Recall Lecture 6 Rectification – transforming AC signal into a signal with one polarity Half wave rectifier Full Wave Rectifier Center tapped Bridge Rectifier parameters Duty Cycles Peak Inverse Voltage (PIV)

Clipper Circuits

Standard Clipper Circuits Clipper circuits, also called limiter circuits, are used to eliminate portion of a signal that are above or below a specified level – clip value. The purpose of the diode is that when it is turn on, it provides the clip value Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on The equation is : V’ – VB - V = 0  V’ = VB + V iii. Then, set the conditions If VI > V’, what happens?  diode conducts, clips and hence Vo = V’ If VI < V’, what happens?  diode off, open circuit, no current flow, Vo = Vi VI V’ = VB + V L1

EXAMPLE For the circuit shown below sketch the waveform of the output voltage, Vo. The input voltage is a sine wave where VI = 5 sin t. VB is given as 1.3 V. Assume V = 0.6 V 1.3 V

Solution Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on The equation is : V’ – 1.3 - 0.6= 0  V’ = 1.9 V iii. Then, set the conditions If Vin > V’, what happens?  diode conducts, clips and hence Vo = V’ = 1.9 V If Vin < V’, what happens?  diode off, open circuit, no current flow, Vo = Vi 1.3 V VI V’ = 1.9V

EXAMPLE For the circuit shown below sketch the waveform of the output voltage, Vout. The input voltage is a sine wave where Vin = 10 sin t. Assume V = 0.7 V + Vout - Vin

Solution Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on The equation is : V’ – 4 + 0.7= 0  V’ = 3.3 V iii. Then, set the conditions If Vin > V’, what happens?  diode off, open circuit, no current flow, Vo = Vi If Vin < V’, what happens?  diode conducts, clips and hence Vo = V’ = 3.3 V VI V’ = 3.3V L1

Parallel Based Clippers Positive and negative clipping can be performed simultaneously by using a double limiter or a parallel-based clipper. The parallel-based clipper is designed with two diodes and two voltage sources oriented in opposite directions. This circuit is to allow clipping to occur during both cycles; negative and positive

Clipper in Series ECE 1201 The input signal is an 8 V p-p square wave. Assume, V = 0.7 V. Sketch the output waveform, Vo. c b a Initial stage: node a = 0V Hence, node b = 1.5 V So, node c of the diode must be at least 2.2 V

Hence, the positive cycle of the square wave has met the condition of  2.2 V Perform KVL as usual: – 4 + 0.7 + 1.5 + Vo= 0 Vo = 1.8 V 4 1.8 -4

What if now the input is change to 4 sin t? It has to wait for the input signal to reach 2.2 V. Before that, the output, Vo is zero as diode is off. Perform KVL as usual: – 4 + 0.7 + 1.5 + Vo= 0 Vo = 1.8 sin t V

Peak value is 1.8 V

So, basically, clipper in series clips at zero So, basically, clipper in series clips at zero. It is similar to half wave where the diode only turns on during one of the cycle.

Clipper in Series Problem 3.11 Figure P3.11(a) shows the input voltage of the circuit as shown in Figure P3.11(b). Plot the output voltage Vo of these circuits if V = 0.7 V a c b P3.11(a) P3.11(b) Initial stage: node a = 0V Hence, node b = - 0.7 V So, node c of the diode must be at least -1.9 V

Hence, the negative cycle of the square wave has met the condition of  |1.9| V Perform KVL as usual: + 3 - 0.7 - 1.2 + Vo= 0 Vo = - 1.1 V -1.1 vo - 3V +

What if now the input is change to 5 sin t ? It has to wait for the input signal (negative cycle) to reach  |1.9| V. Before that, the output, Vo is zero as diode is off. Perform KVL as usual: + 5 - 0.7 - 1.2 + Vo= 0 Vo = -3.1 sin t V

Peak value is approximately -3.1 V