The post Automatic night lamp circuit using LDR and IC 555 appeared first on Mechatrofice.

]]>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.

IC – NE555

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

LDR

Capacitor – C1 – 0.01uf

Diode – D1 – D9 – LED

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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 Alternate light flasher circuit – Dual LED 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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]]>The post Continuity tester circuit using 555 appeared first on Mechatrofice.

]]>Normally continuity testers will have a preset threshold resistance value to determine whether the points are electrically closed or open. The circuit detects the continuity if the resistance between the point is below the threshold resistance value. So if a continuity tester shows any break in the line, it cannot be considered as a complete open connection. Because it means it can be either open or with a higher resistance than 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 and 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. Hence the reset pin will enable and disable the tone generator with the continuity between the probe.

The circuit will be useful during electrical maintenance to spot break in lines. 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.

IC – NE555

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

Capacitor – C1,C2 – 0.1uf

LS – Loudspeaker

Supply – 9V battery

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]]>The post Ripple factor and Efficiency of rectifier appeared first on Mechatrofice.

]]>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 Form factor and peak factor of rectifier appeared first on Mechatrofice.

]]>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 a 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 a 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 a 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 a full wave rectifier = **( V_{m} / √ 2 ) / ( 2V_{m} / π )

= π V_{m} / 2√2 V_{m} = 1.11

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]]>The post RMS value of Rectifier appeared first on Mechatrofice.

]]>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

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]]>The post Average value of full wave and half wave rectifier appeared first on Mechatrofice.

]]>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}.

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]]>The post LED strobe light circuit – Police lights appeared first on Mechatrofice.

]]>The given strobe light circuit consists of two 555 astable multivibrators. One with a lower oscillation frequency (IC1 multi vibrator section) and other with a higher frequency (IC2 multi vibrator section).

The IC2 consists of a simple LED flasher circuit, which continuously flashes the LED. And using the signal from the low frequency multivibrator, the signals from the IC2 are alternatively switched to the RED and BLUE LEDs. Here the LED set consists of a 3×3 Red and 3X3 blue LEDS.

The square wave output of IC2 is connected to the RED and BLUE LEDs through an NPN transistor Q1-BC547 and a PNP transistor Q2-BC557.

The base of the both transistors is biased from the output of low frequency multivibrator. So it switches the transistor Q1 and Q2 alternatively with the high and low state at the output.

On high output state of IC1, the Q2 switches to forward active mode and drives the flashing signals to BLUE LEDs. Similarly, in the low output state of IC2, the Q1 drives the signals to the RED LEDs.

The rate of flashing of LED sets and switching time between the LED sets can be adjusted by varying the time period of the two multivibrator circuits.

Integrated circuit – IC1,IC2 – NE555

Transistor – Q1 – BC557, Q2 – BC547

Resistor – R1, R3, R10, R11 – 1K, R2 – 100K, R4-R9- 220ohms, R12- 470K

Capacitor – C1,C4 – 2.2uf, C2,C3 – 0.01uf, C5 – 220uf

Diode – D1 – D9 – Red LED, D10 – D18 – Green LED

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]]>The post Flashing brake lights for motorcycles appeared first on Mechatrofice.

]]>The circuit can use to make a simple pulsating stop brake light, which continuously flashes as long as the brake has applied. Here we are using a 555 astable multivibrator to generate the flashing. The working of flash brake light circuit is same as 555 flasher circuit. With the given values for R1, R2 and C1, the circuit has an ON period and OFF period of 0.125 seconds and 0.124 seconds respectively.

The LED brake light has a set of 3×3 LED bulbs, 3 parallel set of 3 series LEDs and 220 ohms resistance. The 555 IC has a maximum output current of 200mA. The output current driven by an LEDs are about 20ma – 40 ma. Here in the set, it is about 25ma. As the total current is below 100ma, the LEDs can be connected directly to the output pin of the 555IC. Also, as the brake light is not lighting for a too long the IC does not heat as much. Even though, the circuit can work fine if it kept ON for a long period.

The switch should be placed so as to activate the circuit while applying the brake. Or replace the switch with any another easy mechanism which makes an electrical contact during braking. Also, the circuit can be connected in parallel to the existing brake light. But the power rating of wire and fuse should be considered while connecting along with it.

**Components required**

Integrated circuit – IC – NE555

Resistor – R1 – 1K,R2 – 82K, R3,R4,R5- 220ohms

Capacitor – C1 – 2.2uf, C2 – 0.01uf

Diode – D1 – D9 – Red LED

In flash and hold circuit, the light does not flash continuously as like a normal flasher. Instead, the circuit flashes initially for a few seconds and then hold the ON state until the brake switch has released.

This brake light circuit consists of an additional monostable multivibrator with the above flasher circuit. In the circuit, the output of the monostable multivibrator has been connected to the base of a PNP transistor, which has connected across the capacitor C4. So, during the high state of the IC1, the Q1 is OFF. Thus the astable multivibrator generates periodic signals, which flash the LEDs. After the quasi state period of IC1, the output return to the low state. At this time the Q1 ON and it shunts the capacitor. Thus the voltage at the trigger pin becomes below 1/3Vcc and holds the ON state at the output of IC2, which is actually interrupting the working of the astable multivibrator.

The flashing time of the circuit is equal to the time period of the monostable multivibrator. So the period of flashing can be adjusted by varying the time period of the monostable multivibrator.

**Components required**

Integrated circuit – IC1,IC2 – NE555

Transistor – Q1 -BC557

Resistor – R1 – 470K, R2,R3 – 1K, R4 – 100K, R5-R7- 220ohms

Capacitor – C1 – 4.7uf, C2 – 22pf, C3,C5 – 0.01uf, C4 – 2.2uf, C6 – 220uf

Diode – D1 – D9 – Red LED

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