Recall Lecture 6 Rectification transforming AC signal into

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 i. Find the clip value = V’. To find V’, use KVL at L 1 assuming the diode is on ii. The equation is : V’ – VB - V = 0 V’ = VB + V VI V’ = VB + V L 1 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

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 i. Find the clip value = V’. To find V’, use KVL at L 1 assuming the diode is on ii. The equation is : V’ – 1. 3 - 0. 6= 0 V’ = 1. 9 V 1. 3 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 VI V’ = 1. 9 V

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 + + Vin Vout - -

Solution i. Find the clip value = V’. To find V’, use KVL at L 1 assuming the diode is on ii. 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. 3 V L 1

Parallel Based Clippers Ø Positive and negative clipping can be performed simultaneously by using a double limiter or a parallel-based clipper. Ø The parallel-based clipper is designed with two diodes and two voltage sources oriented in opposite directions. Ø 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 Initial stage: node a = 0 V Hence, node b = 1. 5 V So, node c of the diode must be at least 2. 2 V a

• 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. It is similar to half wave where the diode only turns on during one of the cycle.

Clipper in Series Problem 3. 11 Figure P 3. 11(a) shows the input voltage of the circuit as shown in Figure P 3. 11(b). Plot the output voltage Vo of these circuits if V = 0. 7 V c P 3. 11(a) Initial stage: node a = 0 V Hence, node b = - 0. 7 V So, node c of the diode must be at least -1. 9 V b P 3. 11(b) a

• 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 vo 3 V + -1. 1

• 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