circuit boards

CIRCUIT BOARD 1:

For my first circuit board i was asked to create a compound circuit using 2 NPN type transistors, 4 resistors, a 12V supply, 2 LED's and two 5V supplies to open transistors. this circuit is a fuel injector circuit design with a earth trigger.

I then had to calculate the size of the resistors i needed:

for R14 and R15:   V=IxR
12-1.7=10.3
10.3-0.5=9.8
I then knew that the LED needed 0.2amps to push through it
9.8/0.02=490ohms

i worked out each one separately because they each have their own LED and their own transistor, they are also in parallel which means that they each have an even 12 volts.

For R13 and R16: Ic=BxIb
I checked the data sheet for Beta and found it was 100
I needed to find the amps at the collector so i then divided the amps at the collector by 100 (beta)
Ib=Ic/B
Ib=0.02/100 = 0.0002
i then used V=IxR
i started with 5 volt supply but then needed to minus 0.6V to open diode.
5-0.6 = 4.4
4.4/0.0002 = 22000   

i then checked the data sheet and found out that i needed 0.005amps to push through the transistor
4.4/0.005 =  880 ohms

they both need to be worked out separately because they both have their own transistor and they both have different 5 volt supplies.

Next i had to design my own circuit on the computer with my own layout, this is what i designed:

Once i had planned it out on the computer, i then went ahead and made it on a real circuit board.
 The idea of this circuit is to be used in a fuel injection system.
when this circuit was made and hooked up to the battery supplies (one 12volt and its earth and two 5volt supplies). the LED's were not meant to turn on while the 12 volt supply was on, but when i turn on one of the 5volt supplies aswell it would open the transistor and allow the 12volt supply to go to earth which made the LED go on. The LED's can go on separately depending on when the 5 volt supplies for each LED was on.

below is a video of my circuit working:

 

 When i tried my red bulb it was not working. i then performed a voltage test and realised that i was getting a high voltage at the emitter leg of my transistor, i then looked on the back of it and then soldering had a poor connection and was not ground to earth properly, i then re-soldered the leg and my LED lit up.

CIRCUIT BOARD 2:

In this circuit i had to create a circuit using a voltage regulator, this simulates a ECU output if you wanted to have a steady 5 volt output.

It consists of:
3 resistors (R1 = 165 ohms, R2 = 240 ohms, R3 = 720 ohms)
2 diodes
1 LED
1 zener diode
2 electrolytic capacitors
1 voltage regulator
a 12 volt supply and 5 volt output line.

I had to calculate the size of resistors needed for my circuit:

R1= 5 - 1.7 = 3.3
 3.3/0.02 = 165
165 ohms


R2 = i checked on the data sheet for the resistance and found out it was 240
240ohms

R3 = R2x3 = (R3/R2)xR2

3xR2 = R3
3x240 =   720
720 ohms

Once i had calculated the size of the resistors, i then made a layout of my board on the computer:

note: i forgot to add the diode bridging across the input and output of the voltage regulator but i put it in when i made the actual board.

Once i finished the layout on the computer i went ahead and made my circuit on a real board.

The idea of this circuit is to have a 12 volt out put from a battery supply. The 12 volt supply goes through a diode so it has 0.6 of a volt and then comes out at 11.4 volts it then goes through the input leg of the voltage regulator and comes out at the output leg of the voltage regulator at 5 volts, it then goes through the R2 which is bridged to the adjust leg and loses around 2 volts and then goes through R3 and loses the rest of the voltage but leaves roughly 1.2 milli volts and goes to earth.
The output leg also goes strait to and output which reads a constant 5 volts.



CIRCUIT BOARD 3


This circuit is designed to work as a oxygen sensor. when you turn the dial on the sensor the 3 LED's should turn on and off and different times.

This circuit consists of:
1x four opamp chip
7x resistors (3x 1k ohms, 1x 10k ohms, 2x 330 ohms, 470 ohms)
3x LED's (red, yellow, green)
2x electrolytic capacitors 
1x zener diode
3x diodes
1x input sensor
1x 12volt supply 

When i got my circuit layout i had to calculate my 7 resistor sizes:

R2 =  V = I x R
12-0.6 =  11.4
11.4-1.8 = 9.6
9.6/0.0095 = 1010
1010ohms

R3 = V = I x R
12-0.6 = 11.4
11.4-0.6 = 10.8
10.8-1.8 = 9
9/0.0095 =947 ohms

R4 =  V= I x R
12-0.6 =  11.411.4-1.8 = 9.6
9.6/0.0095 = 1010
1010ohms

R5 = V = I x R
12-0.6 = 11.4
11.4-9.1 =   2.3
2.3/0.0056 = 410 ohms


R6 = 10k ohms

R7 =  V = I x R
0.23-0 = 0.23
0.23/0.0008 = 287 ohms

R8 =  V = I x R
 9.1-.63 = 8.47
8.47/10000 = 0.0008
0.63-0.23 = 0.4
0.4/0.0008 =  500 ohms

I then made a basic outline of my circuit on the computer:

Once i had made that outline on the computer, i then proceeded to make the real circuit on my board.
I was unable to get my circuit going properly. I was only able to make my green LED go on and stay on.
I think this was because i had bad bridge wires going from my 12 input to 9, 6 and 3 because i did a resistance test and i had OL on my multimeter which means they were not connected together so it was not having an input to my red or yellow LED's. If they were connected properly i would've got a reading but it would have been very low resistance.


I also had an extremely low voltage (2.3mV) at my number 5 input, when i should of been getting 0.63volts at this point. I also think this i due too poor connections so it was not getting the proper voltage that it should of,.

Transistors

To begin with, i was asked to check if it is a PNP (positive, negative, positive) type or a NPN ( negative, positive, negative) type.
I was already told that the base of the transistor was the middle and then. i then set my multimeter to diode check so i could check the resistance and then placed my positive prong in the middle leg of the transistor (base) and my negative prong on one of the outer legs(emitter or collector) , i got low voltage which tells me that i had the polarity around the right way, this means that my transistor was a PNP type. If i had my prongs of my multimeter around the other way i would of got a reading of OL (over load). If OL was my reading this would indicate that it is a NPN type transistor.

Below is a graph showing my resistances when my multimeter is in diode test mode. 

Transistor number	Vbe	Veb	Vbc	Vcb	Vce	Vec
NPN	0.719	OL	0.722	OL	OL	OL
PNP	OL	0.722	OL	0.719	OL	OL

If i wanted to check which of the outer legs was the emitter and which one was the collector, one of the outer legs of the transistor would have a slightly higher resistance, this leg will be the emitter. If you do not have multimeter you can always check the information sheet of the transistor on the internet.

For the next experiment i was asked to make a compound  circuit on a breadboard using a NPN transistor and 2 resistors (1000ohms, 10000ohms) and a 15 volt supply.

I was calculated the voltage between the base and the emitter of the transistor. my reading was 0.781V because it take about 0.7V to push through a diode.
I then measured the voltage between the base and the collector and got 0.051V, this was a very small voltage because the resistors are hooked up in a series circuit and this means that all of the voltage gets used, this is why i got a very minimal voltage at the collector.

In the graph below i was then asked to explain it:

the cut off region (B) means the transistor is fully off and not working and not voltage is crossing though it.
the saturated region (A) means the transistor is fully open and active. there is also around 0.5 voltage drop across the collector and emitter.

the power dissipated by the transistor at Vce is:
Pd = Vce x Ic
3 x 13 = 39
Pd = 39W

to work out the Beta of the transistor at Vce 2,3 and 4 volts:
B = Ic/Ib

2 volts:
20/0.8 =  25

3 volts:
15/0.6 = 25

4 volts:
5/0.2 = 25

Beta = 25


For experiment number 8 i was asked to make a circuit on a breadboard using a LED (light emitting diode), a 470ohm resistor, a NPN(negative, positive, negative) type transistor and another resistor which i was asked to change after every test to get different readings.

Rb	Vbe	vce	Ib	Ic
47k ohms	0.67V	0.12V	10mA	4.0mA
220k ohms	0.68V	0.53V	10mA	50.27mA
270k ohms	0.67V	0.99V	10mA	53.13mA
330k ohms	0.67V	1.36V	10mA	78.98mA
1M ohm	0.63V	2.50V	10mA	99.83mA

the voltage drop across the base and emitter stayed  the same because it take roughly 0.6 volts to push the diode in the transistor. the voltage drop across the collector and emitter was low to begin with but slowly increased as the resistor got bigger, this happened because there was more resistance in the circuit which leaves more voltage for the other consumers in the circuit. the amps at the base got stayed the same because the resistor stayed the same and the base is a different of a different circuit to the emitter and collector so the change in resistor wont effect the current at the base.  The current at the collector got higher as the resistor got bigger, this happened because the larger the resistance, the more force needed to push though the circuit. 

I was then asked to plot the points for amps at the collector and the voltage drop across collector and emitter:

The load line in the graph is telling us that when the volts across the collector and the emitter is at a certain voltage, if you look up to the load line, The amps at the collector will be at a certain amperage and then on the opposite side of the graph it will show the amps at the base aswell.




 References:
I got the picture of the transistor from www.tradekorea.com 

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