The common emiter amplification
For the given common emitter transistor amplification circuit, connect the oscilloscope at the in, at the out and at the base ofthe transistor. For the following base resistors values a) $R=470 KOmega$, b) $R=220 KOmega$, c) $R=47 KOmega$ d) $R=27 kOmega$ please comment on the voltage value and phase of the transistor base, and on the out voltage. When is the transistor amplification switched on?
Just switched on
All the measured waveforms are in the graphs at the bottom. The measuring points are specified in the schematics. The explanations follow.
The transitors is switched on but only for a very small time of conduction (see the graphs below). To switch it on completely you need to supply a $Vbe=0.7 V$ all times by using a base divider (between Vcc and GND) and also to assure that the base current is greater than a certain value.
Until the BE junction opens (0.7 V) , the voltage on the base will rise. When the $Vbe=0.7 V$ the base voltage will be capped at this value.
The maximum current going into the base is $Ib = (1-0.7)/470 k =0.3/470 k = 0.64 mu A$
As a result, the maximum collector current will be $Ic=beta*Ib= 64 mu A$
Hence, the maximum voltage on the Rc is $V =Rc*Ic = 64 mu* 1k = 64 mV$.
Therefore altough the transitor is switched on there is no amplification.
The maximum colector voltage (output voltage) is $Vc =Vcc-Rc*Ic = 15-64 mV =14.936 V$
Increasing base current
When you lower the base resitance to 220 K you only increase the maximum available current into the base. The conduction time (time when base voltage is higher than 0.7 V is the same).
$Ib =(1-0.7)/220 k =1.36 mu A$
The maximum collector current is $Ic= beta*Ib =136 mu A$
The maximum voltage on collector resistance is $V= Ic*Rc =136 mu*1 k = 136 mV$
The voltage on the load resitor has increased but still there is no amplification
Transistor Switched on
$R(base) =47 k$
The waveforms remains the same. This time the maximum voltage swing on the collector resitor and on the output (Vce) is even higher.
$Ib =(1-0.7)/47 k=6.38 mu A$
$Ic =beta*Ib =100*6.38 mu =638 mu A$
On the load Rc the voltage is $V=Ic*Rc =638 mu*1000 =638 mV$
On the output (Vce) the voltage is $Vc = Vcc-Ic*Rc =15-0.638 =14.362 V$
Comments: The transitor is at the threshold of amplifying. For some transitors that have
$beta =200-300$ the amplification has already started.
Rbase =27 k
Same waveforms. This time the voltage swing on the collector resitors is greater than the input voltage. The transitor is clearly amplifying.
$Ib = (1-0.7)/ 27 k =11.1 mu A$
$Ic = 100*Ib = 1.11 mA$
On the Rc the voltage swing is $V= Ic*Rc =1.11 m*1 k =1.11 V$
On the output the voltage is $Vc =Vcc-Ic*Rc = 15-1.11 =13.89 V$
Comments: The transitor is clearly amplifying (but still for only a part of the time of the input wave). If you further reduce the Rb, at some moment the voltage swing in collector will be so big, that the transitor will become saturated. This happens when output voltage is $Vc = 0.2 V$
or voltage on Rc is $V =Ic*Rc =15-0.2 =14.8 V$
At this moment no matter if you reduce further the value of the base resitor, the amplification will become capped.
f) The voltage on the load resitor (collector resitor) is in phase with the input voltage. However the ouput voltage (voltage between collector and ground) out of phase with half wavelength (pi radians).