Capacitors

25/3/2011

The first exercise i did, i had to calculate and measure the charge time of a range of different size capacitors and using different size resistors aswell.

 while doing this experiment i found out that the bigger the resistor in the circuit, the longer it takes for the capacitor to charge up. this happens because the more resistance in the circuit, the less amperage aloud to flow though the circuit which makes the capacitor charge up slower.
when i introduced a bigger capacitor, i found it took  longer to charge up because it is a larger space to fill, therefore it takes longer to fill up.
e.g. a bigger bucket will take longer to fill with water than a small bucket.

to calculate the the time you use the formula. (capacitor x resistance x 5)

1 micro farrad = 100 exp-6 x 1000 ohms x 5 = 0.5 secs

to measure the time it took to charge i used a wire on one side of my capacitor and a wire on the other side. when i connected the two wires my capacitor started charging and when i disconnected the wires my capacitor discharged.
i measured to see if my calculations were correct by attaching a oscilloscope to each side of the capacitor and gathered my data:

Each division on the y axis of the graph is 2 volts and each division of the x axis is 100 milli seconds

I recorded my first 3 graphs

This is my 100 micro farrad capacitor with my 1000 ohms resistor.

This shows it took 600 milli seconds to charge up the capacitor.

This is the 330 micro farrad capacitor with the 1000 ohm resistor




This tells us that is took around 800 milli seconds to fully charge the capacitor.

this is the 100 micro farrad capacitor with a 0.1 killer ohm resistor



In this picture you can see that the capacitor was fully charged within 100 milli seconds.

As you can see, a capacitor has 5 different charging stages.
the first stage, the capacitor gets filled up the most up to 63%.
The next stage is 85% then 95%, 98% and 99%.

 note:    the capacitor never fully gets filled 100%




Reference:

I got this image from: www.ehobbycorner.com
I also got the picture of the capacitor from

ca7science.wikispaces.com

Diodes

16/03/2011

For this exercise i was asked to identify the anode and cathode side of a diode and and LED.
Firstly i was asked to get the voltage in forward bias direction of each. To do this i simply put the prongs of my multimeter on each side of my diodes and when i got a reading i knew that it was in forward bias direction because it would of given me the reading of OL (overload)

LED = 1.7V, Diode = 0.5V

I then measured the voltage drop in reverse bias direction and my reading was

LED = OL, Diode = OL
This symbol (OL) stands for over load this means that the multimeter is not getting a reading when the diodes are in reverse bias.

You can identify the cathode side of a diode because there is a band around the closest side to the cathode and on an LED there is a flat spt on it which is the cathode (the cathode is the negative side)

For the next exercise i made a circuit on a breadboard, the components i used were a 1k ohm resistor, 1N4007 diode and a 5v voltage supply.

 



This image is of my breadboard circuit







I then calculated the current flowing through the diode. to do this i subtracted 0.7 (voltage drop from the diode) from 5 (the voltage supply), this gives me 4.3V.
I then divided that by 1000 (resistor)
My answer was 0.0043 Amps
4.3mA

I then measured the current with my multimeter and found out that my calculations were correct by making my multimeter part of the circuit by taking the positive side of the diode out of the bread board and attaching it to my negative negative prong and taking out the positive power feed and attaching it to my positive prong.
The reading i got was 4.4 milli amps


I was then asked to calculate the voltage drop across the diode and i checked the data sheet on the internet and found out it was 0.6-0.7V.
I measured the voltage drop with my multimeter and i got 0.65V, this shows me that my diode was within specifications.

For the next question i was asked to look at the data sheet given to me on the previous page and find information like:
- what is the max value of current that can flow through the given diode = 1amp
- what is the maximum value of  Vs so that the diode operates in a safe region = -55 to 150 degrees Celsius.

I then replaced the the diode with a LED and was asked to calculate the current across it.

I calculated the current by using ohms law = 3.6 m amps
To measure the current i did the same as i did to measure the current flowing through the diode. = 4.2 m amps

i saw that the light emitted briefly when i made my multimeter part of the circuit.

for my next experiment i was asked to create a circuit on a breadboard with, 2 resistors (2x 100 ohms), 1 zener diode (5V1 400mW) and a voltage supply of 12 volts.
 to measure the voltage across the zener diode i put my multimeter prongs on either side of the diode and i got the reading of 4.9 volts. this tells me that there is 4.9 volts running through the zener diode.
 when i varied the voltage supply from 12 volts up to 15 volts, the voltage across the zener diode went up to 5.15 volts.
I found out that the more volts you introduce, the more volts will push through the diode.

when i measured the zener diode in reverse polarity (Negative prong on positive side and positive prong on negative side), i got the reading 0.8 volts, i got this reading because that is the volts needed to push through the diode to make it open.

The next experiment i did, i was asked to make a circuit on a breadboard and use 1 resistor (1 K ohms), 1 zener diode (5V1 400mW) and 1 diode (1N4007). i then had to measure the voltage drop.

these are the results i collected:

10 Volt supply:                                  15 Volt supply:

zener diode - 4.6V                              zener diode - 4.81
diode - 0.63 V                                    diode - 0.67
both diodes - 5.62V                            both diodes - 5.48
resistor - 3.75V                                   resistor - 9.52V
current - 1.1A                                     current - 1.15A

this shows me that the bigger the voltage supply, the bigger the volts running through the circuit.
The zenor diode and the diode get the same amount of voltage drop because it takes about 0.6 of a volt to push through the diode and it takes around 4.7 volts to push through the zenor diode.
the rest of the voltage gets used by the resistor.
the amperage only goes up slightly beacuse everything in the circuit is the same size apart from the supply voltage.

Identifying Resistor Values

09/03/2011

Today i had six different types of resistors, i then looked at the colours on the resistors and matched them up with the numbers on the work sheet that was given to us.
e.g. brown = 1, red =2

One of my resistors was colour coded. Br, B, R and then a gold band at the end.

Br = 1
B = 0
R = 2
Gold = tolerance (5%)

Below is a picture of a 1k ohms resistor:

The first colour was brown so i put a 1 down because thats what it equals on the chart.
The second colour was 0 so i wrote it down next to the 1.
The third colour tells you how many zeros to put down, mine was red which means i put down 2 zeros.

i now have the answer 1000, because i was measuring resistance my answer is 1000 ohms

the gold band is the tolerance of how much my resistor is allowed to be out by, mine was 5%. i then measured my resistor with a multimeter and i was within specification reading 1000 ohms


i was then asked to choose to resistors measure their individual ohm resistance with a multimeter.
Resistor 1 =  98.7ohms, Resistor 2 = 27ohms

Once i did that i had to join them together in series by joining one end of each together, then calculate and measure the combined value.

for the calculated value i just added the two seperate resistances together because in a series circuit the total resistance is the individual resistors added together. Calculated = 125.7ohms
When i did the measured resistance i put the prongs of my multimeter on each end of the joined resistors  and set my multimeter to ohms and measured the resistance. Measured = 125.3

This is easy to remember and is a fast was to find out if your resistors are in spec

When i had finished that, i was asked to join the resistors together in parallel by joining both ends of the resistors together so one was on top of the other. I then had to calculate the total resistance.

For the calculated value i used the formula:

1/Rt = R1xR2 / R1+R2

      = 98.7x27 / 28.7+27

       = 2664.9 / 125.7

       = 21.2 ohms

The total resistance must always be lower than the resistance of the lowest resistor.

I then put the prongs of the multimeter on each end of the resistors and set the multimeter to ohms and measured the resistance. my answer was = 21.4ohms

i find that this equation is good for working out because i can now tell that my resistors are within specifications.



Reference:

i got my link of resistor from: www.thebestcasescenario.com

i got my colour chart from: thebestcasescenario.com