Clipper Clamper Circuits
Clipper Clamper Circuits
Clipper Clamper Circuits
The previous section on rectification gives that diodes used to change the
appearance of an applied waveform.
This section on clippers and next on clampers will expand on the wave-shaping
abilities of diodes.
1. Series
The series configuration is defined as one where diode is in series with the load
2. Parallel
Series
1
Fig. Series clipper
The response is not as obvious because the dc supply can aid or work against
source voltage, and dc supply can be in leg between the supply and output or
in branch parallel to output.
𝒗𝑶 = 𝒗𝒊 –V
For the “off” region, diode is an open circuit, 𝑰𝑫= 0 mA, and output voltage is
𝒗𝑶 = 0 V
EXAMPLE: Determine the output waveform for the sinusoidal input of Fig.
Fig. Series clipper 1
+ve region of 𝑽𝒊 and dc supply are both applying “pressure” to turn diode on
Once supply goes -ve, it would have to exceed the dc supply voltage of 5 V
before it could turn the diode off the transition from one state to the other will
occur when
𝑽𝒊+ 5 V = 0 V
𝑽𝒊 = -5 V
Fig.3 Sketching 𝑽𝑶
EXAMPLE Find the output voltage for the network examined in
example if the applied signal is the square wave of Fig. 1
For 𝑽𝒊 = 20 V (0→ 𝑻/2) the network of Fig. 2 results. The diode is in the
SC state, and 𝑽𝑶 = 20 V + 5 V = 25 V. For 𝑽𝒊 = -10 V the network of Fig. 3
results, placing diode in “off” state, and 𝑽𝑶= 𝒊𝑹 R = (0)R = 0 V, output voltage
appears in Fig. 4.
Clipper not only clipped off 5 V from total swing, but also raised dc level of
signal by 5 V
Parallel
The network of Fig. is simplest of parallel diode configurations . The analysis of
parallel configurations is very similar to series configurations
Fig. Response to a parallel clipper
𝑽𝒊= 4V
Fig. Sketching 𝑽𝒐
𝒗𝒊 + 𝑽𝒌 – V = 0
𝒗𝒊 = V - 𝑽𝒌 = 4V – 0.7V
= 3.3V
∴ 𝑭𝒐𝒓 𝒗𝒊 > 3.3V , diode will be OC
𝒗𝒐= 𝒗𝒊
Fig. Sketching 𝒗𝑶
Note that the only effect of 𝑽𝒌 was to drop the transition level to 3.3 from 4 V.
CLAMPERS
The previous section investigated a number of diode configurations that clipped
off a portion of applied signal without changing remaining part of waveform.
This section will examine a variety of diode configurations that shift applied
signal to a different level.
The chosen resistor and capacitor of network must be chosen and time
constant by 𝑟 = RC is sufficiently large to ensure that voltage across capacitor
does not discharge significantly during interval diode is non-conducting.
Fig 1. Clamper
The capacitor is connected directly b/w input and output signals and resistor
and diode are connected in parallel with output signal
For network of Fig 1. diode will forward biased for +ve portion of applied signal
The SC equivalent for diode will result in 𝒗𝒐 = 0 V for this time interval, as
shown in sketch of 𝒗𝒐 in Fig.2.
𝑟 = RC is very small b/c R “shorted out” by diode & only capacitor quickly
charge to peak value of V volts as shown in Fig. 2.
Fig.3
with OC equivalent for diode determined by applied signal and stored
voltage across capacitor current through diode from cathode to anode.
𝒗𝑶 ∥ with diode and resistor, Applying KVL around input loop results in
-V –V - 𝒗𝒐 =0
𝒗𝒐= 2V
Fig 4 . Sketching 𝑽𝑶
EXAMPLE: Determine 𝑽𝑶for the network of Fig. 1 for the input indicated
Fig. 1
Sol//
f = 1000Hz, t=1ms ,Interval = 0.5ms ,Period 𝒕𝟏 → 𝒕𝟐
For this interval network shown in Fig.
Fig 2 Determining 𝑽𝑶 and 𝑽𝒄 with the diode in the “on” state.
-20 V + 𝐕𝐜 - 5 V = 0
and 𝐕𝐜 = 25 V
The OC equivalent for diode removes 5-V battery ,𝐕𝐎 and KVL around
outside loop of network results in
+10 V + 25 V - 𝐕𝐎 = 0
and 𝐕𝐎 = 35 V
Since the interval 𝒕𝟐 → 𝒕𝟑 will only last for 0.5 ms, capacitor will hold its
voltage during discharge period between pulses of input signal.
Fig. 4
+5 V - 0.7 V - 𝐕𝐎 = 0
and 𝐕𝐎 = 5 V - 0.7 V = 4.3 V
Fig 1
For the input section KVL
-20 V + 𝐕𝐂 + 0.7 V - 5 V = 0
and 𝐕𝐂 = 25 V - 0.7 V = 24.3 V
+10 V + 24.3 V - 𝐕𝐎 = 0
and 𝐕𝐎 = 34.3 V
Fig 3
The resulting output appears in Fig. 4 , that input and output swings are same.
Fig 4