BLOG 4

OPERATIONAL AMPLIFIER

An operational amplifier is a DC coupled high gain electronic voltage amplifier having 2 different inputs which are inverting(-ve) and non-inverting (+ve).  The op amp produces a really high output voltage as compared to the difference between the input terminals.

WORKING OF A OP AMP

The Op amps main work is to amplify the difference between the voltages of the 2 inputs. The output of the amplifier can be negative or positive depending upon which input is bigger than the other one.

Image from wikipedia.com

The op amp also has 2 power supplies:-

·         A positive power supply usually symbolised by +Vss

·         A negative power supply usually symbolised by –Vss

The output is denoted by Vo

While the inputs are denoted by Vp and Vn

Vo = A (Vp – Vn)

In the above formula A is the open loop gain in voltage.

Usually A has a huge value it sometimes reaches to value of 1000000.

The value of Vo is always restricted by the power supplies (in case the output is negative then by –Vcc or if the output is positive then by + Vcc)

APPLICATIONS OF OP AMPS

The OP AMPS are used for the following purposes:

·         Audio  pre-amplifiers and buffers

·         Convertors ( analog to digital and vice versa)

·         Voltage clamps

·         Differentiators  and integrators

·         Precision rectifiers

Op Amps are mainly configured in 3 different ways:

·         Inverting amplifiers

·         Non inverting amplifiers

·         Differential amplifiers

INVERTING AMPLIFIERS

Image from Wikipedia.com

As we know that the gain can be really high but this really high gain is of no real use to us as it makes the amplifier both unstable and hard to control as the smallest of input signals, can make the output voltage to saturate and then swing towards one of the power supply rail and thus leading to complete loss of control.

As we also know that the open loop DC gain of an operational amplifier is pretty high so we would not be affected even if we lose some of the gain. Thus a fraction of output is put back into the inverting input. This is done by using a suitable resistor. By sending back the output to the inverting input we gain control over the amplifier. This process is known as negative feedback.

This negative feedback connection forces the differential input voltage towards 0. This causes a closed loop circuit and thus the gain of the amplifier by this is termed as closed loop gain. This helps us control and reduce the overall gain but at the cost of amplifiers bandwidth.

NON INVERTING AMPLIFIER

Image from play-hookey.com

 In non-inverting amplifier the input signal is applied directly to the non-inverting input terminal. Thus the output becomes positive as the positive input is bigger than the negative input. The output signal in such sort of setup is “in phase” to the input signal.

The feedback control of the non-inverting amplifier is done by sending a fraction of output voltage back to the inverting input terminal. This makes a closed loop configuration which makes a non-inverting amplifier circuit with a great stability and really high input impedance.

DIFFERENTIAL AMPLIFIER

Image from allaboutcircuits.com

A differential amplifier is one in which we connect an input voltage to both of the input terminals (inverting as well as non-inverting). In this type of setup the Op Amp amplifies the difference between the two input voltages. Thus this type of a setup is a subtractor set up. It is different from the others as it subtracts the voltages from the input terminal rather than adding them up. This type of amplifier is known as differential amplifier.

IDEAL AMPLIFIER

An ideal Operational Amplifier is basically a three-terminal device which consists of two high impedance inputs, one called the Inverting Input (-ve) and the other one called the Non-inverting Input (+ve).

References:

Google.com

Moodle.unitec.ac.nz

www.electronics-tutorials.ws

howstuffworks.com

Images have been reffered to in text only.

ELECTRIC COMPONENTS

CAPACITOR

A capacitor also called condenser is a passive two-terminal electrical component which stores energy in an electric field by a layer of insulating film. The forms of practical capacitors vary widely, but all contain at least two electrical conductors separated by a dielectric (insulator). Capacitors used as parts of electrical systems, for example, consist of metal foils separated.

When there is a potential difference (voltage) across the conductors, a static electric field develops across the dielectric, causing positive charge to collect on one plate and negative charge on the other plate. Energy is stored in the electrostatic field. An ideal capacitor is has a single constant value, capacitance, measured in farads. This is the ratio of the electric charge on each conductor to the potential difference between them.

The capacitance is greatest when there is a narrow separation between large areas of conductor; hence capacitor conductors are often called "plates," referring to an early means of construction. In practice, the dielectric between the plates passes a small amount of leakage current and also has an electric field strength limit, resulting in a breakdown voltage, while the conductors and leads introduce an undesired inductance and resistance.

Capacitors are widely used in electronic circuits for blocking direct current while allowing alternating current to pass, in filter networks, for smoothing the output of power supplies, in the resonant circuits that tune radios to particular frequencies and for many other purposes.

Circuit Number	Capacitance (uF)	Resistance (Ohms)	Observed Time (ms)
1	100 (uF)	1000 Ohms	500 ms
2	100 (uF)	100   Ohms	45   ms
3	100 (uF)	470   Ohms	250 ms

The current flow and charge time are directly proportional to each other.

RELAYS

A relay is a device which switches on a high amperage device using a low amperage one. This low amperage circuit is called a control circuit. The control circuit has a coil of wire that creates a magnetic field around it when the circuit is powered and earthed. The switching circuit (higher amperage circuit) will has a set point of contacts that are switched on and off by having the magnetic field pull (attract) the points over to connect with another set of points.

The control circuit of the relay usually gets its power from the battery. It will also have a switch that will turn on and off the circuit. This switch can either be on the positive side of the circuit or the negative side of the circuit. The circuit can be switched by either a switch, a sensor with a switch inside it, or an ECU (electronic control unit) that does the switching based on a logic circuit.

The switching circuit (high amp circuit) also gets its power from the battery and this circuit is connected to the component.

TRANSISTOR

It’s a device which is used to amplify and switch electric signals. It uses a small amount of current to open the gate for a high current and voltage flow. It is composed of a semiconductor material with at least three terminals for connection to an external circuit. A voltage or current applied to one pair of the transistor's terminals changes the current flow in the other pair of terminals. Semi-conductor is a device which has either extra electrons or extra protons. There are 2 types of transistors NPN and PNP. NPN, as the name suggest has 2 layers of –vely charged semi-conductor and 1 layer of +vely charged semi-conductor. The PNP has 2 +vely charged and 1 –vely charged.

MOSFETS

Metal–oxide–semiconductor field-effect transistor

It is a transistor used for amplifying or switching electronic signals. The basic principle of this kind of transistor was first proposed by Julius Edgar Lilienfeld in 1925. In MOSFETs, a voltage on the oxide-insulated gate electrode can induce a conducting channel between the two other contacts called source and drain. The channel can be of n-type or p-type.

OXYGEN SENSOR CIRCUIT

In this we used the following components:-

12v Battery

3 LED's:

1x Red led

1x Yellow led

1x Green led

1x Op Amp  LM324

3 diodes 1N4001

7 resistors (R2=1KΩ, R3=1KΩ, R4=1KΩ, R5=380Ω, R6=10KΩ, R7=270Ω, R8=470Ω)

1 zener diode 9v1

2 capacitors  0.1uF

A O2 sensor is located in your exhaust (just after the exhaust manifold), and as the name states it is senses the amount of oxygen in your exhaust gases.

CALCULATION

R2, R3 & R4

I= 9.5mA =0.0095A

The voltage drop in R2 is at 9.6v, Vd=12 - 0.6 - 1.8 =9.6v

R=V/I =9.6/0.0095 =1010.5Ω

The voltage drop in R3 is at 9v, Vd=12 - 0.6 - 0.6 - 1.8 =9v

R=V/I =9.6/0.0095 =947Ω

The voltage drop in R4 is at 9.6v, Vd=12 - 0.6 - 1.8 =9.6v

R=V/I =9.6/0.0095 =1010.5Ω

R5=

Power supply voltage is at 12v, Vd of crossing diode D2=0.6v, Vd of zener diode D1=9.1v

Vd of R5 = 12-0.6-9.1= 2.3v

I=5.6mA=0.0056A

R=V/I =2.3/0.0056 =411Ω

R7 an R8

R6=10KΩ Voltage drop in crossing R6 is at 8.47v

The volage are at 9.1v and 0.63v before the R6 and after the R6 each.

So, Vd=9.1-0.63 = 8.47v

ohms law I=V/R =8.47/10,000 =0.000847A

R8=

VD= 0.4

Vd= 0.63 - 0.23 =0.4v

R=V/I =0.4/0.000847 =472Ω

R7 consumes the voltage which voltage drop is at 0.23

Vd= 0.23 - 0 =0.23Ω

R=V/I =0.23/0.000847 =271.5Ω

Total resistance RT=R6+R8+R7 =10,743Ω

I=9.1/10,743 =0.000847A

OXYGEN SENSOR

n the early 1980s when oxygen sensor was first introduced, it has only one signal wire. This sensitive wire was designed so that it only takes low voltage signal under 1 volt. This crude design has a major flaw; it takes too much time for the sensor to give signal to the engine computer.

It is the computer’s job to regulate the fuel mixture after the sensor warms up which sometimes takes about 10-15 minutes. This time delay can gets worse when the weather gets cold and would dump a lot of raw fuel to the catalytic converter (CAT). This leads to premature CAT failure and high fuel .But since 1994 to 2006 the oxygen sensor has been upgraded to solve this problem.

The O2 sensor is mounted in the exhaust manifold to monitor how much unburned oxygen is in the exhaust as the exhaust exits the engine. Monitoring oxygen levels in the exhaust is a way of gauging the fuel mixture. It tells the computer if the fuel mixture is burning rich (less oxygen) or lean (more oxygen).

In this first of all I made the circuit on breadboard and then took a pcb board and then made the whole thing on it. It was not too easy. I Faced few problems.

REFLECTION

One thing I learnt from this task was to never overlook any aspect of the wiring diagram. As this can cause in the circuit to not work and can also cause damage to the components of that circuit. Always look carefully for the current flow and follow it as you are making your circuit. I also improved on my soldering. I learnt that you are meant to heat up the component leg as well and not just the area around it. Although you have to be care full not to overheat the leg otherwise you can damage the component you are trying to solder. If i was to do this task again a second time the one thing i would make a change on would be the fault finding exercise. I would want a more challenging fault to diagnose.

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.