Electrical and Computer Engineering Department (ECE)
Effat University
Spring 2015
ECE391: Special Topics in Elect. & Comp. Eng.
Supervisor: Prof. Mohamed Shehata
Contents:
1- The design method and calculation of all parts in the circuit, showing the values and how they are calculated.
2- Simulation using multisim.
3- The practical circuit
Circuit #1
Components:
- Resistor: 1.2k Ohms
- Green LED
- 2 Switches
- Battery: 9V
- Copper wires
Calculations:
power supply = 9v
I(led) = 20mA
R = (9 - 2.6) / 20mA = 320 ohm
Circuit #2
Components:
- One push button
- Transistor BC336-25
- Resistors: 1.2K ohms - 100 Ohms - 120 Ohms
- Potentiometer
- Wires
- LED
- Battery: 9V
Calculations:
Ic = I(LED) = 20 mA
Vd = 2.6v
Vcc = 9v
Rc = (Vcc-Vce-Vd) / Ic
(9-2.6-0.2)/20mA = 310 ohm
Ib = Ic/B
We got B from the datasheet and calculated Ib
From the datasheet of the NPN transistor (BC336-25)
B changes based on temperature and other factors.
When:
B = 160 , Ib = 20mA /160 = 0.125mA
Rb = VCC-Vbe/ Ib
= (9-0.7)/0.125mA
= 66.45 K ohm
When:
B = 250 , Ib = 20mA /250 = 0.08mA
Rb = VCC-Vbe/ Ib
= (9-0.7)/0.08mA
=103.75 K ohm
Circuit #3
Note:
R(LDR) when it is dark = 30KΩ , when there is light = 1KΩ
Light Detector Circuit:
With the LDR at the top (R1), the brighter the light on the LDR, the higher the output voltage.
NPN transistor has VBE = 0.7v, VCC = 9v
VE = 0v.
Using voltage divider to find VB
VB = (R2 /R1+R2)*VCC
0.7 = R2/30*R1*9
R2 = 2.2KΩ
Finding IC:
IC = Is e^(VBE / VT)
Is = 10-15A, VT = 0.026v //Data sheet for 2N2222.
IC = 1.42 mA
LED has forward bias voltage of 1.8v to 2.2v and can withstand a current of 16 to 20mA.From the table the maximum current flowing through the LED = 20mA
V(LED) = 2V
VCE + VRC = 9-2 = 7V, VCE +VRC = 7V
VCE = 9/2 = 4.5V, VRC = 7-4.5 = 2.5V
RC = 2.5/1.42 = 1.8KΩ
Circuit #4
Done By:
Hedayh Menkabo S11102402
Wejdan Fatani S11102460
