Operational amplifier (OPAMP)

What is a OPAMP
OPAMP is an electronic circuit module  made up of components like transistors, resistors and diodes all into one. Originally, OPAMPS were named because they were used to model the basic mathematical operations add, subtract,integrate and differentiate in analogue computers.

Characteristics of an Ideal OP-AMP
The ideal op-amp has:
-open-loop gain infinite,
-infinite bandwidth
-Infinite input impedances so that no current is drawn
-Zero output impedance so that maximum current is transferred to the load.
-input offset zero
-No thermal drift.




Inverting op-amp
Vin goes to (-) inverting input

Calculate Vout
If Vin =14 ,R1/Rf=15000, Rin=180
the Rf(R1) resistor is counted as a minus
formula is Vout=Vin(-Rf/Rin) 


Vout=14(-15000/180)
Vout=-1166 use 1k2



Non-Inverting op-amp 
Vin going to (+) non inverting input

Calculate Vout
Vin=27, Rf=150000,Rin=390
In a non-inverting op-amp we have to add +1 to Rf/Rin
so Vout=Vin(1+Rf/Rin)
Vout=27(1+150000/390)
Vout=10411 use 10k

Metal Oxide Field Effect Transistors (MOSFET)

Transistors has at least 3 terminalsis. Voltage/Current to one pair of the terminal changes the current flowing through the other pair of terminals thus used for amplifying or switching electronic signals. 


Metal Oxide Field Effect Transistors (MOSFET)




















MOSFET
-low voltage gain
-high current gain
-very high input impedance
-high output impedance
-low noice generation
-fast switching time
-easily damage by static!!!(USE OF AN ANTI-STRAP IS RECOMMENDED)
-some require an input to turn it "OFF"
-voltage controlled device
-more expancive than BJT

2 Types of mosfet= 'N' channel and 'P' channel

Data sheet

The Linear De-rating Factor, tells you how many watts must be reduce as the temperature increase
example. If the device was operating at 55oC, what would be the max power the device could handle?

(temp needed to operate at) - (normal operating temp) = X
X x (linear derating factor) = Y
(Power Dissipation(PD) - Y = Z <-- this is now the max PD@TC=XoC

55 oC – 25 oC = 30
30 x 0.59 = 17.7
88 – 17.7 = 70.3W – This is now the max PD @ TC = 55 oC.

Bipolar Junction Transistor (BJT)

Transistors has at least 3 terminalsis. Voltage/Current to one pair of the terminal changes the current flowing through the other pair of terminals thus used for amplifying or switching electronic signals. 


Bipolar Junction Transistor (BJT)

Not Pointing iN (NPN)  Point iN Please (PNP) 

Bipolar
-high voltage gain
-low current gain
-low input impedance
-low output impedance
-medium noise generation
-medium switching time
-robust
-requires zero input to turn it "OFF"
-current controlled device
-cheap



BJT as a switch.
To work out RB in this type of circuit

Calculate the the resistor needed for Rb
If Vcc=14, VBB = 5 volts, Rc=150,Beta= 55, Vce= .5 volts


We first need to get Ic, this is done by dividing Vsupply (Vcc) by Rc, do not forget to minus Vce 
So:Ic= (Vs-Vce)/Rc
Then we use the beta formula : Ic=Beta x Ib,
 we to transpose this formula to get Ib
Once we have Ib we can use it to calculate Rb by dividing VBB by Ib, do not forget to minus Vbe :
Rb=(VBB-Vbe)/Ib


Ic=(14-0.5)/0.09
Ic=0.09A


Ib=0.09/55
Ib=0.00163A


Rb=(5-0.7)/0.00163
Rb=2638 use 2700




Fault finding a BJT

Diode and Zeners

Diode
The function of a diode is to let electrical current flow only in one direction, in a circuit is to stop electrical current from flowing in the wrong direction. Thus diode can be put in place to protect other component from back emf happening.



IMPORTANT: As you can see in the graph below, too much forward current and the diode may be damaged and conduct in both directions. Too much reverse voltage will damage the diode and cause it to conduct in the wrong direction.

If you look closely at VF the diode doesn't start conducting until 0.6-0.7v this is known as the knee voltage, so when calculating voltage after the diode, voltage suplied(vs) - (vd) voltage diode 12-0.7=11.3v

Data Sheet infomation
what to look for so you dont blow the diode
Peak Inverse Voltage: This represents the maximum reverse voltage that can be
applied before the diode breaks down.
Maximum DC forward Current:This is the maximum recommend forward current, if the current exceed the rating it will destroy the diode.

What is a Zener diode
A zener diode is  a  special kind of diode that if voltage is apply in the reverse bias it acts as a voltage regulator, so input might be 12v but the output of the end result would be eg 9.1v if you use 9v1


zener diodes are typically available in the following voltages: 2.4, 2.7, 3, 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1, 10, 11, 12, 13, 15, 18, 20, 22..., and in 300mW, 500mW, 1.3W, 2W, 3.25W, and 5W rated packages.

How to use
To use the zener diode, the diode has to be in parallel with the zener as above in the picture



To work out Zener diode resistor

If Vs=10, Vz=5v1, Iz max= 0.06, RL=470
I=V/R
Rs=Vs min-Vz/Iz min

I=5/470=0.01063A
Rs=(10-5)/0.01063=470



Breakdown voltage
 Breakdown voltage of a zener is  the maximum reverse voltage the zener can handle until it stops working like it should. eg 9.1v  but only a small amount of current can be pass through it eg 1A a resistor is put in series to prevent this.
This is why we use a zener diode for, so that no more than eg 9.1v can be pass through in the reverse bias to protect components 

Forward bias region
This is the region voltage is passing through the zener diode in forward bias anode to cathode.this work just like a normal diode.

Knee voltage

This is the region in the forward bias before it reaches the working voltage of 0.7V.

power circuit

     Circuit diagram 

           A component list

     2x1n4007 Diode

     2x capasitor 33mF  

     1xZener diode 

     1x voltage regulator LM317

     1xR1

     1xR2

     1xR3

     1xLED
1x PC Boards Vero Type Strip

Calculations

           Calculate the values of R3,R2

           Vref=1.25 

           Vout=5v

           Use Vout=Vref (1+R3/R2)

               5 = 1.25(1+R3/R2)

       5/1.25 = 1+R3/R2

               4 = 1+R3/R2

          4 - 1 = R3/R2

               3 =  R3/R2

             R3 = 3 x R2

This mean that R3 is 3 times bigger than R2, So we were given 330R for R2 and 1K resistor for R3.

For R1: LED resistor

    R1 =   (V_S - V_L) / I

    R1 =  (5-2.1)/30

    R1 = 100 R

Lockmaster

      A Technical Explanation 
     This is a normal voltage regulation circuit. As the raw 12V is presented into the circuit, it goes through the diode (These diodes 1N4001 is to  prevent back EMF) using  0.7V so 11.3V goes into the regulator  (The regulator is a LM317T) which is then stepped down to 5V. C1 and C2 are 33 mF , the capacitors job is to  provide stability in the output voltage .The 2 zener is a 2V2 zener. . The R1 100R with the LED is a indicator to tell if there is voltage going through the circuit. R2 is 330 R and R3 is 1 K, R2 and R3 are used to determine how much voltage output will be. 

      Test Procedure 
 12V to the input  and out put should be 12v on the red wire and 5v on the yellow wire and black to ground (negative)

      Reflection 
I wouldn't do anything different if i was to build the circuit again. as my circuit worked  first time, so i dont see how i could make a different. 

   

      Finish result

Injector circuit

 Circuit diagram

A component list 

2 X 330R

2 X 1K3

2 X LED http://www.kpsec.freeuk.com/components/led.htm

2 X BC547 Transistor. http://pdf1.alldatasheet.com/datasheet-pdf/view/11551/ONSEMI/BC547.html

Calculations 

To work out what resistor to use with LED

Rs=(Vs-Vled)/IIed

Rs=(12v-2v)/0.035A (35mA)

Rs=285ohm closes to E12 group resistor is 330R

To work out what resistor to use with transistor

Beta=IC/IB

IB=0.035/110

IB=0.000318  (0.3mA)

Rb= (V-Vtrans)/I base trans

Rb=(5v-0.7v)/0.000318

Rb=13,314 ohm closes to E12 group resistor is 15K

Lochmaster

 A Technical Explanation of how the circuit works.

What this circuit does is convert a 5v E.C.U signal to 12v to power a 12v component eg injector.In other words  it uses the 5v to switch to 12v

In more details 12v is the main power supply (red + and black -lead) this 12 volt passes through the resistor ,led and onto the collector.

Negative is hooked up to black and 5v is then hooked up to the yellow or white lead this connects to the transistor base  which in turn when 5v is pass through the transistor acts like a switch and enable 12v to pass through to the emitter.

 Test Procedure

Both led should be off when hooking it up to 12v but when 5v is pass through one of the signal leads one of the led should light up.

The avalible voltage to collector

Same as Vce.

This is a good result, ideal should be 0v but this is very near. datasheet says 0.5 saturated Vce

 The avalible voltage to base

voltage at emitter

Problems

The circuit did not work the first time. This was trace to a faulty transistors. when test with a diode tester base and emitter displayed O.L both ways this indicates that there is a open circuit in the transistor.

Reflection  

If i was to do it again i would test all the components first .

Also made sure i was drilling the right spot before drilling, i auctily drilled a row across and when i went to put all my components in one of the legs had no copper connection to solder to. i solve this by bending the legs across.

Finish result

 

4841 Transistor

P= P-type (hole)             A= Anode
N= N-type (electrons)    K= Cathode


DIODE

TRANSISTOR

A transistor is a component used to amplify and/or a switch by using small electric signals.

NPN= Not Pointing In
PNP=Point iN Plz


How it works
To work as a amplifier a high voltage can be pass through the collector (input) and emitter (output) the base is the switch a voltage less than 1v can be use (but not more than 15v or else the transistor will over heat and melt) when base gets a signal, this allows voltage to pass from collector to emitter this acts as a amplifier. 
You can also get transistors that will turn off when base gets a signal these are PNP type transistor.


How to test
Tools need :multimeter on diode test 
There are six positions you should place with your multimeter probe to get a conclusion.
Like a diode, current will only go one way the leg that gives a reading with the two other leg is the base. now that you have found base next is to find emitter, keep one probe on base and with the other test the other two leg and the highest reading out of the 2 leg is the emitter.