Activity 1
Voltage Divider Bias Configuration
Name: Reggie Boy R. Sebarios
Course: BSEE-5
Subject:
Objectives:
 To calculate and to show the schematic diagram.
 To analyze the voltage-divider bias configuration.
 To get the actual voltage of the VCE and IC.
Materials:
AC Power Supply (0-20v)
Multi-tester
BC547
Wires
Schematic Diagram:
Results & Calculations:
Icsat = Vcc / Rc+RE
Rc+RE = Vcc / Icsat
= 20v / 10mA
= 2000 ohms
VE = 1/10 * Vcc
= 1/10 * 20v
= 2v
@ CE Loop where Ic = IE
Vcc – IcRc – VCE – IERE = 0
IcQ = Vcc – VCE / Rc+RE
= 20v – 15v / 2000ohms
= 2.5 mA
VE = IERE
RE = VE / IE
= 2v / 2.5 mA
= 800 ohms
Rc = 2000 ohms – 800 ohms
= 1.2 k.ohms
VBE = VB – VE
VB = VBE + VE
= 0.7v + 2v
= 2.7 v
BRE = 10R2 where B = 294.3
R2 = BRE / 10
= 294.3 * 800 ohms
= 23.54 k.ohms
VB = Vcc R2 / R1 + R2
R1 = Vcc R2 / VB - R2
= 20v * 23.524v / 2.7v – 23.54kohms
= 150.83k.ohms
Actual Results:
R1 :
R2 :
RC:
VCE:
R E:
IC:
Conclusion:
I conclude that by merely changing the value of the resistor the base current can be
adjusted to the desired value. And by using the beta (B) relationship, IC can also be found out
accordingly. Hence the Q point can be adjusted just by changing the value of the resistor
connected to the base.