1. Recall Lecture 8 Standard Clipper Circuit – Step 1: Find the clip value by doing KVL at the output branch – Step 2: Set the conditions to know whether diode is on or off – sketch your output waveform Clipper in series – clips at zero. It is similar to half wave where the diode only turns on during one of the cycle.

2. Clamper

3. Clampers ● Clamping shifts the entire signal voltage by a DC level.  Consider, the sinusoidal input voltage signal, v I.  1st 90 0, the capacitor is charged up to the peak value of V I which is V M.  Then, as V I moves towards the –ve cycle,  the diode is reverse biased.  Ideally, capacitor cannot discharge, hence Vc = V M  By KVL, we get NOTE: The input signal is shifted by a dc level; and that the peak-to-peak value is the same

4. Clampers  STEP 1: Knowing what value that the capacitor is charged to. And from the polarity of the diode, we know that it is charged during positive cycle. Using KVL,  V C + V B – V S = 0  V C = V M – V B  STEP 2: When the diode is reversed biased and V C is already a constant value  V O – V S + V C = 0  V O = V S – V C. ● A clamping circuit that includes an independent voltage source V B. Peak value V M

5. EXAMPLE – clampers with ideal diode For the circuit shown in figure below, sketch the waveforms of the output voltage, v out. The input voltage is a sine wave where v in = 20 sin  t. Assume ideal diodes. Vin

6. What if the diode is non-ideal? C + Vo - 5V + Vi - Vi t -10 10 The diode is a non-ideal with V  = 0.7V  Step 1: V C + V  - V B – V i = 0  V C = 10 + 5 – 0.7 = 14.3V  Step 2: V O – V i + V C = 0  V O = V i – 14.3. -4.3 -24.3 -14.3

7. Multiple Diode Circuits

8. Final Exam SEM I 2013/2014

9. DIODEIDID VDVD OFF0 V D < V  ONI D > 0 V D = V  REMEMBER THAT: A pn junction diode will conduct when the p-type material is more positive than the n-type material

10. OR GATE V1V2VO Vo = voltage across R D1 and D2 off; no current flow,000 D1 off, D2 on, current flow, Vo – V2 + V  = 0 05V ( 1 )4.3V D1 on, D2 off, current flow, Vo – V1 + V  = 0 5V ( 1 )04.3V Both on, using both loops will give the same equation 5V ( 1 ) 4.3V

11. V1V2VO Both on, using both loops will give the same equation 000.7 D1 on, D2 off05V ( 1 )0.7 D1 off, D2 on5V ( 1 )00.7V Both are off; open circuit no current flowing through R since no GND destination 5V ( 1 ) 5V AND GATE Vo = node voltage