Along with the wave shaping circuits such as clippers and clampers, diodes are used to construct other circuits such as limiters and voltage multipliers, which we shall discuss in this chapter. Diodes also have another important application known as rectifiers, which will be discussed later.
Another name which we often come across while going through these clippers and clampers is the limiter circuit. A limiter circuit can be understood as the one which limits the output voltage from exceeding a pre-determined value.
This is more or less a clipper circuit which does not allow the specified value of the signal to exceed. Actually clipping can be termed as an extreme extent of limiting. Hence limiting can be understood as a smooth clipping.
The following image shows some examples of limiter circuits −
The performance of a limiter circuit can be understood from its transfer characteristic curve. An example for such a curve is as follows.
The lower and upper limits are specified in the graph which indicate the limiter characteristics. The output voltage for such a graph can be understood as
V0=L−,KVi,L+V0=L−,KVi,L+
Where
L−=Vi≤L−kL−=Vi≤L−k
KVi=L−k<Vi<L+kKVi=L−k<Vi<L+k
L+=Vi≥L+KL+=Vi≥L+K
There are few types of limiters such as
· Unipolar Limiter − This circuit limits the signal in one way.
· Bipolar Limiter − This circuit limits the signal in two way.
· Soft Limiter − The output may change in this circuit for even a slight change in the input.
· Hard Limiter − The output will not easily change with the change in input signal.
· Single Limiter − This circuit employs one diode for limiting.
· Double Limiter − This circuit employs two diodes for limiting.
There are applications where the voltage needs to be multiplied in some cases. This can be done easily with the help of a simple circuit using diodes and capacitors. The voltage if doubled, such a circuit is called as a Voltage Doubler. This can be extended to make a Voltage Tripler or a Voltage Quadrupler or so on to obtain high DC voltages.
To get a better understanding, let us consider a circuit that multiplies the voltage by a factor of 2. This circuit can be called as a Voltage Doubler. The following figure shows the circuit of a voltage doubler.
The input voltage applied will be an AC signal which is in the form of a sine wave as shown in the figure below.
The voltage multiplier circuit can be understood by analyzing each half cycle of the input signal. Each cycle makes the diodes and the capacitors work in different fashion. Let us try to understand this.
During the first positive half cycle − When the input signal is applied, the capacitor C1C1 is charged and the diode D1D1 is forward biased. While the diode D2D2 is reverse biased and the capacitor C2C2 doesn’t get any charge. This makes the output V0V0 to be VmVm
This can be understood from the following figure.
Hence, during 0 to ππ, the output voltage produced will be VmaxVmax. The capacitor C1C1 gets charged through the forward biased diode D1D1 to give the output, while C2C2 doesn’t charge. This voltage appears at the output.
During the negative half cycle − After that, when the negative half cycle arrives, the diode D1D1 gets reverse biased and the diode D2D2 gets forward biased. The diode D2D2 gets the charge through the capacitor C2C2 which gets charged during this process. The current then flows through the capacitor C1C1 which discharges. It can be understood from the following figure.
Hence during ππ to 2π2π, the voltage across the capacitor C2C2 will be VmaxVmax. While the capacitor C1C1 which is fully charged, tends to discharge. Now the voltages from both the capacitors together appear at the output, which is 2Vmax2Vmax. So, the output voltage V0V0 during this cycle is 2Vmax2Vmax
During the next positive half cycle − The capacitor C1C1 gets charged from the supply and the diode D1D1 gets forward biased. The capacitor C2C2holds the charge as it will not find a way to discharge and the diode D2D2gets reverse biased. Now, the output voltage V0V0 of this cycle gets the voltages from both the capacitors that together appear at the output, which is 2Vmax2Vmax.
During the next negative half cycle − The next negative half cycle makes the capacitor C1C1 to again discharge from its full charge and the diode D1D1 to reverse bias while D2D2 forward and capacitor C2C2 to charge further to maintain its voltage. Now, the output voltage V0V0 of this cycle gets the voltages from both the capacitors that together appear at the output, which is 2Vmax2Vmax.
Hence, the output voltage V0V0 is maintained to be 2Vmax2Vmax throughout its operation, which makes the circuit a voltage doubler.
Voltage multipliers are mostly used where high DC voltages are required. For example, cathode ray tubes and computer display.
While diodes are used to multiply the voltage, a set of series resistors can be made into a small network to divide the voltage. Such networks are called as Voltage Divider networks.
Voltage divider is a circuit which turns a larger voltage into a smaller one. This is done using resistors connected in series. The output will be a fraction of the input. The output voltage depends upon the resistance of the load it drives.
Let us try to know how a voltage divider circuit works. The figure below is an example of a simple voltage divider network.
If we try to draw an expression for output voltage,
Vi=i(R1+R2)Vi=i(R1+R2)
i=V−i(R1+R2)i=V−i(R1+R2)
V0=iR2→i=V0R2V0=iR2→i=V0R2
Comparing both,
V0R2=Vi(R1+R2)V0R2=Vi(R1+R2)
V0=Vi(R1+R2)R2V0=Vi(R1+R2)R2
This is the expression to obtain the value of output voltage. Hence the output voltage is divided depending upon the resistance values of the resistors in the network. More resistors are added to have different fractions of different output voltages.
Let us have an example problem to understand more about voltage dividers.
Calculate the output voltage of a network having an input voltage of 10v with two series resistors 2kΩ and 5kΩ.
The output voltage V0V0 is given by
V0=Vi(R1+R2)R2V0=Vi(R1+R2)R2
=10(2+5)kΩ5kΩ=10(2+5)kΩ5kΩ
=107×5=507=107×5=507
=7.142v=7.142v
The output voltage V0V0 for the above problem is 7.14v