This document describes four different diode clipper circuits and three diode clamper circuits. The clipper circuits use diodes to clip portions of an input waveform, resulting in outputs that swing within narrowed positive and negative voltage ranges. The clamper circuits use diodes to set a reference voltage level, causing the output to oscillate symmetrically around this clamped level. Key aspects covered include how the diodes bias in different circuit configurations and the voltage ranges of the output waveforms for each circuit.

1. Diode circuits
Clippers and Clampers

2. First clipper circuit
• Diode is forward biased
during positive half
cycle
– This makes the positive
potential 0.7V

3. • Output swings from
0.7V positive to -50V
negative

4. Second clipper circuit
• Diode is forward biased
during negative half
cycle
– This makes the negative
potential 0.7V

5. • Output swings from
+24V positive to -0.7V
negative

6. Third clipper circuit
• Since diodes are in
series, each branch will
drop 0.7V + 0.7V = 1.4V
– Branches are in parallel

7. • Output swings from
+1.4V positive to -1.4V
negative

8. Fourth clipper circuit
• First, determine
potential at VBIAS
1k
– VBIAS = 15 =
6.8k+1k
0.128 15 = 1.92V
• Since diode requires
positive potential of 0.7V
and adds to VBIAS,
maximum positive
potential becomes 2.62V

9. • Waveform swings from
+2.62V positive to -20V
negative

10. First clamper circuit
• DC potential is
pk – 0.7V = 14.3V
– This becomes the zero
reference line for the AC,
so it will go 15V above
and 15V below this line

11. • Output swings from
+29.3V positive to -0.7V
negative

12. Second clamper circuit
• DC potential is
pk – 0.7V = 29.3V
– This becomes the zero
reference line for the AC,
so it will go 30V above
and 30V below this line

13. • Output swings from
+0.7V positive to -59.3V
negative

14. Third clamper circuit
• This one charges the
capacitor during the
positive alternation,
and this will then
discharge during the
negative alternation.
• This makes the output
at maximum a straight
line of 40V (2Vpk)

15. • However, since the diode is not perfect we
need to use the first approximation (simplified
diode)
– This means we need to subtract 0.7V for the one
diode that charges the capacitor (39.3V)
– And lest we forget, subtract the potential for the
second diode as well (38.6V)

16. • Output is pulsating
wave (ripple) that rises
to a maximum potential
of 39.3V and descends
to a minimum potential
of 38.6V