Cockcroft Walton Voltage Multiplier | Diode Capacitor Multiplier

Diode as a voltage multiplier:

A very sophisticated method of obtaining high voltage with the help of simple circuit is Cockcroft-Walton voltage multipliers.  It consists of diodes and capacitors only. There are many possible ways to design voltage multipliers, with the help of diodes and capacitors. This article is intended to provide basic information about Cockcroft-Walton voltage multipliers.

Voltage multipliers are used to produce high voltages ( hundreds and thousands of volts). They are commonly used in high voltage low current device like cathode ray tubes, Lasers etc.

Diode capacitor voltage multiplier takes AC as an input, and produce DC output which is multiple (2,4,6) of AC peak voltage.

VO = 2*N*VP    Equation 1

VO = 2*N*VP - Vdrop     Equation 2

Where

N = number of stages,

VP = peak input voltage

Vdrop = diode drop

Equation 2 >> we consider the diode drop.

For example from a three stage multiplier, we get

VO = 2*3*VP

VO = 6*VP

A single stage Cockcroft Walton multiplier consists of 2 diodes and 2 capacitors. Each stage produce an output which is multiple of 2. Hence, the number of diodes and capacitors stages increase as the voltage increases.

For example at input if you have 10V (AC) at input, each stage produce 20 V. For example, we have 3 stages we get 60 V (DC) at output. It means each stage produce 20V. (20+20+20 = 60). And the voltage multiplier circuit contains 6 capacitors and 6 diodes.

VO = 2*N*VP

VO = 2*3*10 = 60V (neglect diode drop)

Basically voltage multiplication depends on charging and discharging of capacitors. Each capacitor charges upto 2VP except  C1 , which charges upto VP. The upper rail of capacitors clamps the signal, while the bottom rail of capacitors smooths the DC.

Diode Capacitor Voltage Doubler | Single Stage Cockcroft Walton Multiplier:

During negative half cycle D1 is forward biased and charges the capacitor C1 to peak voltage Vp, while D2 is reverse biased. During positive half cycle D2 is forward biased. C2 will try to charge up to 2Vp, because C1 and source are in series. See figure below, C2 tries to charge upto 2VP, after several cycles it will reach upto 2VP. The output is taken across capacitor C2.

Figure 1 Single stage multiplier

Look a figure 2. Input 10V (AC)

Output 20V (DC) approximately (we are considering practical diodes)

Voltage at capacitor C1

Figure 2 Output of Single stage multiplier (ideally output is 20 V, practically it is a little bit less because of diode drop as shown in the figure)

Voltage Quadrupler | Two Stage Cockcroft Walton Multiplier:

Figure 3 Two stage voltage multiplier

The voltage quadrupler circuit consists of two stages of Cockcroft-Walton multiplier. The upper rail of capacitors stores and clamps the signal. The output is observed across lower rail of capacitors.

During first negative half cycle D1 and D3 are forward biased. D1 charges the capacitor C1 to peak voltage Vp, while D2 and D4 remain reverse biased. During first positive half cycle D2 and D4 are forward biased. C2 will try to charge up to 2Vp, because C1 and source are in series.

During second negative half cycle, again D1 and D3 are forward biased. C1 already charged upto VP. Apply KVL and calculate the voltage across capacitor C3. It is 2VP. Look at figure 4. During second positive half cycle D2 and D4 are forward biased. Capacitor C4 charges upto 2VP. C2 has already charged upto 2VP. The output is taken across the two capacitors.

Figure 4 Working of voltage quadrupler

Look a figure 5: Waveforms obtained at various points. You can visualize easily. Voltage across C1 is shifted or clamped. Also, voltage across C3 is also clamped. I discussed earlier, upper rail of capacitors clamp or shift the waveforms. While, the lower rail of capacitors produce DC at the output.

Input 10V (AC)

Output 20V (DC) approximately (we are considering practical diodes)

Voltage at capacitor C1

Voltage at capacitor C3

Figure 5 Output of voltage quadrupler

Diode Peak | Envelope Detector Circuit

Simple Diode Peak Detector Analysis:

Another application of semiconductor diode. An envelope detector is a simple, highly effective, and low-cost solution for the demodulation of narrow band AM, where the percentage of modulation is less than 100%. It is popular in analog signal processing.

Diode Peak Detector Circuit For AM Signals:

An envelope detector circuit consists of a diode and capacitor. The purpose of the diode is to provide a unidirectional path for the current. Or simply it works as a half-wave rectifier. It allows the capacitor to charge only in one direction. The capacitor itself is a storage place. The voltage across the diode is proportional to the signal amplitude.

Figure 1 An Envelope Detector Using a Diode

At the input of the circuit, we have a high-frequency amplitude-modulated signal. We get the envelope of the modulated signal at the output.

Figure 2 Input and output of peak detector

Now let's understand the operation of the circuit.

Look at the input side, there is a single diode that acts as a half-wave rectifier.

The AM signal at the input, during the positive half-cycle diode, is forward-biased. (Consider diode drop which is 0.7V).

The capacitor charges up to peak value VP- 0.7V (consider practical diode).

Negative half-cycle clips. See the figure below.

Figure 3 Output observed after diode. Negative half-cycle clips

See figure 4. This is diode equivalent, the diode has a small internal resistance rd. rd and capacitor constitute a low pass filter. 

Figure 4 Replace diode with equivalent resistance rd

During the forward-biased condition, this small internal resistance followed by a capacitor. It forms a low-pass filter. It allows only the low-frequency component of the rectified signal. It means high-frequency carrier blocks and only the envelope appears at the output.

The capacitor should have a fast charging time but a slow discharging time.

The peaks of the pulsed DC represent the modulating signal.

In the end, we get only the envelope of the modulated signal. See figure below

Figure 5 Final output

Flyback | Freewheeling Diode

This is another application of diode. After carefully reading this topic you can understand

• What is a freewheeling diode?

• What is inductive flyback? How to prevent it?

• What is the purpose of diode connected across inductive load?

I want to explain the terms flyback or inductive flyback or inductive kickback. The term flyback relates to the sudden voltage spike occur in inductive load, when it's supply current suddenly interrupted. The freewheeling diode is there to suppress unwanted effects. Freewheeling diodes are commonly used in controlled circuits, where SCR controls inductive loads.

First we examine the circuit without freewheeling diode and then with freewheeling diode. Let's get started

Circuit Without Flyback Diode:

Here is a simple circuit, battery is connected across an inductive load.

When switch is in closed position, it means there is a voltage across inductor, which is given below.

v = L ( di/dt )

At constant voltage v the current across inductor L tries to increase at a constant rate di/dt.

Figure 1 Circuit without freewheeling diode

Now if switch is opened, the voltage across inductor becomes zero, while the current flows through the coil. Inductor needs to discharge, the current can not zero instantaneously. The inductor needs a path for discharge. The rate at which current change through inductor depends on inductor time constant, which is τ = L/R.

Sudden change in voltage results in a voltage spike,  or arcing across the switch contacts. This is known as inductive flyback. This voltage spike results in damage of your circuit components.

Figure 2 Circuit with an open switch

Circuit With Flyback Diode:

Figure 3 An inductive load with a freewheeling diode

How do we solve the inductive flyback problem? We use a freewheeling diode, connected across inductive load.

When switch is in closed position, the diode is reverse biased. And the rest of the circuit works normally.

When switch is opened, the inductive flyback is not going to be happened because there is flyback diode. It provides the path for current when inductor is disconnected.

The current flows through the circuit when switch is opened is due to the stored energy of the inductor. The diode becomes forward biased.The output current flows through diode and then flows back in the load itself. In this way the stored energy dissipates through freewheeling diode. That's why it is called freewheeling diode.

Applications:

• It is used to suppress the voltage spikes occur in inductive load. I hope you can easily recognise a freewheeling diode present in any circuit. 

Reverse Protection Diode

Let me explain about reverse polarity protection or reverse voltage or reverse current condition. Actually, sometimes it may happen, output voltage rises higher than input voltage, and hence current may start to flow from output terminal to input terminal. This is reverse current condition and may damage your circuit. In electronic devices, swapping of negative and positive leads of battery may result in a damage. Because current flow in opposite direction. Similar situation also occurs in case of an accidental short circuit.

So, what is the solution? What you should do to prevent your circuit. There are various reverse polarity protection circuits available, differ in efficiency and and operation.

To drive electronic devices safely, you need to have a proper input voltage and current polarities. You can get reverse polarity protection with the help of diodes. It has characteristic property of flowing current only in one direction makes it popular in reverse protection circuits.

The simplest form of reverse polarity protection is a single diode connected in series with the battery. Look at the circuit below, in case of reverse polarity, reverse current from output terminals to input terminals, the diode remains reverse biased.

Reverse Protection Diode

Full Wave Rectifier | Bridge Rectifier With Smoothening Capacitor

In the half-wave rectification circuit, you have seen, negative half cycle wastes. In this circuit, we use bridge rectifier configuration to obtain full wave at the output.

During the positive half cycle, diode D2 is forward biased. The current flows through D2 then the load resistor and finally through D3. Hence positive half cycle appears at the output.

During the negative half-cycle, diode D1 is forward biased. Current flows from D1 then load resistor and finally through D4. Hence negative half cycle also rectifies and appears at the output.

Figure 1 The Bridge Rectifier

Figure 2 Output of bridge rectifier

We are considering practical diodes, peak output voltage (VP (out)) is given by

VP (out) = VP (in) -0.7

Average DC Value:

VDC = 0.636 VP

It means the output DC is 63.6% of the peak value.

VRMS = 0.707*VP

Adding A Smoothing Capacitor:

You have seen I  the above figure, the output DC is pulsating which is not smooth. It is undesirable in most cases. Adding a capacitor of a suitable value is used to smooth the output at the load resistor.
 The capacitor charges to peak value and discharges between peaks. The rate at which capacitor discharges is exponential. To minimize the ripples or pulses in the DC we need to choose a proper capacitance value. Now it's time to evaluate the equation.

Figure 3 The Bridge Rectifier With Smoothening Capacitor

Figure 4 Output of bridge rectifier. Fewer ripples, but still needs some improvements (capacitance value 50uF)

Figure 5 Output of bridge rectifier The result is better than above (capacitance value 500uF)

VP = peak voltage

Vr = ripple voltage

T = time period

R = load resistor

C = Smoothing capacitor

\[V_O = V_P e^{\frac {-t}{RC}}\]
 
At t= T >> Vo = VP - Vr

Apply Taylor series and expand e-t/RC

Neglecting higher powers we get

Set  Vr
With the help of the above equation, we can get the value of the capacitor.