If we forget to turn off the lights in the car or headlights on all
night but in the morning you could not even start the car Because
batteries lose some voltage, we lets to build the warning forgot off
lights circuit better.
As perfect circuit in Figure 1. Use CMOS-IC no-MC14023B. Which is a NAND gate 3 input, Quantity: 3, a small transistor and a Buzzer only.
When rotate the car ignition key switch to stop the engine. If turns off the small headlights and opening the door out of the car, warning buzzer will sound immediately.
The working principle.
The working of circuit in Figure 2. This circuit use all NAND gate, We see the conditions that the buzzer has three reasons.
1. Rotate the Car ignition key switch to stop the engine. If engine not works also no voltage to charges the car battery. The voltage at switch should of 0-volts or state is “0”.
2. Forgot turns off the small headlight show that headlight still current and voltage at it should has about of 12-volts or state is “1”.
3. Opening the door to get out the car. The voltage may be 12 volts or 0 volts as each brand cars. Suppose the voltage is “0″, or state “0″ and conditions third item is the input signal to our circuit.
Then we see the 3 NAND gate input that we use. This gate is NAND, so if input is all “1” state, will be output equals “0” and it is the only case to the output is “0″.
If others input will has output is “1” all the times.
Use this features we will use this conditions cause the buzzer loud.
In when we need “1” so we need to reverse the logical state “0″ to “1″ by inverter. Which in this case we use others NAND into inverter form to saving IC.
See in Figure 2. We will use NAND gate- I2 inside MC14023B is are a decision making logic circuit. For I1 and I3 will be use as inverter gate to reverse the logical state of an incoming signal.
In Figure 1 from terminal A, which will wiring to headlight switch. If forgot turns off switch will has voltage is 12 volts. Therefore, is signal “1”. This signal will be connected to pin 1. Which is directly input of I2-NAND gate. The terminal B, which is connected to the door switch, the signal this terminal is “0”, if the door is opened and the lamp in car is brightness, this signal will be connected to through the I1-NAND gate, to reverse signal from “0” into “1”, Then so entered in the input of I2 next.
In addition,the door switch of some brands may be connected with a 12 volts or state “1”. We need to modify our circuit by connect this signal to input of I2 as Figure 2. Therefore, your car in which form, should check it before assemble this project.
For the the car ignition key switch will connect to the terminal C, this signal will has voltage is 0 volts. If stop engine. When is signal “0” so connect through I3 to reverse signal into “1” then in to I2 again.
The output of I2, which is pin 8 that is As a result, the decision conditions to be used to the buzzer loud. If we want to buzzer loud when output is “0”, so use how to connecting through resistor to drive the 2SC561- PNP-transistor.
When signal is “0” will has current flow from the power supply through emitter – base ,resistor in to the gate. This current will be amplify as collector current to drive buzzer.
If output is “1” or voltage is 12 volts. Which equals the power supply voltage also will noting have current flow in this section the buzzer so not loud.
In circuit has connecting as accessories for example: Resistor, Diode , R1 are connected to protects a high voltage which may be the start switch.
The R2-R4 are resistors that connects to protect when Input is left alone Because the switch Prevent electrostatic influence by reduce resistance input of CMOS down to same the resistors only.
The 1N4001-diode is connected to protect a noise signal while running which Which may be interrupting the operation of the circuit.
The case of Figure 1, there is NAND I1, which is not in use. Usually the gate is not in use, do not leave it and let the input. because the CMOS has very high input resistance. May be static or interference Itti’s easy. Should be connected to ground or the power supply.
The circuit assembly.
Figure 3 is the component layout on the perforated PCB and wiring, For ICs, transistor, and diode will use others that has any features comparable, Resistors size ¼ watts Precision, 5, 10 or 20%. The buzzer is small size at 12 volts and current less than 100 mA.
Soldering the wires on pcb should be tight.If the this handset equipped with a vibration in the car all the time. Wires may be detached easily. Should be soldered IC on PCB directly over the IC socket.
Creating a project It is not difficult But is installed on the car is the more difficult, which would be a function of your friends, if you want to do this project.
The circuit is successful
The component list.
IC1____MC14023B or TC4023 or CD4023____= 1 pcs.
Q1_____ 2SA561 or BC327 45V 800mA PNP Transistor___1 pcs.
D1_____1N4001____1000V 1A Diodes _______________= 1 pcs.
22K = 2 pcs.
100K____= 3 pcs
Buzzer 12V
The perforated board
Terminal,wires,and others.
As perfect circuit in Figure 1. Use CMOS-IC no-MC14023B. Which is a NAND gate 3 input, Quantity: 3, a small transistor and a Buzzer only.
When rotate the car ignition key switch to stop the engine. If turns off the small headlights and opening the door out of the car, warning buzzer will sound immediately.
The working principle.
The working of circuit in Figure 2. This circuit use all NAND gate, We see the conditions that the buzzer has three reasons.
1. Rotate the Car ignition key switch to stop the engine. If engine not works also no voltage to charges the car battery. The voltage at switch should of 0-volts or state is “0”.
2. Forgot turns off the small headlight show that headlight still current and voltage at it should has about of 12-volts or state is “1”.
3. Opening the door to get out the car. The voltage may be 12 volts or 0 volts as each brand cars. Suppose the voltage is “0″, or state “0″ and conditions third item is the input signal to our circuit.
Then we see the 3 NAND gate input that we use. This gate is NAND, so if input is all “1” state, will be output equals “0” and it is the only case to the output is “0″.
If others input will has output is “1” all the times.
Use this features we will use this conditions cause the buzzer loud.
In when we need “1” so we need to reverse the logical state “0″ to “1″ by inverter. Which in this case we use others NAND into inverter form to saving IC.
See in Figure 2. We will use NAND gate- I2 inside MC14023B is are a decision making logic circuit. For I1 and I3 will be use as inverter gate to reverse the logical state of an incoming signal.
In Figure 1 from terminal A, which will wiring to headlight switch. If forgot turns off switch will has voltage is 12 volts. Therefore, is signal “1”. This signal will be connected to pin 1. Which is directly input of I2-NAND gate. The terminal B, which is connected to the door switch, the signal this terminal is “0”, if the door is opened and the lamp in car is brightness, this signal will be connected to through the I1-NAND gate, to reverse signal from “0” into “1”, Then so entered in the input of I2 next.
In addition,the door switch of some brands may be connected with a 12 volts or state “1”. We need to modify our circuit by connect this signal to input of I2 as Figure 2. Therefore, your car in which form, should check it before assemble this project.
For the the car ignition key switch will connect to the terminal C, this signal will has voltage is 0 volts. If stop engine. When is signal “0” so connect through I3 to reverse signal into “1” then in to I2 again.
The output of I2, which is pin 8 that is As a result, the decision conditions to be used to the buzzer loud. If we want to buzzer loud when output is “0”, so use how to connecting through resistor to drive the 2SC561- PNP-transistor.
When signal is “0” will has current flow from the power supply through emitter – base ,resistor in to the gate. This current will be amplify as collector current to drive buzzer.
If output is “1” or voltage is 12 volts. Which equals the power supply voltage also will noting have current flow in this section the buzzer so not loud.
In circuit has connecting as accessories for example: Resistor, Diode , R1 are connected to protects a high voltage which may be the start switch.
The R2-R4 are resistors that connects to protect when Input is left alone Because the switch Prevent electrostatic influence by reduce resistance input of CMOS down to same the resistors only.
The 1N4001-diode is connected to protect a noise signal while running which Which may be interrupting the operation of the circuit.
The case of Figure 1, there is NAND I1, which is not in use. Usually the gate is not in use, do not leave it and let the input. because the CMOS has very high input resistance. May be static or interference Itti’s easy. Should be connected to ground or the power supply.
The circuit assembly.
Figure 3 is the component layout on the perforated PCB and wiring, For ICs, transistor, and diode will use others that has any features comparable, Resistors size ¼ watts Precision, 5, 10 or 20%. The buzzer is small size at 12 volts and current less than 100 mA.
Soldering the wires on pcb should be tight.If the this handset equipped with a vibration in the car all the time. Wires may be detached easily. Should be soldered IC on PCB directly over the IC socket.
Creating a project It is not difficult But is installed on the car is the more difficult, which would be a function of your friends, if you want to do this project.
The circuit is successful
The component list.
IC1____MC14023B or TC4023 or CD4023____= 1 pcs.
Q1_____ 2SA561 or BC327 45V 800mA PNP Transistor___1 pcs.
D1_____1N4001____1000V 1A Diodes _______________= 1 pcs.
22K = 2 pcs.
100K____= 3 pcs
Buzzer 12V
The perforated board
Terminal,wires,and others.
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