Transistor

Transistor just like combination of 2 diodes share one end.NPN transistor share the anode, and PNP transistor share the cathode.

Identify the Transistor

Here is the transistor, there are 3 wires of it. Base, collector and emitter.




identify the wire of transistor by multimeter

Touch any 2 wires by multimeter and check the reading
Transistor BC547

0.732V if middle wire is + and left wire is -

0.725V if the middle wire is + and right wire is -









Because transistor just like 2 diodes share one terminal, there are reading when the middle wire is +, left and right wires are -, there is two diodes share + terminal. So BC547 is NPN transistor,and the middle wire is base. Vbe>Vbc, so left wire is emitter and right wire is collector.

Transistor BC557

0.73V if middle wire is - and right wire is +

0.735V if middle wire is - and left wire is +










There are reading when the middle wire is -, left and right wires are +, there is two diodes share - terminal. So BC557 is PNP transistor,and the middle wire is base. Vbe>Vbc, so left wire is emitter and right wire is collector.

Diode test:
NPN
Vbe=0.732V Veb=N/A Vbc=0.725V Vcb=N/A Vce=N/A Vec=N/A
PNP
Vbe=N/A Veb=0.735V Vbc=N/A Vcb=0.730V Vce=N/A Vec=N/A


Built the circuit on breadboad and measure Vbe and Vce

I build the circuir show as below

Measure Vbe=0.791V
Vbe is the voltage between base and emitter. It means the transistor is active.

Measure Vce=0.055V
Vce is voltage between collector and emitter.
This reading means the transistor is saturated.

Saturated:
If the transistor in saturated region, that means it fully works. The Vbe is high enough to fully open active the transistor, the current flow through collector to emitter has no resistance, so Vce is very low.
Cut off
If the transistor in cut off region, Vbe is too low to active the transistor, the current can not flow through collector to emitter. Between collector and emitter, transistor like a insulator, Vce is OL.
Active:
The region between cut off and saturated is active. Vbe is in the range between cut off and saturate. The current can flow through collector to emitter but still have some resistance, Vce has some reading.

If Vce=3V,  P=VI. more Ib, the transistor dissipate more power.
i.e. Vce=3V Ib=0.2mA, then Ic=5.5mA. P=3V x 0.0055A=0.0165W
     Vce=3V Ib=0.4mA, then Ic=10mA. P=3V x 0.01A=0.03W

When Vce=2V, Ib=0.2mA, Then Ic=5mA. Beta=Ic/Ib=25
When Vce=3V, Ib=0.4mA, Then Ic=10mA. Beta=Ic/Ib=25
When Vce=4V, Ib=0.6mA, Then Ic=15mA. Beta=Ic/Ib=25
Beta the current gain is constant, can not be changed for same transistor.


Set up the following circuit on a breadboad . Use a 470R for Rc and BC547 NPN transistor.

The breadboad circuit shows below

Measure:
Rb=47KOhm
Vbe=0.712V Vce=0.091V Ib=90uA Ic=6.1mA
Rb=220KOhm
Vbe=0.688V Vce=0.549V Ib=19.3uA Ic=5.2mA
Rb=270KOhm
Vbe=0.682V Vce=0.996V Ib=15.8uA Ic=4.26mA
Rb=330KOhm
Vbe=0.675V Vce=1.372V Ib=12.9uA Ic=3.51mA
Rb=1MOhm
Vbe=0.643V Vce=2.52V Ib=4.3uA Ic=1.2mA

In this circuit. Vbb and Vcc are always 5V. V=IR. Vbb=5V. Increase Rc, the total resistance in base circuit is increased, so the current Ib decreased, and Vbe decrease. Current gain for the transistor increased when Ib decrease, if the transistor in saturated, Ic will be the maximum, and beta=Ic/Ib, when increase Ib, Beta decrease, and if the transistor in cut off. decrease Ib, beta will increase. When increase the current gain In collector circuit. Ic decreased. Vcc=5V and Rc is 470Ohm. That means, the voltage drop across Rc is decreased. Vcc=Vc+Vce. Vc decreased, so Vce increased.

Ib=90uA Beta=Ic/Ib=6.1mA/90uA=67.8
Ib=19.3uA Beta=Ic/Ib=5.2mA/19.3uA=269.4
Ib=15.8uA Beta=Ic/Ib=4.26mA/15.8uA=269.6
Ib=12.9uA Beta=Ic/Ib=3.51mA/12.9uA=272.1
Ib=4.3uA Beta=Ic/Ib=1.2mA/4.3uA=279.1

When Ib decreased, current gain beta increased slowly.
The load line shows if Vce drops to 0, transistor has the maximum Ic and if Ic drops to 0, then Vce goes to maximum