Post Four: Oxygen Sensor Display Unit

My task for the past one and a half weeks was to build an oxygen sensor display unit. This display unit consisted of three LED's, each one a different color. This unit when attached to your oxygen sensor (O2 sensor) on your car, can tel you whether your car is running rich or lean. The LED's were used to indicate this.

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. When a engine is running it draws in a mixture of air and fuel combined. this mixture is then compressed and then ignited by the spark plug. This ignition causes a mini explosion inside your combustion chamber which then pushes down on the piston forcing it to go down, giving you your power. When a car runs rich (more fuel being used) the oxygen levels in the exhaust gasses will decrease as now the more fuel particles will cause a bigger explosion, burning off most of the air. As a result the oxygen level goes down. When a car runs lean (more air) the fuel particles are more spread apart and therefore the explosion that takes place isn't very effective in burning up all the mixture. As a result of this the oxygen levels increase. These readings are given out by the O2 sensor in the form of voltages. The O2 sensor works on a voltage scale of 0 - 1V. These readings are sent to the ECU (Electronic Control Unit) which then alters the air/fuel ratio. It the voltage received by the ECU was 0.2V, this would indicate a lean mixture (more air, less fuel). The ECU will then inject more fuel into the next intake stroke to try and make the air/fuel ratio perfect. The same applies when the ECU gets a high voltage reading of 0.9V. This means that the car is running rich (more fuel, less air). The ECU will then lower the fuel content for the next intake stroke, trying to perfect the air/fuel ratio.  

Component List
A few components were used for making this display unit. I used seven resistors, each with a different value. Two 0.1uF capacitors, one 9v1 zener diode, three 1N4001 diodes, three LED's (one green, one yellow and one red) and a LM324IC operational amplifier (Op amp).

Calculations

Before we could make the circuit we were to calculate the resistors that needed to be used. We were given the value for one of the resistors; R6 = 10K. Before i started calculating the resistance i needed to know the available voltage for the circuit. I noticed that straight after the 12V power supply there was a diode. As we know a diode requires 0.6V to let the current through so therefore the current available to the circuit was 11.4V (12V - 0.6V). To calculate R5 i needed to know two things voltage drop (over R5) and current. Current was given to us, it was 0.0056A. Now the zener diode is rated at 9v1, which means it requires 9.1V to push the current through. So 11.4V minus 9.1V equals the voltage drop over R5. 11.4V - 9.1V = 2.3V. To find R5 i used the formula R = V/I. Therefore  2.3V divided by 0.0056A equals 410.71Ohms. R5 is 410Ohms. Now to find R8 and R7 we needed to know the current flow through that circuit, as we were already given the voltage readings at different point of the circuit. Firstly i found out the voltage drop over R6. The available voltage before R6 was 9.1v and after R6 was 0.63V. Therefore  9.1V minus 0.63V equals the voltage drop over R6, which is 8.47V. 8.47v divided by 10000ohms equals the current flowing through the circuit, which was 0.000847A. To find R8 i needed to know how much voltage it uses. the voltage available before the resistor is 0.63V and after the resistor is 0.23V. Therefore 0.63V minus 0.23V equals voltage drop over R8 which was 0.4V. So R8 equals 0.4V divided by 0.000847A which is 472.26Ohms. R7 was easy to calculate as the available voltage after it is 0V and before it is 0.23V. So R7 equals 0.23V divided by 0.000847A which is 271.55Ohms. The formula I am using is R = V/I.

Resistors 2, 3 and 4 are on a different circuit. They are placed after the LED's. Now as we know LED stands for light emitting diode and as all diodes it requires a voltage to let the current through. This voltage value is stated at 1.8V. This means that now we have to minus 1.8V from 11.4V to get the available voltage and the voltage drop over resistors 2, 3 and 4. 11.4v minus 1.8v equals 9.6V. Now the current required to turn the LED's on is 0.0095A. So to find resistor 2 and four i used the rule R = V/I. So 9.6V divided by 0.0095A equals 1010.53Ohms. So therefore resistor 2 and 4 are  1010.53Ohms. There is a diode in the circuit of resistor 3. This means that we have to subtract an additional 0.6V from 9.6V. This leaves us with 9V. So R3 equaled 9V divided by 0.0095A which was 947.37Ohms.

Now I had all the values for my resistors.

The Circuit   

This circuit looks complicated but in fact it is actually very simple. The triangles you see represents the op amps. All these four op amps were in a chip we got. The numbers on the ends of the op amps represents the leg number on the chip. As we know current requires a positive terminal and an earth terminal to flow through a circuit. Without this the current wont flow and the components on that circuit wont work. The same rule applies with this circuit. The LED's get their positive end from the battery terminal, and it requires the op amp to supply it with earth for it to turn on.

The red LED represents the car running rich. As we can see pin five which is the positive of the op amp, gets a constant voltage of 0.63V. Pin six which is the negative of this op amp is connected to the sensor input.         When the car is running lean or at perfect air/fuel ratio the sensor input wont deliver a voltage that it higher than 0.63V. As a result of this the op amp wont provide ground and the red LED will remain off. When the car does start to run rich, the sensor input will start to reach over 0.63V. This is when the op amp will supply ground for the red LED circuit and the red LED will turn on, indicating a rich air/fuel mixture.

The green LED works in a very similar way to the red LED, except this time the negative pin of the op amp receives a constant voltage of 0.23V (pin 13). The positive pin of this op amp (pin 12) is connected to the sensor input. When ever the sensor input is higher than 0.23V the green LED circuit wont have ground and therefore it will not light up. When the car starts to run lean and the O2 sensor is sensing high levels of oxygen in the exhaust gases. The sensor input will go below 0.23V. This is when the op amp will provide ground for the green LED circuit and the green LED will turn on, indicating a lean air/fuel mixture. 

The yellow LED circuit is a bit tricky as it requires the output from two op amps. It needs pin one and pin eight to output ground. This will only happen it the sensor input is between 0.23V and 0.63V. If the sensor output goes lower than 0.23V then pin eight wont output ground and therefore the yellow LED wont turn on. If the sensor output goes above 0.63V pin one wont output ground. It will output a positive voltage. This will mean that yellow LED will positive voltage coming to it from both sides. This will also cause the yellow LED to not turn on. When the yellow LED turns on it indicates that the car is running at perfect or very close to perfect air/fuel mixture.

The capacitor circuits on the two sides at the top were there to make the current flow smoother.    

Breadboard 
Before we can build this circuit on a proper board we first had to build it on a breadboard. This was done to see if the resistors we chose were right for the job. This was a very tricky task. But to overcome this i followed each individual circuit on the wiring diagram and built it circuit by circuit. Following the individual circuits made it a lot easier to understand what goes on in this circuit, and also helped me understand on how it works. This is where i encountered my first problem. I had not given my LED's the 11.4v power supply. This was then easily fixed as i used jumper wires to deliver the 11.4V after diode one. My second problem i faced while building this circuit was that i had placed my LED's the wrong way around. I had the negative leg going towards the 11.4V power source and not the positive leg. I noticed this when i wired up my circuit to test it. The warning light came on so i quickly turned it off. I checked all my wires and connections but couldn't notice anything wrong. But then as I was following how the current would travel i noticed that the positive leg of the LED has to be facing the 11.4V power source. I then checked my LED's and swapped the legs around. This mistake was made by me because i overlooked the position of the LED's in the wiring diagram. I quickly learned my mistake and checked all my components and wiring before reconnecting it. When I reconnected my circuit to the power sources, it worked!

This is the circuit that I made on the breadboard.

Fault Finding
After our O2 sensor display unit was completed, we were to create a fault on it and let a friend assist it. I assisted Richard's one. he had cut off on of the legs of resistor five. This meant that now resister six, eight and seven didn't receive any voltage at all. After i found the cause i connected the faulty display unit to the power source to see what was happening. No mater what the sensor input changed to only the red LED was on. And it stayed on. I then took down the voltage readings from each of the pins of the LM324IC op amp chip. I wrote these figures down on a spare wiring diagram and then followed the current flow to see what was going on and why the red LED was permanently on. Since resistor six, eight and seven weren't receiving any voltage the pin seven would always output ground as it was very close to zero 0.0022V. I also noticed that the lowest the sensor input voltage went was 0.0086V. This is why the red LED was always on as it always had ground. But what about the other two LED's. Since the lowest the sensor input went was 0.0086V it meant that pins 14 and 1 would always put out positive voltage. Pin 14 is required to put out ground for the green LED to turn on and since it couldn't, the green LED stayed off. Pin 8 did supply ground to the yellow LED but since pin 1 was supplying the yellow LED with positive voltage, it didn't turn on. As we know the yellow LED requires both pin 1 and 8 to be supplying ground to work. Since pin one was supplying positive voltage; the yellow LED got no ground but only positive voltage from both sides; hence it didn't turn on either.  After the fault was found and diagnosed the resistor leg was soldered back on and the display unit worked again.

This Picture shows my findings.

These were the voltage drop readings taken from each pin of the LM324IC operational amplifier.

This picture shows the current flow in the faulty circuit. The arrows represent positive current flow. As you can see from the picture that only the red LED circuit had a positive and a negative.

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 over heat 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.          

Post Three: Electronic components

Capacitor  

A capacitor is an electronic component that stores electrical charge. It does this by providing ground when there is an open circuit (switch is open). This storage of electrical charge prevents voltage spikes from happening. A capacitor consists of two metal plates very close together. They are separated by an insulator. When connected to a battery or a power source electrons flow into the negative plate and charge up the capacitor. This charge still remains when the battery or the power source is removed. The amount of charge a capacitor can store depends on the capacitance of the capacitor (measured in Farads F). Capacitors are also used in circuits to smooth-en out the current flow. This allows for the circuit to have a constant smooth flow of current. 

In practical we looked at the relationship between the amount of current flowing through a circuit and the amount of time it took for the capacitor to charge. This was done on the breadboard. We had a 12V power supply and we used a bridging wire as the switch. The amount of current flow was controlled by the resistors (resistance in the circuit). The capacitor used was measured at 100uF. We did three tests, each with a different value of resistor (different amount of current flow). Here's a table showing our recordings.

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

As we can see the relationship between the current flow and the charge times is very simple. The more current flow in a circuit the quicker the charge time of the capacitor. This is because more current means greater electron flow, hence the capacitor fills up quicker. 

Relays   

A relay is an electronic component that uses a low amperage circuit to switch on a higher amperage circuit. This low amperage circuit is called a control circuit. The control circuit will have 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 have 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 (Bipolar)
A transistor is an electronic component that uses a small amount of current to open the gate for a high current and voltage flow. A bipolar transistor is constructed of three semiconductor plates and are either a PNP transistor or a NPN transistor. As we know protons and electrons attract each other. The semiconductor plates in a transistor either consist of extra protons (+) or extra electrons (-). The PNP transistor has two plates with extra protons and only one plate with extra electrons. The NPN transistor does the opposite of that. Now as we know current consists on electrons. Say you put the P plate in a circuit. The extra protons will attract the electrons flowing in the circuit. This attraction will cause electrons to flow through this plate. The transistor uses this principle. The transistor has three legs the Base, the Collector and the Emitter. (The base is the gate). The base is connected both to the emitter and the collector but neither of those two (the emitter and the collector) are connected with each other. The base is like the control circuit in a relay. When current flows through the base of the transistor it then connects the collector to the emitter allowing for high voltage and high current flow. We call this opening the gate.

Here is a picture of the transistor symbol and the semiconductor construction

How to Check the legs of a transistor
Not all the transistors out there come with their legs named. So we need a way to determine them without mixing them up. Now as we know that the base leg is connected to both the emitter and the collector. This means that we will only have voltage readings when we connect the base terminal to either the emitter or the collector terminal. These voltage readings show us how much voltage is required to push the current through those points. Hence showing us that we have a circuit. Now we know that there are two types of base terminals in a transistor. P-type and N-type. This is determined by which lead of the multimeter is touching the base terminal. 
Here are the multimeter check readings for testing both a PNP transistor and a NPN transistor


* E and C reverse: 1(+) and 2(-): "OL"
* E and C reverse: 1(-) and 2(+): "OL"
* E and B forward: 1(+) and 3(-): 0.655 V
* E and B forward: 1(-) and 3(+): "OL"
* C and B forward: 2(+) and 3(-): 0.621 V
* C and B forward: 2(-) and 3(+): "OL"

This shows us that leg number three in this transistor is the base leg. Now as we can see in both the combination of the conductive readings we got our negative (-) lead on the base terminal hence indicating that it is a PNP transistor. Collector to base junction will always have a lower voltage reading to the emitter to base junction. Therefore with these set of reading we can determine that leg two is the collector leg and that leg one is the emitter leg. For the NPN transistor we would get voltage readings when the positive lead of the multimeter is on the base leg.


Voltage Dividers
A voltage divider is a simple linear circuit that produces an output voltage that is much lower than its input voltage. The voltage divider divides the voltage among the components of the divider. A simple construction of this will be three resisters wired up in series as shown below. 

Now as we know from series circuits the available voltage after each resistor will be lower. Same principle applies over here. Lets say that V in is 12V and you was V1 to be 7V, V2 to be 2V and V3 will naturally be zero or very close to it. Now to create these outputs you need to add resistance to the circuit. Lets say that you decide to set R1 at 800Ohms. To find the current flowing through the circuit you will have to use ohms law (I = V/R). V in this case is 5V (12V minus 7V = 5V). 5V is the voltage drop across resistor one. So V = 5V and R = 800Ohms. (5V/800Ohms = 0.00625A) now we know the current running through the circuit. This makes it easier to set resistor two and three. To set resistor two you will use ohms law again (R=V/I). In this case I is 0.00625A and V is 5V (7V minus 2V). So 5V divided by 0.00625A equals 800Ohms. Therefore resistor two is set at 800Ohms. We repeat this step to find out the value of resistor three. 2V divided by 0.00625A equals 320Ohms. Therefore resistor three is going to be set at 320Ohms.
This is how a voltage divider works. V1 can be used to power up low beam head lights and V2 can be used to power up park lights in a vehicle.


MOSFET   
A MOSFET is another type of transistor and it is used for amplifying or switching electronic signals.  MOSFET stands for Metal Oxide Semiconductor Field Effect Transistor. This transister differs quite a lot from the bipolar transistor (mentioned above). A MOSFET is voltage operated and not current. It is capable of handling much higher voltages and amperage compared to a bipolar transistor. It uses the same principle tho. It has three legs just like a bipolar transistor and they work similar to the ones of the bipolar transistor. The three legs are Gate, Drain and Source. Gate acts like the base leg. It turns on and off the flow between the drain and the source. But unlike a bipolar transistor the Gate is voltage operated. It does not need current flow to turn on. It requires voltage. The metal oxide gate electrode is electrically insulated from the main semiconductor by a thin layer of insulating material usually glass. This insulated metal gate can act like a capacitor and therefore it can be damages easily by static electricity. Since the gate is isolated there is no  current flow through he gate.

Here's a link of a youtube video that explains more about MOSFETs and how it works.
"http://www.youtube.com/watch?v=j47Yk7bJbxw&feature=related"












   
 

Post Two: Fuel Injector Circuit

The task at hand this week was to design and wire up a circuit which mimics a fuel injector circuit. Two LED's and transistors were used for this circuit. Voltage was supplied to the base of the transistor, this then turned on the connection between the collector and the emitter of the transistor. This action grounded the LED circuit and allowed the LED to turn on. The LED turning on indicates that the fuel injectors are firing.

Component List
There were not many components needed for this circuit. We had two pairs of resisters with different values for each pair. The values for these resistors were calculated (calculations are shown later in the post). We had two NPN BC547 transistors. Two LED's (measured with a multimeter to be 1.7V). Lastly we had a PCB board. All these components were soldered onto a PCB board. 

The Circuit & its Calculations

As you can see from the wiring diagram we have two separate circuits. One is the control side circuit and the other is the main LED circuit. Lets talk about the transistor. The transistor has three legs, the base leg (B), the collector leg (C) and the emitter leg (E). The transistor works like a relay. It needs a low amperage to open the gate for the high amperage flow. In this case when we supply current (amperage) to the base leg, the transistor will ground that circuit with the emitter leg, as the emitter leg is wired up on the ground side of the circuit. This will then open up the gate and connect the collector leg with the emitter leg. This will allow current to flow throw the LED as the circuit is now grounded. Now as you know different components require different amount of current to operate. Using the data sheet we found out that the saturation current flow for the base side of the transistor was 0.5mA (0.005A). The base side of the transistor circuit has a 5V power supply and a resistor. The value of the resistor had to be calculated using ohms law (R = V/I). So my calculation to find out the resistor value was 4.4V divide by 0.005A. This equaled to 880Ohms. So the resistor value is 880Ohms. The reason why I used 4.4V and not 5V is because the base side of the transistor uses up 0.6V to turn on and therefore the voltage available to the circuit is 4.4V and not 5V. Now lets find out the resistor values for the LED circuit. We were given the current required by the LED to operate. It was 20mA (0.02A). The power supply on the LED circuit was 12V. But we got to keep in mind that the LED requires 1.8V to turn on and the collector side of the transistor requires about 0.2V to turn on. This means that the voltage available to the circuit is 10V. We use the same formula as above to calculate the resistance (R = V/I). My calculation was 10V divided by 0.02A, which equaled to 500ohms. The resistor values are 500ohms.  

The Two Laws Relating To This Circuit 
Ohms Law:
"Ohms Law states that the current through a conductor between two points is directly proportional to the potential difference across the two points, and inversely proportional to the resistance between them." (Wikipedia quote) The formula for this is I = V/R. This formula can be rearranged to find any of the values as long as you know two of them. 
Kirchhoff's Law:
Kirchhoff's Law states that the voltage supplied to a circuit is used up within the circuit before it reaches back to the power source (ground). Voltage supplied = Voltage used by the circuit.


Breadboard Circuit Test
Before we were allowed to make our circuit on the PCB board we had to first prove that it actually works. That's why we first build our circuit on a breadboard to see whether it actually works. This is when i encountered my first problem. The class didn't have the exact value of the resistors i needed so therefore i had to compromise. Instead of a 500ohm resister and a 880Ohm resister, i was supplied with a 470ohm resister and a 800ohm resister. This meant i had to re-calculate my current readings. The 30ohm difference from 500ohms to 470ohms only made a difference of plus 0.1mA. This meant that my LED was now going to receive 0.021A instead of 0.02A. This was a very small change and therefore was nothing to worry about. The calculation was 10V divided by 470ohms, which equaled to 0.021 (formula used was I = V/R). The same calculations were carried out for the 880ohms resistor to the 800ohms resistor. The calculation was 4.4V divided by 800ohms, which equaled to 0.0055A. The lower resistance added 0.05mA to the current flow. This change would not make a difference as it is too small to cause any damage. Also the saturation value means what is the minimum current required to turn the base side of the transistor fully on. That means a little more current is fine and wont cause any harm to the transistor. 

PCB Board
Once we proved that our circuit works on the breadboard we were allowed to make it for real on the PCB board. The PCB board is something that we haven't used before. It has rows of holes where you components can sit into and all the holes on each row are connected together (but row to row are not connected). This is why we had to put cuts in the rows to stop it from connecting to all the components. The purpose of the PCB board is for us to use these rows and and make a circuit and not use wires. Before we could lay out our components on the PCB board we had to plan it on the computer, using a program called LochMaster. This program allows you to place your components any way you want and it then allows you to check for continuity between each component. If say my resister and LED were connected together (had continuity between them), both of them would then receive 10V power supply and the voltage wont be shared. This is why we put cuts in between the components (middle of my resister and in the middle of my LED).

After our design was checked and ticked off we were allowed to place the components on the PCB board and solder the components down. When we solder the solder has to be small and shinny. A dull soldering represents a bad connection and it wont have a long life. I encountered my second problem while soldering. I bridged two components together, creating a short circuit. This would mean that my LED's wouldn't turn on and the circuit wont work. To fix this problem was easy. I used the soldering iron to break the solder in between the two components to unbridged them. This technique was showed to me by Vijay, my tutor.

Here is the pictures of my completed task

 

This is the frontal view

This is the backside view

Problems & Reflection

My main problem in completing this task was soldering. I just couldn't do a good joint soldering. My soldering sometimes even bridged two components together creating a short circuit. This was fixed as explained above. Soldering a component can be the thin line between a good connection and a bad connection. A good connection will have little solder. It will be covering the whole leg and it will be shinny. A bad connection will be having lumps of dull solder that don't even cover the entire surface of the components leg. If given the opportunity i would improve my soldering skills by practicing on a dummy PCB board. This will improve my soldering skills and it will result into a better joint connection. A bad connection can also cause resistance in the circuit. This resistance can create restriction to current flow. This will mean that the circuit will now experience a lower current flow and the voltage available to components will also decrease by a small amount, as this voltage is now required to push the current through this bad connection. This will not make much difference to the circuit but it can corrupt the circuit readings. This is why it is vital to have good connections. 

   

     

Automotive Electronics

Resisters
Resisters are electronic components that are used in various circuits. A resistor's job is very simple. It restricts the amount of current flow in a circuit. This restriction to current flow is called resistance and the value of resistance is measured in Ohms. A resistor has no positive or negative end, and therefore it can be wired up either way in a circuit. The amount of resistance caused by a resistor is determined by the colour bands on it.
Colour bands
The first couple of bands on a resistor are the units you write down.
The second to last band is the multiplier.
The last band is the tolerance of that resistor.
E.G  Say a resistor has these bands; Red, Violet, Brown and Gold. Now following the colour code chart the units are 2 (Red), 7 (Violet), 10 (Brown, also the multiplier) and 5% (Gold). The resistance of this resistor would be 265Ohms.
Here is the resistor colour code chart

 

What we did in class was first get ourselves familiar with recording the resistance of individual resistors by using the colour code chart and by using a multimeter. The multimeter was set on Ohms. The next thing we did was we chose two resistors and put them in series (one after another). We then had to calculate the total resistance and then measure it using a multimeter. When you put two resistor in a series circuit they act like one big resistor. Their resistance are added together. This was proved by out measured resistance of the two resistors. The resistors used were an 9940Ohms resistor and a 5520Ohms resistor. The total resistance measured with a multimeter was 15,490Ohms. This is the same as adding 9940 to 5520.

The third thing we had to do was to wire up the same to resistors in parallel. In a parallel circuit the total resistance of the circuit is lower than the resistance of its lowest resister. This means that the circuit's total resistance to current flow will be less than 5520 Ohms as this is the value of the circuits lowest resister. The formula for this is RT = 1/R1 + 1/R2 + 1/RN... Using this formula we calculated a resistance of 3549Ohms. But when we measured it with a multimeter the resistance came out to be 3420Ohms. The slighly lower resistance can be caused by the tolerance in the resistors.  

Diodes  

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 our practical class we 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 we 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 our normal diode. We 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. We then measured the current flow using our multimeter. To do this you have to set eh 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. We 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.