + 0.7 = 3.7 V
C charges to: Vc = vin − vo = 6 − 3.7 = 2.3 V
(left plate +)
Negative half (vin = −6 V): the diode is reverse biased (open). The capacitor keeps its 2.3 V, so
vo = vin − Vc = −6 − 2.3 = −8.3 V
Check: peak-to-peak output = 3.7 − (−8.3) = 12 V = input peak-to-peak ✓.
vin vo
+6 |‾‾‾| |‾‾‾| +3.7 |‾‾‾| |‾‾‾|
0 +---+---+---+-- 0 +---+---+---+--
-6 | |___| |___ | | | |
-8.3 | |___| |___
Answer: the output is a square wave switching between +3.7 V and −8.3 V (shifted down by 2.3 V), same shape and 12 V peak-to-peak.
Note: if the series element is really a resistor (as drawn), the circuit acts as a biased clipper, not a clamper: vo = +3.7 V during the positive half (diode on) and vo = −6 V during the negative half (diode off).
✓ Answer checked by Claude Opus 5.5 (high effort)
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