EXPERMENT NO.8

. Use a Set up the following circuit on bred board470 R for Rc and a BC547 transistor.

In this circuit i used five resistors for Rb. 1M, 47K, 220K, 270K & 330K ohms. Why we doing that , because change the risitors we see when the transistor saturated switch rigion and when is active amplifier region. Also i had record voltage drop across Vce & Vbe & recorded the current for Ic & Ib.
Every change the risitor for Rb i have measure & recorded in below chart.

Rb 270K    Vbe: 0.66V       Vce: 1.73V         Ib: 0.02mA        Ic: 0.425mA
Rb 47K      Vbe: 0.67V       Vce: 0.72V         Ib: 0.10mA        Ic: 4.38mA
Rb 2.2K      Vbe: 0.70V       Vce: 0.16V        Ib: 1.97mA        Ic: 4.46mA
Rb 1M        Vbe: 0.63V       Vce: 2.25V        Ib: 0.01mA        Ic: 0.02mA
Rb 330K     Vbe: 0.66V       Vce: 0.25V        Ib: 0.02mA        Ic: 4.13mA

Let see some of my experiment pictures, I had take of every measarment that,s too much just afew i will put in this blogg.

In this picture voltage drop across Vbe, for Rb 270K.

Mesurment of current Ic, Rb 47K ohms in circuit.

   In this picture measuring current flow Ib for Rb 47K.

Measuring Voltage drop a cross Vce for Rb 1M in the circuit.

    

In this picture measuring current of Ic, Rb 330K ohms in the circuit.

Now discussing what happed for Vce during expeeriment & what caused of change.
In this circuit i had used diffrent resistor the big risistor on Rb Voltage a cross Vce is high small risistor voltage drop Vce is low. We can see when high voltge drop on Vce this cauuse we have put big risstor on Base that way high risstnce on base means less current flowing on base (Ib) and Ic is also low current. In this stage the transitor in cutt off or active region not saturatioin region, because not enough current for Ib to switch on of transisor. The small risistor voltage drop across Vce is low and the current Ib is high and also the curret of Ic is also high in this the transistor is the saturarated region.
If we see on record on top the smallest risistor i used in this circuit in Rb is 2.2K ohms the Vce is 0.16V , Ib 1.97mA & Ic 4.46mA that is the transistor is saturated region fully swith on , because more current flowing on base. The voltage drop across Vce from below 0.3V transistor is in saturaed region and above 2.0V transistor is in active region.

Discuss what happened for Vbe during this experiment . What change took place, if any and what caused the change?
The voltage drop in Vbe in this circuit not changed in any diffrent risistor, because transistors are designed two diode back to back voltge drop Vbe sammelar like normal diode.Also risistor only control the current not voltage taht way changing the risistor only change the current flow Ib & Ic voltage drop Vbe is same.
We allways in our calculation voltage drop Vbe is 0.7V.
For example: if we want to find the valve of risitor Rb incircuit we used this formulla.
Rb= Vb - Vbe / Ib  (Rb= 5v -0.7Vbe /Ib).

Discuss what happened for Ib duuring in this experiment. What took place, and what caused the change     
During this experiment current Ib is changing the cause of cahnging Ib is risistor high risstance on base less current flow to base to emitter and Ic is also low. The big risistor on circuit voltage drop a cross the Vce is high means high voltage dorop on Vce that way we can get less current on base, Also in less current or not enough current the transistor is cutt off rigion or active region. Small risistor current Ib is high and vce is low transistor is saturated.

Discuss what happened for Ic during this experement. what change took place , and what caused the change?
Ic current flow to collector is also related to risistor, high risitance more current . In this circuit i have noticed high voltage drop across vce get less curent Ic . When i used a small ristor on base we can get more current in Ib .When Vce is less the Ib is high .If we lock the plot the Vce geting smaller Ic is going up and also the Ib and transitor is saturated . You can see difrence of Ic in ristor in my measarment rrecod .

Calculate the Beta (Hfe) of this transistor.
B= IC/Ib     B= 0.425/Ib 0.02mA= 21.25  OR (Ib= IC/Ib  0.425mA/B21.25= Ib 0.02mA
                 B= IC4.38mA/Ib0.10mA= B 43.8   OR (Ic4.38mA/B43.8= Ib 0.10)
                 B= Ic4.46mA/ Ib 1.97= B 22
                 B= Ic0.002A / Ib 001A = B 20
                 B =Ic 0.o413A / Ib 0.002A = B 20.65

Explain what the load line graph is telling you.. Discuss the region of graph where the transistor is Saturated, cut off, or in the Active area.

The load line the graph is telling cut off and saturated region. the end of the load from Ic to is the large supplay voltage , if the saturated area when the base current has incresed The emmeter base junction is forward biase infact if the base current it can cause Ic current flow to increase.
In saturation the transistor
Volt 0- 0.2V
Vce<=0.2V tis knew the saturate volt or Vce(sat)
Ib >0.and Ic >0
Vbe>= 0.7V
Cutt off region the transisor inaactive. In the cutt off the following behave is noted
Ib = 0 (no base current)
Ic = 0 (no collecor emmeter )
Vbe = 0.7V (emetter , base junction is not forward)
Active region the transistor can act as fairly liner amplipir in this region
0.2V>Vce<Vce
Ib>0 and Ic>0
Vbc> = 0.7V
The transistor in ON and  the collector to emetter voltage is some where between the cutt off and saturated stage. this the transistor is able to amplyfy smalll voltage present to the base the act is extracted if the collector in forward active stage the collector current is proprtional to the base current by a constant multiplir Beta (Ic= B* Ib).
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