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