Set up the following curcuit on a bread board. Use a 470R for Rc and a BC547 NPN transistor.
The 5 resistors I will use at Rb are 47k, 220k, 270k, 330k, and 1MΩ
Rb 47kΩ Vbe 0.68v Vce 91.2mv Ib 90μA Ic 6.55mA
Rb 220kΩ Vbe 0.66v Vce 0.7v Ib 20.4μA Ic 5.32mA
Rb 270kΩ Vbe 0.69v Vce 1.16v Ib 16.7μA Ic 4.40mA
Rb 330kΩ Vbe 0.69v Vce 1.53v Ib 13.7μA Ic 3.62mA
Rb 1MΩ Vbe 0.68v Vce 2.7v Ib 4.5μA Ic 1.21mA
Discuss what happened for Vce during this experiment. What change took place, and what caused the change?
The Vce went from very low 91.2mv up to 2.7 v which is half the supply voltage. At Vce 2.7v Ib and Ic were very low meaning transistor is in cut off region. As the Ib increased so did the Ic but the Vce was decreasing until Vce 91.2mv Ib 90μA Ic 6.55mA this was the highest base current and highest collector current so this is when the transistor was fully switched on or saturated. Between Vce 91.2mv and 2.7v the transistor was in active region.
Discuss what happened to Vbe during this experiment. What change took place if any, and what caused the change?
The Vbe stayed the same within a couple of milli volts because the base and emitter act as a diode in forward bias which used 0.7 volts
Discuss what happend for Ib during this experiment. What change took place, and what caused the change?
With the lowest resistor 47kΩ the base current was high so transistor was saturated. As the resistor size increased the Ib decreased switching the transistor through the active region untill it got to cut off region.
Discuss what happened for Ic during this experiment. What change took place, and what caused this change?
As Rb increased collector current decreased. The base current had a direct effect on the Ic so as Ib decreased so did Ic.
Plot the points for Ic and Vce on the graph below to create a load line.
Calculate the Beta (Hfe) of this transitor using the above graph.
Beta = Ic/Ib
Vce at 1.8 = Ic 3.2mA and Ib 37μA so β = 0.0032/0.000037 = β 86.4
Explain what the load line graph is telling you. Discuss the regions of the graph where the transistor is saturated, cut off, or in the active area.
The load line goes up from the highest Vce voltage where Ib will be zero and goes up to the highest collector current where Vce volts will be zero, and gives you a place to calculate the Beta at different Vce.
The saturated region is at top of the load line where collect current is highest and Vce is 0 to 0.7 this is where the transistor is fully switched on.
The cutt off region is where Ic is below 1mA and Vce is 2.7v this is where transistor is switched off, so inbetween the cut off and saturated region is the active region which is used for amplification.
