EXPERMENT NO. 7 Transistor as aseitch

Componints; 1 * Small signal NPN transistor, 2 resistors.
Exercise; Connect the circuit  Vs 15V, R1= 10K ohms, R2 = 1K ohms

I had build this circuit and used 1 * 1K ohms, 1 * 10K ohms transistor, 1 * PNP(C557) transistor and 15Vs.
Connect the multimeter between base and emitter.
Note the voltage reading reading and explain what this reading is indicating.

The voltage drop between base and emitter in my test 0.78V.
The voltage drop across the base to emitter junction is also called in datasheet Vbe, this also for silicon transistor 0.6V to 0.7V is the valve of the voltage drop.
If we calcullate the current that will flow in the base must subtract this valve from the total voltage.
for example: We have in this circiut 15Vs the voltage across the R1 = 14.2V (15Vs - 14.2 - 0.7=0
Vcc=IC*Rc+Vce or Vcc- Ic * Rc - Vce =0.

Connect the meter beteween collector and emitter .
Note the voltage reading and explain wwhat this reading is indicating

The voltage drop between collector and emiter is 55.8 mv.
The voltage drop between collector and emetter even is saturated means not perfect conducter. This voltage is also call Vce (sat). BJT two transistor connect back to back . If the transistor saturated voltge .55v in this circuit the current that mean the current will fully switch on Vs - Vce / R.

In the plot given below what are the regions indicated by the arrows A &B?

Arrow A is SATURATION region in this region emitter and collector are in forward biased , which is the Vce is will be zero and the supplay volatage will be across the load means ON .

Arrow B is represiting CUTOFF region the emitter and collector junction are be in reverse bisssed , and this region the current Ib is zero and the collector current will not fllowing. means (OFF)
Cutoff and saturation regions the miniumum power dissipation of transistor are :
Furmulla: PD=Vce * Ic  in both regions Vce are =0 then the Pd = 0 if we flow  the ohms law P=VI.

What is the power dissipated by the transistor at Vce of 3V?

P= IV
Pd = Vce * Ic
Pd = 3Vce * 14Ic = 42mW( 4.2W)

What is the Beta of this transistor at VCe 2, 3 and 4 Volts?

B= IC / Ib
Vce2, B= Ic/Ib   20Ic/0.8Ib = B25
Vce3, B = Ic/Ib   14Ic/0.5Ib = B28
Vce4, B = Ic/Ib    5Ic/0.2Ib = B25