Parts UsedLED 5MM YLW 35000MCD
Emitted Colour: Yellow
Vf (V) Max : 2.6
if mA : 75
if mA (peak) 100
Vf (V) Typ: 2.2
LED 5MM WTE 35000MCD
Emitted Colour: White
Vf (V) Max: 2.6
if mA : 75
if ma (peak) 100
Vf (V) Typ: 2.2
R1 x2
470R +/- 5% 0.5 Watts
R2 x2
R2 x2
47R +/- 5% 0.5 Watts
BC547 NPN Type Transistors x2
Jumper Leads 2cm x2
How it works:
Connected to a Vs of 12 across the Positive lead and Negative Lead, It appears that this circuit does not work. That is because an additional power supply of +0.7 V needs to be connected to Vbe . The additional power supply is needed because with no voltage across Vbe, the transistor switch will not close. With the additional supply added, The user can switch what light to turn on by alternating the positive lead on the inputs.
Fault Finding:
In order to make sure my components would not be destroyed, i needed to calculate a suitable Resistor to restrict the current flow. With a Vs of 12 and 25mA flowing through the LED's, I could then use Ohms Law to calculate the Resistance needed to prevent the LED's blowing up.
But I also had to take into acount the Voltage Drop across Vce, which is around .5V. So 11.5/0.025= 460R. The closest resistor I could get to 460 was 470, so I decided to use a 470R.
Then I had to calculate the resistor I would need for the base of the transistor. As there is a 0.7 voltage drop across Vbe, I had to subtract this from 5V. My first mistake was when I divided 4.3 by the amount of current flowing through Vbc, which was 25mA, when infact I needed to divide 4.3 from the current gain of the transistor. I needed to obtain the Beta by loooking online for the datasheet. The current gain of the BC547 transistor had a range of 110-800. I used the minimum of 110 and divided this from the 4.3V supply and got 39. I had the option of using either a 33 or 47 Ohms resistor and I chose to use the 47 Ohms. After constructing my circuit I noticed my white LED would turn on when connected to the positive and negative supply, which it shouldn't because the transistor is preventing current to flow through it until a second power supply has been added to the circuit. After doing some fault finding I realisised my transistors were faulty and needed replacing. Then I constructed on a new board due to solder being everywhere from trying to pull the transistors out, but this time I only constructed half of the circuit. This way I could test my circuit quicker, and identify faults with less components. When I connected my power supply to the positive rail and to earth and noticed the LED were not on I knew the transistor was working. Then, connecting the 5V supply to the circuit I could make the transistor turn on and let current flow from Ic-Ie. This would turn the LED on, and it did. I then constructed the other half of the circuit and got the circuit running properly.
BC547 NPN Type Transistors x2
Jumper Leads 2cm x2
How it works:
Connected to a Vs of 12 across the Positive lead and Negative Lead, It appears that this circuit does not work. That is because an additional power supply of +0.7 V needs to be connected to Vbe . The additional power supply is needed because with no voltage across Vbe, the transistor switch will not close. With the additional supply added, The user can switch what light to turn on by alternating the positive lead on the inputs.
Fault Finding:
In order to make sure my components would not be destroyed, i needed to calculate a suitable Resistor to restrict the current flow. With a Vs of 12 and 25mA flowing through the LED's, I could then use Ohms Law to calculate the Resistance needed to prevent the LED's blowing up.
But I also had to take into acount the Voltage Drop across Vce, which is around .5V. So 11.5/0.025= 460R. The closest resistor I could get to 460 was 470, so I decided to use a 470R.
Then I had to calculate the resistor I would need for the base of the transistor. As there is a 0.7 voltage drop across Vbe, I had to subtract this from 5V. My first mistake was when I divided 4.3 by the amount of current flowing through Vbc, which was 25mA, when infact I needed to divide 4.3 from the current gain of the transistor. I needed to obtain the Beta by loooking online for the datasheet. The current gain of the BC547 transistor had a range of 110-800. I used the minimum of 110 and divided this from the 4.3V supply and got 39. I had the option of using either a 33 or 47 Ohms resistor and I chose to use the 47 Ohms. After constructing my circuit I noticed my white LED would turn on when connected to the positive and negative supply, which it shouldn't because the transistor is preventing current to flow through it until a second power supply has been added to the circuit. After doing some fault finding I realisised my transistors were faulty and needed replacing. Then I constructed on a new board due to solder being everywhere from trying to pull the transistors out, but this time I only constructed half of the circuit. This way I could test my circuit quicker, and identify faults with less components. When I connected my power supply to the positive rail and to earth and noticed the LED were not on I knew the transistor was working. Then, connecting the 5V supply to the circuit I could make the transistor turn on and let current flow from Ic-Ie. This would turn the LED on, and it did. I then constructed the other half of the circuit and got the circuit running properly.
No comments:
Post a Comment