Diode

A diode is an electronic component which conducts and lets current flow through in only one direction. From Anode (positive) to Cathode (negative). There are a few different types of diodes. For my practical class i used a normal basis diode plus a LED (Light Emitting Diode). The first exercise was to measure the voltage drop over the diode. A voltage drop over a diode tells us how much voltage is required to open the diode's gate to let the current through. The voltage drop over the diode then stays constant no matter how much or how little load is applied to the circuit. The voltage drop over our diode was 0.564V and for the LED it was 1.783V. These readings were taken in forward biased direction (Anode to Cathode). In reverse biased direction (Cathode to Anode) the voltage drop reading was 0, as current cant flow through a diode backwards. To take these readings our multimeter was set on 'Diode Test Mode', The red lead was on the Anode leg of the diode and the black lead was on the Cathode leg of the diode.

In the next exercise i had to wire up the diodes in a simple circuit. The circuit had a Vs (voltage supply) of 5V, R (resistance) of 1000 Ohms and a diode. First was the normal diode. I then had to use Ohms law and calculate the current through the circuit. The formula for this is I = V/R. Therefore the calculation was 4.4/1000 = 0.00449A (A stands for amps. Amps is the unit for current). The reason why voltage is 4.4V and not 5V is because 0.6V is used up by the diode to let the current through and therefore is subtracted from the voltage supply. The voltage available for the circuit to use is now 4.4V and not 5V. I then measured the current flow using our multimeter. To do this you have to set the meter onto mA and then place it in series in the circuit. The measured reading was 0.0045A. We then had to measure the voltage drop over the diode. This is done the same was as explained above. Meter set on diode test mode, red lead on anode leg and black lead on cathode leg. The measured voltage drop was 0.601V. 

                                                           

The Diode was then replaced by our LED. I then had to record the current flow in the circuit. The current flow had reduced from 0.0045A to 0.0030A. This is because an LED requires a higher voltage to let the current through. Therefore the voltage available to the circuit has been reduced and that's why so has the current flow. This LED required 1.8V to let the current through and that meant that the circuit only had a available voltage of 3.2V. This is why current flow was reduced in the circuit.

INJECTOR CIRCUIT

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.

After that i went on to lochmaster and made a circuit over there and then took a board and made the circuit on that board and then i soldered the circuit on it. 

EXPERIMENT ONE

IDENTIFYING, TESTING AND COMBINING RESISTORS

WHAT IS A RESISTOR?

A resistor is a two terminal, passive electronic component that implements electronic resistance as a circuit element. It limits or regulates the flow of current in a circuit.

In the experiment I took different types of resistors and measured there resistances by two methods.

One was color coding and another one being using an ohm meter.

I took 5 different resistors of different resistance values and then first using a coloring code chart I found out the value of all the resistors.

                         

Then I took a multimeter and set it to ohms and measured the value of each resistor.

                

Then I took two resistors of different values of resistance and measured the value of each.

The value of each resistor was 266 ohms and 98 ohms respectively.

Then I connected the two resistors in series and measured the combined value. It was 364 ohms.

          

According to ohm’s law the value of resistors get added when connected in series and thus the value of the combination should be 266 + 98 ohms which is equal to 364 ohms. Thus this verifies the ohm’s law as both the measured and calculated values are equal.

Then I connected both the resistances in parallel and then measured the combined value.

                

The measured value came to be 72 ohms and the calculated value from the formula 1/R = 1/R₁ + 1/R₂

It came to be 71.61 ohms which and thus this also verifies ohm’s law.

REFLECTIONS

Resistors are used because in circuits some devices are very vulnerable if exposed to a bit high voltage. The devices can get damaged. Resistors control the flow of current and just supply that much voltage that the devices do not get damaged.

We use the resistors in series and parallel to vary the amount of resistance using the same resistors. If we want to increase the value of resistance then we will connect the resistors in series and if we want lesser value of resistance then we would connect the resistors in parallel.

This experiment also showed that we can use any of the two methods to measure the resistance and we will get the same value.