Components: 1x small signal NPN transistor, 2x resistors
working :
vce2=21ma/0.8ma 26.25 beta
vce3-14ma/0.58ma=28beta
vce4=5ma/0.2ma=25beta
beta amplificationgain differnece = ic/ib
notes":
-Using a multi meter i did a voltage drop test across be, base and emitter.
this reading at .80volts indicates the amount of voltage flowing between the two points.
-the reading 0.06v was the voltage reading across the ce, collector and emitter of the transistor
the three regions of a working transistor, active, saturated, and cutoff.
active region is the main operation zone for amplifying small siganls digtal and analog . one problem with the active region is the lose in power dissipation mailto:T@=25%20oc
there is a current gain in this region also BETA, this is the current ratio between the base and collector
saturated and cutoff regions act like a switch across the emitter and collector junctions ,when the bjt is fully saturated it is when cloised and voltage reading will be emitent, and cutoff like an open circuit having no flow .
p-v*i
3v *0.13ma=0.39watts
working :
vce2=21ma/0.8ma 26.25 beta
vce3-14ma/0.58ma=28beta
vce4=5ma/0.2ma=25beta
beta amplificationgain differnece = ic/ib
notes":
-Using a multi meter i did a voltage drop test across be, base and emitter.
this reading at .80volts indicates the amount of voltage flowing between the two points.
-the reading 0.06v was the voltage reading across the ce, collector and emitter of the transistor
the three regions of a working transistor, active, saturated, and cutoff.
active region is the main operation zone for amplifying small siganls digtal and analog . one problem with the active region is the lose in power dissipation mailto:T@=25%20oc
there is a current gain in this region also BETA, this is the current ratio between the base and collector
saturated and cutoff regions act like a switch across the emitter and collector junctions ,when the bjt is fully saturated it is when cloised and voltage reading will be emitent, and cutoff like an open circuit having no flow .
p-v*i
3v *0.13ma=0.39watts
experiment no.8
components :circuit on bread board ,using 470R for Rc and a bc547 npn transistor.
using 3 differnet resistors changing Rb doing this to find beta difference between the resistors.
RB 270k ohm Vbe 750 vce 1296 ib 15.6 mA ic 3.18 mA
RB 330K ohm vbe 753 vce 899 ib 19.6 mA ic 5.18 mA
RB220k ohm vbe 753 vce 578 ib 13mA ic 3.35 mA
1) 3.18/15.6= 4.90 beta ratio
2) 5.18/19.6=3.75 beta
3) 3.35/13=3.88 beta
Your explanation for experiment 7 needs to be checked it is very hard to understand your meaning. You need to put your units for all your readings e.g you have Vbe 750 so this is 750 volts? Also Ib is 15.6mA and Ic is 3.18 mA this is not correct as Ib should always be smaller. You need to explain your reading and come to a conclusion as well
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