Welcome back to my workbench,
The Big Idea – Using Shunt Clippers to Protect Circuits
Biased Shunt clippers are a simple way to protect inputs from excessive positive or reverse voltage.
Drill Down:
From our previous discussion, we know that clipper circuits eliminate either the positive or the negative portion of a waveform – the unwanted output is limited to a maximum of Vf (forward voltage drop) above or below ground. The biased shunt clipper is normally used to protect a device or circuit that has both positive and negative input signals. The bias voltage is selected to prevent the input from exceeding a maximum safe level.
In Figure 15-1, diode D1 has its cathode connected to a bias of +2.4 V and D2 has its anode connected to -2.4V. As shown, D1 will be reverse biased while the output of the clipping circuit is below 2.4V.The positive output will be limited to a maximum of (2.4V+Vf). Similarly, D2 will be reverse biased until the output is more negative the -2.4v. The negative output will be limited to a maximum of – (2.4V+Vf)
Figure 15-1 Biased diode Shunt Clipper
The Zener clipper shown in Figure 15-2 does not require separate bias voltages. When the input signal becomes positive, D1 operates like an ordinary forward biased diode while D2 goes into breakdown. The output voltage at this time is (Vf1+Vz2). When the input is negative, D1 is in Zener breakdown and D2 is forward biased. The output voltage is now – (Vf2+Vz1).
Figure 15-2 Zener shunt Clipper
Example 1:
Assume we require a shunt clipper (Fig.15-1) to protect a circuit. The input voltage to the circuit cannot exceed Vo = +/- 3.1V. The input to the clipper will be +/- 8V (square wave) and the output current is to be 1mA. Specify the diodes and calculate R1.
Given Ifwd (forward current) = 10mA.
Vo = Bias voltage Vb + diode voltage drop Vf
Vb = Vo-Vf
Vb = 3.1V – 0.7V (assume a Si diode)
Vb = 2.4V
Voltage across R1 = (Ifwd + Io) X R1
(Ifwd + Io) X R1 = E – Vb – Vf
R1 = (VG_out – Vb – Vf) / (If + Io)
(8 – 2.4 – 0.7) / (10mA + 1mA) = 445 ohms
Use standard value 470 ohms
Figure 15-3 Diode Shunt Clipper Simulation
The diodes selected will be low current devices with a Vf (forward voltage) of approx 0.7V at If (forward current) = 10ma, and a peak reverse voltage greater than 10V
Example2:
Assume we would like to use a Zener shunt clipper (Fig.15-2) to protect a circuit. The input voltage cannot exceed Vo= +/- 7.5V. The input to the clipper is a +/- 24V square wave and the output current is to be 100 mA. Specify the diodes and calculate R1.
Given output is Vo = +/- 7.5V
Vo = Vf+Vz
Vz = Vo-Vf = 7.5V – 0.7V = 6.8V
Assume a 1N2804. From the spec sheet Vz = 6.8V.
Therefore:
VR1 = VG_out-Vo
= VG_out – (Vf+Vz)
= 24V – (0.7V + 6.8V) = 16.5V
To ensure that Iz > Izk
Iz is approx 1/4 Izt = 1/4 X 1.86A = 465mA
IR1 = Iz+Io = 465mA+100mA = 565mA
R1 = (VR1) / (IR1) = (14.3V) / (565mA)=253 ohms
Nearest standard value is 270 ohms
Figure 15-4 Zener Shunt Clipper Simulation
What Have We Learned?
1) Biased shunt clippers are a simple way to protect circuits from excessive positive and/or negative input voltages. Inputs are limited to an amount equal to the applied bias voltage plus the forward diode drop.
2) Zener clippers do not require the bias voltage.
What’s Next?
We will examine positive and negative diode clamping circuits
