Circuit One (Injector Circuit)

Circuit One (Bipolar Junction Transistor Circuit)

From the data sheet:

20 mA : LED
1.8V : LED (Voltage Drop)
NPN :  547 Transistor
Ic :  100mA
Ib :  5mA
Vce : 0.2V
Vbb = 5V
Beta = 100


Calculation

Saturation
Rb = Vce - Vd be / Ib
= 5V - 0.9V / 0.005 = 4.1 / 0.005 = 820 Ohms

Active Current Gain
Ib = Ic / B
= 0.02 / 100 = 0.0002A
Rc = Vcc - VD - VD LED / Ic
= 12V - 0.2 - 1.8 = 10 / 0.02 = 500 Ohms
Rb = Vbb - VD / Ib
= 5 - 0.9 / 0.2 = 20.5 K Ohms

The Voltage Drop Reading around the Circuit is Shown in the Figure Above

The circuit illustrated above is a Bipolar Junction Transistor circuit which is consisted of two LEDs in parallel, 4 resistors and 2 transistors which we will be focusing on its function. BJT has 3 terminal electronic devices constructed of dope semiconductor materials which is used in switching application, this BJT circuit needs only 5V to switch and a 20mA lamp which means that 5V will go through base to emitter to open the gates for Vce which is a higher voltage to cause the signal or the LED to flash and this process takes place in active region of transistor, Normally transistors has 3 regions.

1. Active Region: where power dissipation is very high.
2. Saturation Region: with both junctions forward base BJT saturation mode facilitates high current conduction from emitter to the collector which is a closed switch.
3. Cut off Region: this is the opposite to saturation, both junctions are reverse base, there is very little current flow which corresponds to an open switch.

The easy and more efficient way to find three regions and calculate the current gain is the graph of Vce vs Ic for different levels of Ib and the common Emitter circuit B which is the ratio between collector current and base and representing Beta = Ic / Ib

The two pulsing terminals which are only 5 volts going through base to Emitter (Vbe) and this will opens the gates for the collector to emitter (Vce) the path way for the current to flow and cause the LED to flash, this flash of LED depends on the frequency of pulses from the oscilloscope, the higher the frequency of pulses the faster the LED flashes.

Test

Supplying 12V to the circuit with two pulsing terminals connected to 5V to open the gates for Vce.

Checked the available voltage as illustrated in the figure below:

Problems

The only problem which i have came across while making this circuit was that i have placed the resistors on the board on straight line which were connected to Vs.

After finding the problem i have resolved the issue.

Reflection

Making the circuit was not a difficult task, however i have still made few minor mistakes which i have learned from those mistakes and i am confident that i would be able to make a much better circuit in future.