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NDSU 9: Clippers ECE 321 - JSG Clippers & Clampers Clipper Circuits: Problem: Design a circuit which clips the voltage output at <6V using an ideal silicon diode. Solution: R + + Vin Vout 5.3V - Assume the diode is off. In this case, Vout = Vin. Now, assume the output is large (say, +10V). The diode has more than 0.7V across it, so it turns on. When it turns on, the output is clipped at +6V (0.7V + 5.3V = 6.0V). Problem: Design a circuit which clips the voltage at the output at >1V. Solution: R + + Vin Vout 1.7V - Assume the diode is off. Again, Vout = Vin. Now, assume Vout is small (say, -10V). Now the diode is forward biased and turns on. The voltage at the output is then 1.0V (-0.7V + 1.7V = 1.0V). Problem: Design a circuit which clips a signal at +1V < Vout < +6V. Solution: Put the previous two circuits in parallel: 1 January 26, 2016 NDSU 9: Clippers ECE 321 - JSG R + + Vin Vout 5.3V 1.7V - When Vout is large (+10V), D1 turns on, D2 turns off, and the output clips at +6V. When Vout is small (-10V), D2 turns on, D1 turns off, and the output clips at +1V. When 1V < Vout < Vin, both diodes are off and Vout = Vin. Zener Diodes An easier solution is to use a Zener diode. Zener diodes are special diodes which are designed to break down at a certain voltage when reverse biased. For example, a 6V Zener diode will have a fixed, 6V drop across it if you try to reverse bias it by more than 6V. (i.e. a zener diode will dump whatever current is required to hold the voltage at 6V. ) If you want to clip an output at <6V, use a 6V Zener diode as follows. If the output tries to exceed 6V, the Zener diode turns on and keeps the output at +6V. Note that a zener diode is still a diode. If the output tries to go below -0.7V, the diode turns on like any silicon diode would and clips the output at -0.7V as well. R + 6V Zener + Vin Vout - - If you want to clip an output at > -8V, flip the diode around and use a 8V zener diode. If you want to clip an output between -8V < Vout < +6V, use two zener diodes: 2 January 26, 2016 NDSU 9: Clippers ECE 321 - JSG R + 6V Zener + Vin - Vout 8V Zener - Note that the output will actually clip at -8.7V and +6.7V. If the output tries to become large (say, +10V), the 6V zener turns on and clips at +6.0V. The 8V zener is now a forward biased diode, which has another 0.7V drop across it, for a total of 6.7V. This configuration is fairly common. If you want to protect an electronic device from abuse, two zeners are often placed across the input pins as shown above. In normal operation, one of the zener diodes is off and the diodes have no affect. If the operator tries to hit your device with +24V, the zeners turn on. clip the input at +6.7V, and protect your device. Function Approximation With a clipper circuit, you can approximate different functions. The heart of this circuit is as follows: 1k Vout R1 Vin R2 R3 Load 1M + Vz1 Vz2 Vz3 Function Approximation: Vz1 < Vz2 < Vz3 Here, the circuit is to drive a load of 1M or more. So that you can ignore the 1M load, assume a 1k resistor is connected to the input. When Vin < Vz1, all three zener diodes are off. In that case, ignoring the 1M resistor load. 3 January 26, 2016 NDSU 9: Clippers ECE 321 - JSG Vout = Vin When Vout > Vz1, the first zener turns on. At that point, you have a voltage divider and R1 ⎞ V in + c V out ≈ ⎛⎝ R 1 +1000 ⎠ or the slope (gain) of the circuit drops. When Vout > Vz2, the second zener turns on. At that point, you now have R1 in parallel with R2 (think Thevenin) and R R2 ⎞ V in + c 2 V out ≈ ⎛⎝ R R1 +1000 ⎠ 1 2 When Vout > Vz3, the third zener turns on. R R R3 ⎞ V in + c 2 V out ≈ ⎛⎝ R R1 R2 +1000 ⎠ 1 2 3 Example: Design a circuit to approximate the following Vin / Vout relationship. Input: 0 .. 10V source, capable of driving 20mA Output: 0 .. 10V, capable of driving 1M Ohm Relationship: Follow the following curve with a tolerance of +/- 0.5V Vout 5 4 3 2 1 0 0 1 2 3 4 5 6 Vin 7 8 9 10 11 12 Step 1: Draw a straight-line approximation to this curve. Start with a slope of one (all zener diodes off) Each time the slop drops, you turn on another stage The slope determines the net resistance: 4 January 26, 2016 NDSU 9: Clippers ECE 321 - JSG Vout 5 Vz3 = 5V slope = 0 slope = 0.31 Vz2 = 4V 4 3 slope = 0.6 Vz1 = 2V 2 1 0 slope = 1 0 1 2 3 4 5 6 Vin 7 8 9 10 11 12 Assume a 1k resistor initially (draws less than 10mA from the source). The initial slope (all zener diodes off) is 1M ⎞ slope = ⎛⎝ 1M+1k ⎠ = 0.999 ≈ 1 Stage 1: The slope changes when Vout = 2V. Vz1 = 2V. The slope becomes 0.6 R ⎞ slope = ⎛⎝ R+1000 ⎠ = 0.6 R 1 = R = 1500Ω Stage 2: The slope changes when Vout = 4V. Vz2 = 4V The slope becomes 0.31 R ⎞ slope = ⎛⎝ R+1000 ⎠ = 0.31 R = R 1 R 2 = 449Ω R 2 = 640Ω Stage 3: The slope changes when Vout = 5V. Vz2 = 5V The slope becomes zero 5 January 26, 2016 NDSU 9: Clippers ECE 321 - JSG R3 = 0 The net circuit is then 1k Vout 1500 Vin 0 640 Load 1M + 2V 5V 4V Note: If you don't have alot of zener diodes, you can approximate these using silicon diodes in series: each diode increases the voltage drop by 0.7V. If the initial slope is more than one, add an amplifier to the circuit. 6 January 26, 2016