INJECTOR CIRCUIT
In this experiment I was given a circuit diagram and I had to create the circuit.
COMPONENTS USED IN THE CIRCUIT.
In the circuit 2 LEDS, 2 NPN transistors and 4 resistors are used. The circuit is supplied 2 voltages, one 12 Volts DC and another 5 Volts digital voltage.
First of all I calculated the value of each resistor being used in the circuit.
The resistors R14 and R15 would have same values as they were being used in the similar positions.
The voltage coming to R14 and R15 is 12V each.
The voltage drop across LED is 1.8V and across the collector of a NPN transistor is 0.2V. The current flow across LED and the collector should be 20mA.
Thus the voltage drop across R14 should be
12V – (1.8 + 0.2) V = 10V
Using ohm’s law (V = I x R) across the resistor R14 with 20 mA as the current because that’s the appropriate amperage for the circuit.
10V = 20/1000 A x R
Thus,
R = 10 x 1000/20
= 500 ohms
Hence the resistance of R14 and R15 would be 500 ohms each.
Now for R13 and R16 there value also would be same as they are placed in identical positions in the circuit.
The voltage supply to both of these is 5V each. The voltage drop across the base of the NPN transistor is 0.6V. I took the data sheet to see the appropriate value of amperage across the base side circuit of the NPN transistor.According to the datasheet when the current on the collector side of the transistor is 10mA the value of current on the base side would be 0.5mA.
As in this case the current on the collector side was 20mA thus the value of current on the base side would be 0.5mA times 2 i.e. 1mA.
Then the voltage drop across the base would be 0.6V hence the voltage drop across the resistor should be 4.4V.
Thus using ohm’s law again,
4.4V = 1/1000 A x R
Hence,
R = 4.4 x 1000
Therefore
R = 4400 ohms
But because of buffering the value of resistance used should be half
Thus
R13 used would be 1/2 x 4400 ohms
I.e. 2200 ohms
Thus the resistors R13 and R16 would be 2200 ohms each.
Then I took the following things:-
1. Bread board
2. 2 LEDS
3. 2 NPN 547 Transistors
4. 2 x 500 ohms resistors
5. 2 x 2200 ohms resistors
6. A 12v DC supply and a 5V digital voltage supply
7. Few connecting wires
First of all I took the breadboard and by seeing the circuit I made the same on the breadboard.
I faced few problems while doing it. First there were a few loose connections in the circuit. Secondly sometimes I put wrong resistors on wrong places. I wasn’t able to recognise the terminals of the transistor. The rectify my mistakes and problems I first of all took a multimeter and by connecting it to different terminals of the transistor I found out the value of voltage and thus came to know about the terminals of the transistor. About the wrong resistor placement I measured the value of resistor and then again saw the circuit more carefully.
After facing few problems I was able to make the whole circuit on the breadboard and then I connected the power supply to it and then the circuit was working properly.
The picture shows the circuit on the board and the LEDS emitting the light.
As we were supplying 5V of digital voltage the LEDs were blinking because the voltage was fluctuating between 5V and 0V
In this picture the LEDs are in the off position as the voltage is 0 at this point.
WORKING OF THE CIRCUIT
The current starts flowing from the 12V power supply to the resistors R14 and R15 and there occurs a voltage drop on each of these of around 10V. Then the current passes through the LEDs and a voltage drop of around 1.8 V takes there and then the current enters the transistors through the collector terminal of the NPN transistor. A voltage drop of around 0.2V takes place on the collector terminal. Then the remaining current flows through to the earth via emitter terminal.
The digital 5 volts voltage supplied to the bases of the two NPN transistors through the resistors R13 and R16 sends the signals to the LEDs and thus the LEDs Blink and when we change the frequency the speed of blinking changes.
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