Wiring Diagram

I've drawn the circuit with a single pole relay. But you can use a multi-pole relay if it suits your application. Only one half of the Cmos 4013 is used. So you could construct two independent toggle switches with a single IC. The circuit will work at anything from 5 to 15-volts. All you need do is select a relay with a coil voltage that suits your supply.

The LED provides a visual indication that the relay is energized. In effect - it tells you whether the switch is on or off. It's not necessary to the operation of the circuit. If you wish you may leave out R3 and the LED.

The circuit on the left is a common collector or emitter follower and has the advantage of one less part since a resistor is not needed in series with the transistor base. However the voltage across the relay coil will be two diode drops less than the supply voltage, or about 11 volts for a 12.5 volt input.

The common emitter configuration on the right offers the advantage of the full supply voltage across the load for most of the delay time, which makes the relay pull-in and drop-out voltages less of a concern but requires an extra resistor in series with transistor base. The common emitter (circuit on the right) is the better circuit since the series base resistor can be selected to obtain the desired delay time whereas the capacitor must be selected for the common collector (or an additional resistor used in parallel with the capacitor).

The time delay for the common emitter will be approximately 3 time constants or 3*R*C. The capacitor/resistor values can be worked out from the relay coil current and transistor gain. For example a 120 ohm relay coil will draw 100 mA at 12 volts and assumming a transistor gain of 30, the base current will be 100/30 = 3 mA. The voltage across the resistor will be the supply voltage minus two diode drops or 12-1.4 = 10.6. The resistor value will be the voltage/current = 10.6/0.003 = 3533 or about 3.6K. The capacitor value for a 15 second delay will be 15/3R = 1327 uF. We can use a standard 1000 uF capacitor and increase the resistor proportionally to get 15 seconds.

As you may have notice, the 741 is connected as a voltage comparator. Two voltage dividers are easy to be found: The first one is the10K resistor and the LDR . The second one is composed by the two 470 Ohms resistors and the potentiometer. Both the outputs of the dividers are connected as inputs to the voltage comparator.

The second voltage divider will settle the reference voltage. The first voltage comparator that contains the LDR, will change it's voltage according to the light level. When the voltage across the negative input of the comparator is less than the voltage to the positive input of the comparator, the output is held low. When the voltage on the negative input rises, there will be a time that it becomes greater than or equal to the positive (pre-selected) voltage, and then the output becomes high and the relay through the 2N2222 is actuated.

When the switch is opened, the timer produces an approximate 1 second clock signal, decrementing the counter until the 0 count is reached. When the zero count is reached, the 'carry out' signal at pin 7 of the counter moves low, energizing the 12 volt relay and stopping the clock with a low signal on the reset line (pin 4). The relay will remain energized until the switch is again closed, resetting the counter to 9. The 1 second clock signal from the 555 timer can be adjusted slightly longer or shorter by increasing or decreasing the resistor value at pin 3 of the timer.

This circuit will energize and de-energize a relay at the push of a button. Any type of momentary action push-to-make switch can be used. Pushing the button once - will energize the relay. And pushing it a second time - will de-energize the relay

I've drawn the circuit with a single pole relay. But you can use a multi-pole relay if it suits your application. Only one half of the Cmos 4013 is used. So you could construct two independent toggle switches with a single IC. The circuit will work at anything from 5 to 15-volts. All you need do is select a relay with a coil voltage that suits your supply.

The LED provides a visual indication that the relay is energized. In effect - it tells you whether the switch is on or off. It's not necessary to the operation of the circuit. If you wish you may leave out R3 and the LED.

CONTROL RELAY WITH INFRARED CIRCUIT SCHEMATIC DIAGRAM

Normally, home appliances are controlled by means of switches, sensors, etc. However, physical contact with switches may be dangerous if there is any shorting. The circuit described here requires no physical contact for operating the appliance. You just need to move your hand between the infrared LED (IR LED1) and the phototransistor (T1).

The infrared rays transmitted by IR LED1 is detected by the phototransistor to activate the hidden lock, flush system, hand dryer or else. This circuit is very stable and sensitive compared to other AC appliance control circuits. It is simple, compact and cheap. Current consumption is low in milliamperes. The circuit is built around an IC CA3140, IRLED1, phototransistor and other discrete components. When regu lated 5V is connected to the circuit, IR LED1 emits infrared rays, which are received by phototransistor T1 if it is properly aligned. The collector of T1 is connected to non-inverting pin 3 of IC1. Inverting pin 2 of IC1 is connected to voltage-divider preset VR1. Using preset VR1 you can vary the reference voltage at pin 2, which also affects sensitivity of the phototransistor. Op-amp IC1 amplifies the signal received from the phototransistor. Resistor R3 controls the base current of transistor BC548 (T2). The high output of IC1 at pin 6 drives transistor T2 to energise relay RL1 and switch on the appliance, say, hand dryer, through the relay contacts. The working of the circuit is simple.

Control | Pengendali Relay Menggunakan Transistor

The following series of functions to control the relay kutup. With an input signal greater than 0.3V pk-pk (100mV RMS) the positive half of the waveform will switch on transistor Q1, and Q2 and the relay. As the input signal switches to its negative transition, Q1 will switch off, but the base current in using-multimeter-to-measure-transistor.html">transistor Q2 continues to flow via the C2, so Q2 and hence load relays remain on. This will happen for any ac signal within 50 to 1000Hz. R1 prevents excessive base current flowing in transistor Q2, if required a series resistor of 100 ohms can be included with C1 to reduce excessive current flow, though this may decrease sensitivity.

C2 has a dual purpose; as well as smoothing the input signal, it adds a delay to the on / off operation. The delay is dependent on the value of C2 and the coil resistance of the relay. Instead of a relay, a LED and series resistor of 1k could be used instead, however the relay has the advantage of being able to switch large loads on and off. C2 has a dual purpose; as well as smoothing the input signal, it adds a delay to the on / off operation.