- 2072 Magh · 1+3 marks
Clamper adding +5 V DC level
Short answer
A clamper shifts an AC waveform by adding a DC level. To add +5 V, use a biased positive clamper with a 5 V battery in series with the diode (anode to battery positive, cathode to output). The capacitor charges to Vm+5 V, and output becomes Vin+Vm+5, with negative peak clamped at +5 V.
What is a clamper circuit? Draw the clamper circuit that adds +5 volts DC level on AC voltage.
A clamper circuit (dc restorer) is a circuit that shifts an ac waveform up or down by adding a dc level to it, without changing the waveform's shape. It is made of a capacitor, a diode, a resistor and, when needed, a dc battery.
Clamper that adds +5 V
Assumption: "adds +5 V" is taken to mean that the negative peak of the output is clamped at +5 V (ideal diode), so the whole waveform is moved upward.
C
Vin o---------| |--------+--------+---o Vout
| |
----- |
/ \ D R
/___\ |
| |
+ | |
----- |
--- 5 V |
- | |
GND o--------------------+--------+---o
The diode anode connects to the positive terminal of the 5 V battery (battery negative to ground) and its cathode to the output node. The resistor R is across the output.
Working (Vin = Vm sin ωt, ideal diode)
- The diode turns ON whenever Vout tries to go below +5 V. In the first negative half cycle, the capacitor charges to Vc = Vm + 5 V.
- Afterwards the diode is OFF and the capacitor holds this voltage (RC ≫ T).
- Output: Vout = Vin + (Vm + 5)
For example, for Vm = 10 V, Vc = 15 V and Vout ranges from +5 V to +25 V.
If the aim is to add exactly a +5 V dc component to the signal (mean level = +5 V, i.e. Vout = Vin + 5), the battery value should be 5 − Vm (a negative value, i.e. a reversed battery, when Vm > 5 V).
For a real silicon diode, the clamped level is reduced by 0.7 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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