26-09-2013, 03:27 PM
To verify the characteristics of CLAMPERS
INTRODUCTION
The circuits in Figure above are known as clampers or DC restorers. These circuits clamp a peak
of a waveform to a specific DC level compared with a capacitively coupled signal which swings
about its average DC level (usually 0V). If the diode is removed from the clamper, it defaults to a
simple coupling capacitor– no clamping.
What is the clamp voltage? And, which peak gets clamped? In Figure above the clamp voltage is
0 V ignoring diode drop, (more exactly 0.7 V with Si diode drop). In Figure above, the positive
peak of V(1) is clamped to the 0 V (0.7 V) clamp level. Why is this? On the first positive half
cycle, the diode conducts charging the capacitor left end to +5 V (4.3 V). This is -5 V (-4.3 V) on
the right end at V(1,4). Note the polarity marked on the capacitor in Figure above. The right end
of the capacitor is -5 V DC (-4.3 V) with respect to ground. It also has an AC 5 V peak sinewave
coupled across it from source V(4) to node 1. The sum of the two is a 5 V peak sine riding on a -
5 V DC (-4.3 V) level. The diode only conducts on successive positive excursions of source V(4)
if the peak V(4) exceeds the charge on the capacitor. This only happens if the charge on the
capacitor drained off due to a load, not shown. The charge on the capacitor is equal to the
positive peak of V(4) (less 0.7 diode drop). The AC riding on the negative end, right end, is
shifted down. The positive peak of the waveform is clamped to 0 V (0.7 V) because the diode
conducts on the positive peak.