Board1 Injector Simulator.

 Image. Wiring diagram.

Red LED and  Green LED were chosen for this circuit. According to the LED datasheet forward current for RED LED 10ma<I<30mA
GREEN LED 10mA<I<25mA
Let'assume that working current for both LED's is 20mA.

Vdrled- voltage drop for Red LED is 1.785 V
 Vdgrled-voltage drop for Green LED is 1.755 V
 Transistor BC547A is chosen for this circuit. From this transistor datasheet:
  Vbe=700 mV
  Vce=90 mV
  Thus: R14=(Vcc-Vdrled-Vce)/Ic=(12V-1.785V-0.09V)/0.02A=506 Ohms
From resistor value table choose the resistor with 5% tolerance. This is E24 series.
Therefore, R14=510 Ohms
R15=(Vcc-Vdgrled-Vce)/Ic=(12V-1.755V-0.09V)/0.02A=507.75 Ohms
For the same reason as it is mentioned above in R14 calculation, choose resistor 510 Ohms > R15=510 Ohms
At the next stage let's calculate Ib=Ic/b, where b=Hfe- current gain
b varies from 110 to 800 for this type of transistors. To be on the safe part of the road choose b=110. That means transistor will work under minimum conditions. If b is higher the transistor will work for sure.
Ib=Tc/b=0.02A/110=0.18 mA
However, returning back to the datasheet, we can see Ib=0.18 mA is not enough to get the transistor hard saturated. According to the datasheet for BC547A Ib must be 5mA.
Thus, R13=R16=(Vpwm-Vbe)/Ib=(5V-0.7V)/0.005 A= 860 Ohms
Choose R13=R16=910 Ohms from the resistor value table
Limiting resistors which were calculated will provide saturation mode.
The results are :  R13=R16=910 Ohms + - 5%
                          R14=R15=510 Ohms + - 5%
                          T1=T2= BC547A


Board for this circuit is on the photo :
This is a working board examle. Soldering has revealed that power supply wire was covered by oxide . Ideally, all these wire must have been treated with flux in order to remove oxidation before soldering.
Other components were soldered with common bar solder.
The measurements on the diagramm show volatge drop accross each device.
It is obvious that voltage drop accross current limiting resistors provides accord to the calculation above with reference to the components tolerance. Both transistors are saturated. That's why for the transistor T1 Vbe=0.817V and Vce=0.041V. For the transistor T2 Vbe=0.818V and Vce=0.042V. Voltage drop accross LED1=2.129V and LED2=2.136V what allows them to glow brightly.

 

Reflection: In this particular case the elements layout is a bit irrational. If I could rebuild it I would do it in more compact way. There is no high demand for heat exchange because the transistors work in the saturation mode. 

Practical Workbook Experiment #6 BJP TRANSISTORS

It's clear from all measurements above how to identify the transistor's electrodes.Vbe is always higher than Vbc. Once you pay attention what is the polarity of multimeter lead you attach to the base and collector or emitter you can easily find what type is the transistor. For NPN type if positive lead is attached to the base and negative to collector or emitter you have Vbe and Vce reading. For PNP transistor picture will be opposite.Negative lead is attached to the base and positive to collector or emitter. Thus, you can check if the transistor is working and also identify its type and mark the electrodes.

 Vbe shows that transistor base to emitter junction voltage drop slightly exceeds so-called 'knee voltage'.
Vce is that small because the transistor is fully opened.

In the A-region transistor is saturated. That means it's fully opened and Ic reaches its maximum while Vce is in minimum.
In the B-region, which is called cut off, transistor is closed.
To calculate power dissipation we can draw  lines to the load line on the graph. Then, we can find the Pd by multiplying Vce and Ic.
Beta that refer as Hfe as well, can be find from this graph also.
Hfe=Ic/Ib. The same technique of drawing lines to the vertical and horizontal axes is used to determine the Ic and Ib.

In Experiment No 8
we changed the Rb and checked what were changes for Vbe, Vce, Ib, Ic.

Rb is the limiting resistor in this case. By increasing Rb, we limit Ib from 90uA to 10uA. The lower Ib the closer transistor to the cut off region. That's why Vce rises during this experiment. We know that Ic is beta times Ib. So that, it goes down. To simplify, we experimented how transistor can be controlled by Ib.

The graph that is drawn above on the basis of the measurements from this experiment allow us to see how all these transistor's characteristics are interrelated. Another advantage of this graph is that it makes easier to understand how transistor change its working mode.The load line also allows to model amplifier and to see input and output signals relation.

Practical Workbook. Experiments on resistors, diodes and capacitors.

Experiment1
 Resistor Colour Code helps to identify resistors; and several simple circuits in series and parallel connection refreshed the Ohm's Law application. I've found that even funny proverb helps to keep in mind row of numbers and write down tha table of resistors value . It's easy to remember and can be usful.

From the table above which is based on the resistor value calculations and experimental measurements I concluded that only one resistor failed because of poor quality.


Experiment2-Experiment4
All our experiments on diodes from the practical workbook showed that voltage drop across diodes is essential feature of this electronics components.Applying this for the calculation from the data sheet or measuring this in particular circuit and studying diagram we can read the whole picture about what is happening with electrical signal at the moment.
Another fundamental rule which is known as  Kirchhoff's current law we applied in Experiment 3.
Along with Kirchhoff's voltage law and voltage drop these are three the most offten applicable rules for circuit reading in electrical and electronics engeenering.








R= (Vs-Vz)/Is; Is=Iz+Il  where Iz from the data sheet. Il=Vz/Rl.




V1=Vz; V2=Vknee(0.7v or 0.6 v depending on the diode properties) and V4=Vs-Vz-V2.


Experiment5 The experiment on the capacitors not only reminded us that this device is used for timing but also demonstrated how the charging time can be changed. The smaller resistor rating the faster capacitor will be charged. Once the capacitor was changed onto the bigger one the charging time increased. The patterns on the oscilloscope screen demonstrate how the charging process goes. It's easier to see 5 stages as well as percentages of charge. The hihger charge the longer it takes because it's harder to move electrons.
Depending on the capasitors construction charge can be stored either for a long time or for a while due to leakage.
R*C*5=T where R is resistance in Ohms, C is capacitance in Farads and T is time in seconds.

experiment1