Tuesday, August 24, 2010
Monday, August 23, 2010
experiment 2
Components: 1 x diode, 1 x LED
Exercise: Using a multimeter , identify the anode and cathode of the diode and the LED
Voltage drop in forward Biased Direction. | Voltage drop in reverse biased direction | |
LED | 1.83V | 2.5V(open circuit) |
Diode | 0.65V | 2.5V(open circuit) |
Explain how you could identify the cathode without a multimeter?
Diode: There is a line on the cathode side of a common rectifier diode.
LED: Shorter side of the LED is the cathode, the taller one is anode.
Table: Data sheet of 1N4007 is as follows
Calculate first the value of the current flowing through the diode, now measure and check your answer?
Calculated: Measured:
I = Vs – voltage drop ÷ R1
I = 5 - 0 .7 ÷ 1000
I = 0.0043 Amps or 4.3 mA
Is the reading as you expected; explain why or why not?
The answer is pretty close since we got the resistance and the voltage supply and voltage drop its quiet easy to calculate and the answer is close enough.
Calculate the voltage drop across the diode, now measure and check your answer?
Calculated: 0.65 V Measure: 0.65 V
Using the data sheet
What is the maximum value of the current that can flow through the given diode?
1 Amp
For R = 1kW. What is the maximum value of Vs so that the diode operates in a safe region?
Because of the gold strip at the end of the resistor it shows +5% or -5% of the 1000v by using ohms law we got the 1000V which is V= R * I .
Max Voltage = R * I max
Vmax=1000*1
Vmax=1000 Voltes
By using the table we get the max voltage that can be put in the reverse direction before destroying the resistor.
Replace the diode by an LED & calculate the current, then measure and check your answer?
Calculated
I = Vs – VD ÷ R
I = 5 – 1.8 ÷ 1000
I = 0.0032 A or 3.2 mA
Measured
3.1 mA
What do you observe? Explain briefly
The current changes slightly due to the different voltage drop of the components. When we have the LED connected we have around 3.2 mA in the circuit but when the diode has less voltage drop then the current is a little bit higher as shown below.
I = Vs –VD ÷ R
I = 5- 0.7 ÷ 1000
I =0.0043 A or 4.3 mA
Experiment 1:
Calculating and measuring resistors:
Value (Calculating colour codes) | Measuring(Multimeter) |
Brown- Black- Red- Gold 1 - 0 - * - 100 - ±5% = 1000W(1kW) | 997W |
Brown- Black- Brown- Gold 1 - 0 - * - 10 - ±5% = 100W | 098.1W |
Brown – green – Orange - Gold 1 - 5 - * - 1000 - ±5% = 10000W(10k) | 9.98kW |
Brown – Black – Yellow - Gold 1- 0 - * - 10000 - ±5% = 100000W(100k) | 99.5kW |
Orange – White – Brown - Gold 3 - 9 - * - 10 - ±5% = 390W | 380.5W |
Yellow – Violet – Black – Black - Brown 4 - 7 - 0 - * - 1 ±1% = 470W | 467W |
Choice of two resistors and their individual resistance values measured with a multimeter:
Resistor 1: 380.5 Ω Resistor 2: 98.1 Ω
Put these two resistors together in Series (end to end, one right after another) calculate and then measure their combined value
Calculated value 1 and 2 in series: 378.6 Ω
Measured value 1 and 2 in series: 482 Ω
Put these two resistors together in Parallel (connect both ends when they are side-by-side). Calculate and then measure their combined value. Show working:
Calculated value 1 and 2 in Parallel: 79.59 Ω
Measured value 1 and 2 in Parallel: 78.7 Ω
What principal of electricity have you demonstrated with this? Explain:
This principal is the Ohm’s Law:
Series: Two resistors connected in series, their combined resistance is equal to each resistance added together.
Parallel: Two resistors connected in parallel, their combined resistance is less than the smallest resistor.
Monday, August 2, 2010
Parallel
and the second formula is :
Resistors in Series
