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]]>The post Voltage Doubler Circuit schematic appeared first on Mechatrofice.

]]>A voltage doubler is a circuit which produces an output voltage which is double of the applied input voltage. The below diagram is an AC to DC voltage doubler circuit. That is the input supply is an AC source which is converted to a DC output with a voltage value of two times the input AC voltage.

In this circuit, the capacitor C1 and C2 charges alternately in the two half cycles of the AC source. So, the total voltage across the two capacitors becomes equal to the sum of the voltage of two capacitors.

On the positive half cycle, the diode D1 is forward biased and D2 is reverse biased, thus the current drive through the path from A to B via Diode D1 and capacitor C1. Now the capacitor C1 charges and the voltage across the capacitor C1 becomes equal to the voltage across AB.

On the Negative half cycle, the D2 forward biased and D1 reverse biased, thus the current flows from B to A through capacitor C2 and D2. Hence the voltage across C2 also becomes equal to the VAB.

So, the total voltage taken across the C1 and C2 becomes V=VAB+VAB=2VAB.

To maintain the output voltage as twice as its input, use a load with high resistance or use a larger capacitance. So the capacitor can be charged in each cycle before it discharges to a lower voltage level.

Here circuit is a voltage doubler or DC to DC converter using a 555 IC. The circuit consists of an astable multivibrator circuit which generates a square wave at the output terminal 3, with a 50% duty cycle. The capacitors C1 and C2 charges alternately with the ON and OFF state of the output. When the output is low the capacitor C1 charges and at the high state capacitor C2 charges. The Diode D2 and D3 prevents the discharging of the capacitors.

The applied input voltage should be at the limit of IC NE555 operating voltage 4.5 V- 16V. If the input voltage is V (4.5 V- 16 V) then the output voltage taken across the capacitors will be 2V, that is double of the supply voltage. The output values may have slight variations; it should measure properly before applying to any components or device. Also, the output load should have minimum resistance to prevent over discharge of the capacitor.

IC – NE555

Resistor – R1, R2 – 39K

Capacitor – C1, C2 – 100uf, C3, C4 – 10nf

Diode – D1, D2, D3 -1N4007

Supply – 4.5V -16V

This voltage doubler circuit also consists of an astable multivibrator, using an Op-amp IC 741. Similar to a 555 voltage doubler, the here circuit also generates a square wave output of a specific time period which switches the output states HIGH and LOW. The base of both NPN and PNP transistors, BC547 and BC557 respectively are connected to the op amp output.

So, during the high state of the op-amp o/p, the transistor Q1 will be active and the capacitor C3 gets charged. And in the low state, the transistor Q2 will be active and capacitor C2 charges. Therefore the total potential across the both capacitors will be double of the supply voltage. Because the value equal to the supply voltage is stored in each capacitor alternatively during each high and low state of the o/p.

The capacitor C1 and Resistor R3 should connect to the ground or zero voltage point of the supply.

IC- 741

Resistor – R1 – 15k, R2, R3 – 10k, R4 – 1k

Capacitor – C1 – .01uf, C2, C3 – 100uf

Transistor – Q1 – BC547, Q2 – BC557

Diode – D1, D2 – 1N4007

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]]>A touch switch operates with a touch action of an object. The given touch switch circuit works with an arrangement of two NPN transistors, which can obtain a high current gain for the input base signal.

In the circuit, the emitter of transistor Q2 is connected to the base of the transistor Q1. So the emitter current which is the sum of the input signal (base current) and the amplified signal of Q2(collector current) is again amplified by the Q1. Thus the resultant output has extra sensitivity to the input signal.

The given circuit is a basic design to just operate a single LED with a 3V supply. The circuit can be further modified for additional applications.

Transistor – Q1,Q2- BC547

Resistor – R1 – 1k, R2 – 220Ω

LED – Red

Supply – 3V battery

In this touch switch, a piezoelectric sensor is used to sense the touch action. It can generate an electrical charge from any change in physical parameters such as pressure, strain, force, acceleration and temperature.

A decade counter IC 4017 is used to toggle the output for successive touch actions.

The clock input of the Decade IC has a positive edge triggering. Thus the output pin holds the state till another edge trigger occurs. The output pin has taken from output 2. It shifts the ON state from output 1 to output 2 on the positive edge trigger. The reset pin 15 of the IC has connected to the pin 4 (output 3 of the decade counter). So, it resets the counter on next triggering. Thus, the ON and OFF state repeats for consecutive touch actions on the sensor.The Output from the IC has connected to a transistor, which drives the Relay.

The sensitivity of the touch can be adjusted by varying the resistor connected across the pin 14.

IC – 4017

Piezoelectric sensor

Resistor R1 – 1M, R2 – 10k

Transistor – Q1 – BC547

Diode – D1 – 1N4007

Relay – SPST – 12V,200ohms

Supply – 12V

Here the circuit diagram for a piezo-sensor based touch alarm. Unlike from normal 555 touch plate switch, the piezoelectric touch switches are more sensitive and accurate in action.

The circuit consists of 7555 IC, which is the CMOS version of the 555 IC. The buzzer connected to the output pin 3 gets ON with a triggering pulse at pin 2. The triggering in a 7555 IC occurs when the voltage at the pin 2 drops below 1/3 of the supply voltage. Here on touching the piezoelectric sensor the vibration or pressure generates a proportional voltage, which triggers the circuit since it drops the voltage.

The alarm gets turned OFF after a short time period. Once the circuit triggered, the output will be ON, until the voltage at pin 6 reaches 2/3 of the supply voltage.

The ON time can be adjusted by varying the 1M potentiometer in the circuit. The ON time can be calculated as,

Time period = 1.1 R C.

Here this circuit can obtain a maximum ON time of 55 seconds, by adjusting the resistance to a maximum of 500KΩ.

**Components required**

IC1 – 7555

Piezoelectric sensor

Resistor – R1 – 1M, R2 – 1M pot

Capacitor – C1 – 100uf, C2 – 100nf

Buzzer

Battery-6V

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]]>The below circuit consists of an astable multivibrator using a 555 timer IC. It generates a continuous square wave output. The circuit has an oscillation frequency about 670-680 HZ. The audio frequency of the circuit can be changed to wide ranges by varying the values of any of the resistance R1, R2 or the capacitance C1.

To calculate the resultant frequency from the values of R1, R2, C1,

The frequency, f = 1 / (0.69*(R1 + 2*R2)*C)

For frequent tuning, it’s better to replace the resistance R2 with a potentiometer. In the position of R2, connect the wiper terminal and either one of the end terminals of the pot.

Choose a Loudspeaker with appropriate impedance and power, an ordinary 8 ohms Loudspeaker can work fine. The output resistance should be adjusted with the power rating of the selected Loudspeaker.

To improve the output quality, amplify the output signal using a transistor. The circuit can perform well even without amplification.

IC – NE555

Resistor – R1 – 15k, R2 – 100k, R3 – 100

Capacitor – C1 – 10nf, C2 – 100nf

Loudspeaker

Supply – 9v battery

This circuit is a tone generator using an op amp. Here also the circuit is an astable multivibrator which generates a square wave signal around 3KHZ frequency.

The output current which can deliver by a 741 op amp is very small. So, if it has connected directly to a speaker, the generated sound will not have enough volume. Thus, the output signal has amplified using a transistor.

Astable multivibrator of an op amp will contain both positive and negative half cycle. But here the emitter of the transistor is connected to the negative of the supply. So it only contains 6V for high state output. And a 0V for low state or during the negative half cycle.

The capacitor C1 and R3 have connected to signal ground or Equipotential. If the supply is taken from a single supply unit, the equipotential for the supply can be obtained from a voltage divider network. Or by using two series battery or supply, a reference node can take as signal ground or zero volts.

IC- 741

Resistor – R1 – 15k, R2,R3 – 10k, R4 – 3.3k

Capacitor – C1 – .01uf

Transistor -Q1 – BC547

LS – Loudspeaker

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]]>The post what is shift keying modulation ? appeared first on Mechatrofice.

]]>Amplitude shift keying is a modulation technique in digital communication. It is a form of amplitude modulation that represents the digital data in discrete amplitude levels.

For a binary data, there are two levels 1s and 0s. In ASK modulation for binary symbol 1, a fixed-amplitude carrier wave is transmitted with constant frequency and phase. And the for symbol 0 no signal will be transmitted.

An ASK system is similar to an ON and OFF switching of sinusoidal transmission for the state of 1 and 0 respectively.

clear all close all clc g=[0 0 1 0 1 0 1 1]; f=50; n=1; while n <= length (g) if g(n) == 0 t1 = (n-1) * 0.1:0.1/100:n*0.1 s1 = (0)*cos(2 *pi*f*t1) plot(t1,s1) grid on hold on else t =(n-1)*0.1:0.1/100:n*0.1; s=(5)*cos(2*pi*f*t); plot(t,s); grid on; hold on; end n=n+1; end title('ASK') xlabel('Time') ylabel('Amplitude')

Phase shift keying (PSK) is a digital communication method in which the phase of a transmitted signal is varied to represent digital states.

For the binary form of PSK called binary phase-shift keying (BPSK), uses signal phase difference of 0 and 180 degrees to represent the two binary states.

clear all close all clc g=[1 0 1 0 1 0 ]; f=10; n=1; while n <= length (g) t = (n-1) * 0.1:0.1/100:n*0.1; if g(n) == 0 p = cos(2 *pi*f*t+0); plot(t,p); grid on; hold on; else p1=cos(2*pi*f*t+(pi)); plot(t,p1); grid on; hold on; end n=n+1; end title('PSK') xlabel('Time') ylabel('Amplitude')

Frequency shift keying (FSK) is a frequency modulation technique in digital communication. FSK transmits through discrete frequency changes of a carrier signal at a constant amplitude and phase.

In Binary frequency shift keying (BFSK), the two binary symbols 0 and 1, are each represented by using a specific frequency for logic 0 and another frequency value for logic 1. Mostly a lower frequency (space frequency) representative for state 0 and a higher frequency (mark frequency) for state 1.

Frequency shift keying MATLAB code

clear all close all clc g=[1 0 1 0 1 0]; f=10; n=1; while n <= length (g) t = (n-1) * 0.1:0.001:n*0.1; if g(n) == 0 p = cos(2 *pi*f*t); plot(t,p); grid on; hold on; else p1=cos(2*pi*f*10*t); plot(t,p1); grid on; hold on; end n=n+1; end title('FSK') xlabel('Time') ylabel('Amplitude')

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]]>The post How to make a Continuity tester circuit appeared first on Mechatrofice.

]]>Here circuit is an audible continuity tester with a beep indicator using 555 tone generator circuit.

Normally continuity testers will have a preset threshold resistance value to determine whether the points are electrically closed or open. If the resistance between the probe is below the threshold, the circuit indicates the existence of a connection with a beep sound. The circuit detects a continuity if the resistance between the point is below the threshold resistance value. So on testing the null indication of a continuity does not mean that the circuit is completely open. Because it means it can be either open or with a resistance above the threshold value.

The threshold resistance value of the circuit can preset by adjusting the potentiometer R4. To set a threshold value, connect a resistance equal to the required threshold value between the probes. Then adjust the pot to a point that just stops the tone. That is, the voltage at pin4 is below enough to reset the 555 IC; IC 555 has an active low reset. The circuit will remain in reset as long as the voltage at the reset pin 4 is below 1-0.4 V. So whenever the resistance between the test probe has value just below the threshold resistance, the voltage at reset pin reaches above than its minimum rest value. At this state, the circuit enables and generates the tone. That is the reset pin will enable and disable the tone generator with respect to the continuity between the probes.

IC – NE555

Resistor – R1 – 1k, R2 – 4.7k, R3 – 220, R4 – 10K pot

Capacitor – C1,C2 – 0.1uf

LS – Loudspeaker

Supply – 9V battery

Here the circuit for a simple audible continuity tester with a buzzer. The circuit consists of an op-amp comparator which compares the voltage values at the inverting and non-inverting terminals.

The voltage at the inverting terminal (pin2) has adjusted with a potentiometer. So the output switches to an active high state when the resistance between the probe reaches a lower value, that gives a voltage at the non-inverting terminal which is just above than at the inverting terminal.

The circuit is a variable continuity tester so the threshold value can set by adjusting the potentiometer R1.

Supply Voltage = 6V

By voltage division, V * R2 / (R2 + R

V * R2 / (R2 + R_{threshold}) = Voltage across R2

For desired threshold values, substitute the values in the equation on variable R_{threshold.}

Calculation for a threshold value, R_{threshold} = 100ohms

6 * 1000 / (1000+100) = 6 * 0.909 = 5.454V

The potentiometer should set so as to obtain a voltage just above 5.454 volts at the inverting terminal (pin 2). Thus the circuit triggers or indicates as the non-inverting pin obtains a voltage above 5.454V. That is whenever the resistance between the probe reaches a value below 100ohms.

IC – LM741

Resistor – R1 – 10k pot, R2 – 1k

B1- Buzzer

Supply – 6V battery

Most continuity testers are with a sound indication only. For a visual indication, an additional LED with appropriate resistance can be added in parallel to the buzzer.

The circuit will be useful during electrical maintenance to spot break in lines without a multimeter. But while using in high voltage lines ensure the circuit has no charge. If the circuit contacts directly to a live line, it can cause electrical shock, damage to the circuit or any other accidents.

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]]>An automatic night light circuit controls the switching of light by sensing the surrounding light intensity. The circuit ON the LED light, when the incident light intensity falls below a particular limit. And OFF the light when the incident light is above the limit.

In the circuit, the threshold and trigger pin of the NE555 has connected with a voltage divider network. It can obtain a trigger voltage which is half of the voltage at threshold pin. In a 555 IC, when the threshold pin has a voltage above 2/3Vcc, the output switches to a low state. And output switches to a high state when the voltage at trigger pin falls below 1/3Vcc.

The LDR (Light dependent resistor) has a negative co efficient of resistance with the light intensity. So, when the intensity of the incident light is high, the voltage drop across the LDR decreases. Thus, the voltage across the threshold pin reaches the threshold value and the Light OFF. Similarly, with the decrease in light intensity, the voltage across the threshold pin and the trigger pin decreases and the Light ON.

The light sensitivity of the circuit can be adjusted by the varying the potentiometer R1.

To operate external lamp with the same circuit, interface a relay with the output pin3 of the 555 IC. So it can control 230V AC lights or similar externally powered lights with the same circuit. For relay interface, refer the circuit diagram of the automatic power cut off circuit. The relay can be connected similarly as shown in that circuit.

IC – NE555

Resistor – R1 – pot 407k, R2,R3 – 82K, R4, R5, R6- 220Ω

LDR

Capacitor – C1 – 0.01uf

Diode – D1 – D9 – White LED

This circuit adjusts the output light with respect to the ambient light intensity. That is as the incident light from the surroundings reduces, it automatically increases the brightness of the LED. Similarly, inverse as the light increases. Hence this circuit can consider as a simple light level regulator. Because it show’s a gradual variation in light, not a steady ON – OFF switching.

In the circuit, the sensing part of the circuit is a photo resistor or LDR, (Light Dependent Resistor) which have a negative coefficient of resistance with respect to the intensity of light. That is the value of resistance varies inversely proportional to the light intensity.

In the above circuit, the LDR is connected across the Base-Emitter terminal of the transistor. So, with the variation in the intensity of light, the base current inversely changes in accordance to it. Thereby adjusts the collector current or current through the LED.

Transistor – Q1- BC547

Resistor – R1 – 39k, R2 – LDR -5mm

LED -White

Supply – 6v battery

Here unlike from a transistor circuit, the output has only steady ON state or OFF state for varying ambient light intensity. The LED ON and OFF when the amount of incident light rise or fall with respect to a threshold intensity value. Whereas the transistor circuits lights LED gradually.

In the circuit the op-amp act as a comparator. In a comparator circuit, the output will be at either positive saturation or at negative saturation. The comparator output will be high (positive saturation) when the input voltage of non-inverting terminal is greater than the inverting terminal. And low output (negative saturation) state for a higher inverting voltage. When both terminals kept open, the output will be positive saturation voltage (due to offset voltage of practical op-amp ICs).

As the light falls on the LDR, its resistance decreases and a voltage drop created across the resistor R1. In the circuit the non-inverting terminal is open. And thus for a small rise in the inverting terminal voltage, the output switch to negative saturation and the LED will reverse biased. In full darkness, the resistance of LDR will be high and the voltage drop across is maximised. Whenever the full voltage drops across the LDR, the op-amp switches to positive saturation and ON the LED. The circuit has a high sensitivity to light, a small variation in LDR resistance can switch the circuit OFF.

The sensitivity of the circuit can be varied by adjusting the resistance R1. Because variation in the value of R1 can alter the threshold point to a small extent. For a wide adjustment, a reference voltage value is required to be added to the non-inverting terminal. A potentiometer or voltage divider circuit can be used to obtain a variable reference voltage.

Op-amp IC – LM741

Resistor – R1 – 39k

LDR -5mm

LED -white 6V

Supply – 6v battery

For here circuits, the LDR should properly be placed to sense the incoming light.

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]]>The post Rectifier circuit : Half wave and Full wave appeared first on Mechatrofice.

]]>A Direct current flows only in one direction, which means it has a constant polarity across its terminals. Where an Alternating current periodically changes its direction of the current, that is an alternating polarity at its terminals.

A rectifier is a circuit which converts AC to DC and this conversion process is called as rectification.

In simple words, a rectifier converts the bi directional flow of current to a uni directional flow, which maintains a constant polarity across the load. It can be done by either blocking the reverse flow of current or by redirecting the reverse flow to one direction.

A half wave rectifier clips the negative half cycles and allows only the positive half cycles to flow through the load. Thus it utilises only the one-half cycle of the input signal.

During the positive half cycle (A- Positive & B- Negative) of the signal, the diode will be forward biased and conducts the current through the load resistance. And on negative half cycle (A- Negative & B- Positive), the diode will be reverse biased and prevents the current to flow in the opposite direction. So the polarity of the output terminals keeps unchanged and obtains a unidirectional current through the load.

Average voltage, V_{average} = V_{m}/π | Average Current, Iaverage = I_{m}/π

Rms Voltage, V_{rms} = V_{m}/2 | Rms Current, I_{rms} = I_{m}/2

Ripple factor = 1.21

Maximum efficiency = 40.6%

Transformer utilization factor (TUF) = 0.287

Form factor = 1.57

peak factor = 2

It converts the complete cycles of AC signal to DC.

A center tapped rectifiers work only with a center tap transformer or a similar common potential point across the terminals. The center tap act as a common zero potential terminal in both half cycles.

On the positive half cycle (A- Positive & B- Negative), the diode D1 is forward biased and diode D2 is in reverse biased. Hence the current flow through D1 and the load resistance, from terminal A to center tap.

On the negative half cycle (A- Negative & B- Positive), the diode D2 is forward biased and diode D1 is reverse biased. Current flows through D2 and load resistance, from the B terminal to the center tap of the transformer.

The biasing of diodes changes alternately with respect to change in polarity of terminals.

In center tap rectifier the output DC voltage will be half of the total output voltage of secondary winding. Because the load is always on the half of the secondary winding.

The bridge rectifier consists of 4 diodes in a bridge circuit configuration. From a center tap rectifier, the bridge rectifier has difference only in the circuit arrangement. The efficiency, ripple factor, average value, RMS value all are same except transformer utilisation factor(TUF). Because in a center tap rectifier the transformer winding should consider separately.

The bridge rectifier has an advantage over center tap, that is it works without a center tapped transformer or a common ground.

On the positive half cycle, (A- Positive & B- Negative) diode D2 & D3 are in forward biased. D1 & D4 are in reverse biased, thus the conduction path forms through diode D2, load resistance and diode D3.

Similarly, on the negative half cycle (A- Negative & B- Positive) diode D4 & D1 is in forward biased. Diode D3 & D2 are in reverse biased. Current flows through D4, load resistance and D1.

The biasing of the diodes alternates in each half cycle and creates a same polarity across the load. Hence, in both half cycles, the load resistance has the same direction of the current.

Average voltage, V_{average} = 2V_{m}/π | Average Current, I_{average} = 2I_{m}/π

RMS Voltage, V_{rms} = V_{m}/√2 | RMS Current, I_{rms} = I_{m}/√2

Center tap rectifier, Transformer utilization factor (TUF) = 0.693

Bridge rectifier, Transformer utilization factor (TUF) = 0.812

Ripple factor = 0.482

Maximum efficiency = 81.2%

Form factor = 1.11

peak factor = √2

The arithmetic average of all the instantaneous values of a signal is called as its average value.

* Average value = area under curve / base*

The average of a sinusoidal wave form can be calculated as,

*Average value = Area of a unit cycle/ base length of a unit cycle*

Derivation to mathematically find the resultant average value for a unit cycle of a sine wave,

V = V_{m}sinωt, V_{m} – Maximum Voltage or peak voltage, V – Instantaneous voltage.

The average value of a function f(x) on an interval [a, b] = (1/b-a) _{a} ∫^{b} f(x) dx.

The area under the curve is the Integral of function f(x) in the interval from a to b. And the base length is the difference between the limits b and a.

For a unit cycle of a sine wave, the area of the region has obtained by integrating the sine wave equation and the base length from the difference of limits 0 and 2π.

Hence the average voltage, Vavg = V_{m}/2π _{0}∫^{2π} sinωt dωt | V_{m} is a constant value.

= V_{m}/2π ( _{0 }∫^{π} sinωt dωt + _{π} ∫^{2π} sinωt dωt ) = V_{m}/2π [ – cosωt]_{0}^{π }+ V_{m}/2π [ – cosωt]_{0}^{π }.

= V_{m}/2π [- cosπ + cos0] + V_{m}/2π [- cos2π + cosπ]

Therefore, Vavg = V_{m}/2π [1+1] + V_{m}/2π [-1-1] = 2V_{m}/2π – 2V_{m}/2π = 0

The average value of a sinusoidal alternating quantity for a complete cycle will be equal to zero. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation.

Negative half cycles are absent in the output wave form of a half wave rectifier. So, in order to find the average value of the rectifier, the area under the positive half cycle has divided by the total base length.

The area under the positive half cycle is the integral of sinusoidal wave equation from the limits 0 to π. The total base length is the difference of limits of a complete cycle (2π – 0 = 2π), which includes the base length of both the positive and negative cycles.

The average output voltage of a half wave rectifier can be derived as,

A**verage voltage,** V_{DC} = V_{m}/2π _{0}∫^{π} sinωt dωt

= V_{m}/2π [ – cosωt]_{0}^{π} = V_{m}/2π [- cosπ + cos0]

= V_{m}/2π [1+1] = 2V_{m}/2π_{ }= V_{m}/π

The average voltage equation for a half wave rectifier is V_{DC} = V_{m}/π.

In a full wave rectifier, the negative polarity of the wave will be converted to positive polarity. So the average value can be found by taking the average of one positive half cycle.

Derivation for average voltage of a full wave rectifier,

The **average voltage,** V_{DC} = V_{m}/π _{0}∫^{π} sinωt dωt

= V_{m}/π [ – cosωt]_{0}^{π} = V_{m}/π [- cosπ + cos0]

= V_{m}/π [1+1] = 2V_{m}/π_{ }

Average voltage equation for a full wave rectifier is V_{DC} = 2V_{m}/π.

So during calculations, the average voltage can be obtained by substituting the value of maximum voltage in the equation for V_{DC}.

RMS (Root mean square) value is the square root of the mean value of the squared values.

The RMS value of an alternating current is the equivalent DC value of an alternating or varying electrical quantity. RMS value of an AC current produces the same amount of heat when an equal value of DC current flows through the same resistance.

RMS value of a signal = √ Area under the curve squared / base length.

For a function f(x) the RMS value for an interval [a, b] = √ (1/b-a) _{a} ∫^{b} f^{2}(x) dx.

RMS Value = √ Area of half cycle squared / half cycle base length

The RMS value of a sine wave can be calculated by just taking the half cycle region only. Because the area of positive half cycle squared and negative half cycle squared have the same values. So the derivation will be same as it for a full wave rectifier.

The RMS Voltage of a sine wave, V_{RMS} = V_{m}/ √2, Vm – Maximum voltage or peak voltage.

In a half wave rectifier, the negative half cycle will be removed from the output. So, the total base length(2π) should be taken from the interval 0 to 2π.

The RMS voltage, V_{RMS} = √ V_{m}^{2}/2π _{0}∫^{π} sin^{2}ωt dωt

= √ V_{m}^{2}/2π _{0 }∫^{π}(1 – cos2ωt) / 2 ) dωt = √ V_{m}^{2}/4π [ωt – sin2ωt / 2]_{0}^{π }

= √ V_{m}^{2}/4π [ π – (sinπ) / 2 – (0 – (sin0) / 2)] = √ V_{m}^{2}/4π ( π ) = √ V_{m}^{2}/ 4

Therefore the RMS voltage, V_{RMS} = V_{m}/ 2

The RMS voltage, V_{RMS} = √ V_{m}^{2}/π _{0}∫^{π} sin^{2}ωt dωt

= √ V_{m}^{2}/π _{0 }∫^{π}(1 – cos2ωt) / 2 ) dωt = √ V_{m}^{2}/2π [ωt – sin2ωt / 2]_{0}^{π }

= √ V_{m}^{2}/2π [ π – (sinπ) / 2 – (0 – (sin0) / 2)] = √ V_{m}^{2}/2π ( π ) = √ V_{m}^{2}/ 2

RMS voltage, V_{RMS} = V_{m}/ √2

The peak factor is defined as the ratio of the maximum value to the RMS value of an alternating quantity.

Peak factor = Peak value / RMS value

RMS voltage of a half wave rectifier, V_{RMS} = V_{m}/2. Where V_{m} is the Maximum or peak voltage.

Then the** Peak factor of half wave rectifier **can be calculated as,

V_{m} / V_{RMS }= V_{m} / ( V_{m} / 2 ) = 2 V_{m} / V_{m} = 2

Similarly, For a full wave rectifier, the RMS voltage V_{RMS} = V_{m} / √2

Therefore, the** Peak factor value of full wave rectifier = **V_{m} / V_{m}/ √2

= V_{m} √2 / V_{m} = √2 = 1.414

The ratio of RMS value to the average value of an alternating quantity is known as its form factor.

Form factor = RMS value / Average value

RMS voltage of a half wave rectifier, V_{RMS} = V_{m} / 2 and Average Voltage V_{AVG}= V_{m}/π, V_{m }is the peak voltage.

**Form factor of half wave rectifier** = V_{RMS }/ V_{AVG }= ( V_{m} /2 ) / ( _{Vm} / π )

= π V_{m} / 2 V_{m} = 1.57

For a full wave rectifier, the RMS voltage V_{RMS} = V_{m} / √2 and the average Voltage,

V_{AVG} = 2 V_{m }/ π

**Form factor value of full wave rectifier = **( V_{m} / √ 2 ) / ( 2V_{m} / π )

= π V_{m} / 2√2 V_{m} = 1.11

The ratio of the RMS value (root mean square) of the AC component to the DC component of the output is defined as the Ripple factor and is denoted by γ.

Ripple factor, γ = V_{AC}/V_{DC }| V_{DC} is the average value of the DC output.

V_{RMS} = √ V_{DC}^{2} + V_{AC}^{2} or I_{RMS} = √ I_{DC}^{2} + I_{AC}^{2}

V_{AC }= √ V_{RMS}^{2} – V_{DC}^{2}

Therefore the ripple factor equation is, γ = √ (V_{RMS}^{2} – V_{DC}^{2}) / V_{DC}^{2 } = √ ( V_{RMS} / V_{DC})^{2 } – 1

To calculate the ripple factor of a half wave and full wave rectifier, just substitute the RMS and Average value of the respective rectifier in the above equation.

RMS Voltage of a half wave rectifier, V_{RMS} = V_{m} / 2 | Vm is the peak voltage.

Average Voltage of a half wave rectifier, V_{AVG} = V_{m} / π

Ripple factor, γ = √ ( [ (V_{m}/ 2) / (V_{m} / π) ]^{2} -1 ) = √ ( π / 2 )^{2} – 1 = **1.21**

RMS Voltage of a full wave rectifier, V_{RMS} = V_{m} / √2

Average Voltage of a full wave rectifier, V_{AVG} = 2 V_{m} / π

r = √ ( [ (V_{m}/√ 2 ) / (2 V_{m} / π) ]^{2} – 1 ) = √ ( π / (2 √ 2) )^{2} – 1 = **0.48**

The ratio of DC output power to the AC input power of a rectifier is called as its efficiency. It is denoted by η.

Rectifier efficiency, η = DC output power/input AC power = P_{DC} / P_{AC}

I_{RMS} = I_{m} / 2 , P_{AC }= I_{RMS}^{2} (r_{f} + R_{L}) = ( I_{m} / 2)^{2} (r_{f} + R_{L})

V_{m} is the peak Current |

P_{DC }= I_{AVG}^{2} R_{L} = ( I_{m} / π)^{2} R_{L}

For a half wave rectifier, the Efficiency η = P_{DC} / P_{AC }= ( ( I_{m} / π)^{2} R_{L} ) / ( ( I_{m} / 2)^{2} (r_{f} + R_{L}) )

= 4 R_{L} / π^{2} (r_{f} + R_{L}) = 0.405 R_{L} / (r_{f} + R_{L})

Therfore the maximum efficiecy = **40.5%**

Similarly for a full wave rectifier,

I_{RMS} = I_{m} / √ 2 , P_{AC }= ( I_{m} / √ 2)^{2} (r_{f} + R_{L})

P_{DC } = ( 2 I_{m} / π)^{2} R_{L}

For a half wave rectifier, the Efficiency η = P_{DC} / P_{AC }= ( ( 2 I_{m} / π)^{2} R_{L} ) / ( ( I_{m} / √ 2)^{2} (r_{f} + R_{L}) )

= 8 R_{L} / π^{2 }(r_{f} + R_{L}) = 0.810 R_{L} / (r_{f} + R_{L})

Maximum efficiecy = **81.0%**

Hence we can see that, the efficiency of a full wave rectifier is double than that of a half wave rectifier.

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]]>The post Why fuse wires are always thin? if it is thick? appeared first on Mechatrofice.

]]>As the Fuse is not just an electrical contact, it should connect the circuit; at the same time, it can melt when the current exceeds the limit. That is why conductors with low melting point are used as a fuse.

To burn out itself a fuse wire should generate a sufficient amount of heat to raise the temperature of the material to its melting point. The heat produced in an electrical conductor is proportional to the product of its resistance and the square of the current. That is, for a rated current a particular value of resistance is required to produce a temperature value equal to the melting point of the fuse.

The resistance of a conductor is proportional to its length and inverse proportional to its cross sectional area. So, If the fuse wires are thick, the larger cross section decreases the resistance across the fuse wire. So, even a high current flows through the fuse, it doesn’t blow out. Because it just acts as a normal electric contact in the circuit or wiring.

Resistance, R = ρ L / A

ρ – Resistivity per unit length

L – length

A – Area

In order to reduce the area and to maintain a minimum resistance to generate heat, fuses are made as thin.

A fuse wire should not have high resistance or low resistance. It should have enough resistance to carry its rated current without unwanted disconnection and melt instantly for a small excess of current. That is the fuse thickness increases with the current rating.

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]]>The post Alternating led flasher circuit appeared first on Mechatrofice.

]]>In the circuit, the LED1 lights when the output pin has a HIGH state and LED2 lights when the output is at a LOW state.

The rate of flashing can be adjusted by varying the ON period and OFF period of the astable multivibrator. Refer link of LED flasher circuit.

IC – NE555

Resistor – R1, R2 – 6.8k, R3,R4 – 1k

Capacitor – C1 – 100uf, C2 – 100nf

LED – 2 Nos

Supply – 9v battery

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