Thursday, July 29, 2010
Experiment 4
Zd=5.1 volts
D=0.7 volts
----------------At 10 volts------------------------------At 15 volts
Volt drop V1=4.68 volts_______________________4.83 volts
Volt drop V2=0.67 volts_______________________0.69 volts
Volt drop V3=5.34 volts_______________________5.52 volts
Volt drop V4=5.22 volts_______________________9.83 volts
Current amps at 10 volts Vs
I=V/R=(10-5.1-0.7)/1000=4.2 mili amps
Current amps at 15 volts Vs
I=V/R=(15-5.1-0.7)/1000=9.2 mili volts
When zener diode installed in reverse bias it takes 5.1 volts to run and let the rest through and another diode installed in forward bias, takes 0.7 volts to run and let the rest through so voltage drop across two diodes would be 5.1+0.7=5.8 volts, it doest matter how much voltage applied as they will act with the same voltage drop, until voltage would be so high it just push thorough the diodes and damage them.
Voltage drop across resistor indicates how much voltage let for it to use.
Tuesday, July 27, 2010
Experiment 3
Vz value
Vz=4.98 volts at Vs=12 volts
Vary Vs from 10V to 15V
At 10V, Vz=4.81 volts
At 15V, Vz=5.1 volts
Explanation:
In this circuit zener diode in reverse bias takes maximum 5.1 volts to work and let the rest through to negative side of the circuit, so only 5.1 volts available for the second resistor in parallel with zener diode.
Such circuit can be used for voltage regulator.
Thursday, July 22, 2010
Experiment 2
---------------------Voltage drop in forward -------------Voltage drop in reverse
LED-----------------1.761 volts---------------------------0 volts
DIODE--------------0.541 volts---------------------------0 volts
LED cathode side got shorter wire and flat side, and diode cathode side marked with grey stripe.
R=1000 ohm __________ V=5.31 v _________ Vd=0.657v
Calculated amps____________________________Measured
5.31-0.657=4.653___________________________4.6 mili amps
I=4.653/1000=4.653 mili amps
Its a voltage drop of 0.657 volts at diode as it use this voltage to run.
Calculated voltage drop--------------------------Measured voltage drop
Va=5.31-----------------------------------------0.664 volts
I=4.653 ma
R=1000 ohm
V=IxR=4.563*1000=4.563 volts
Vd=Va-V=5.31-4.653=0.657 volts
Max value of current that can flow through diode
1 amp at 75 C degrees
For R=1000 ohms, what is max Vs
Vs=1amp*1000ohms=1000 volts
Now diode replaced with LED
Calculated current amps-----------------------------Mesured current amps
Vs=5.31 volts----------------------------------------3.3 mili amps
Vd=1.967
V=Vs-Vd=3.343
I=V/R=3.343/1000=3.343 mili amps
LED light voltage drop was hiher then diode, and as it takes more voltage to run, then less amps flow through LED.
Wednesday, July 21, 2010
Experiment 1
orange, orange, black, black, red 330 ohm 2% ______329 ohm ________323.4 - 336.6 ohm
grey, red, brown, gold 821 ohm 5% _______________810 ohm _________780 - 862 ohm
grey, red, green, gold 8200 Kohm 5%_____________8295 kohm________7790 - 8610 kohm
orange, orange, green, gold 3300 kohm 3%________3300 kohm________3135 - 3465 kohm
orange, orange, black, orange, brown 330 kohm 1%__331 kohm_______326.7 - 333.3 kohm
brown, black, red, gold 1000 kohm 5%___________993 kohm_________950 - 1050 kohm
Resistor 1 ---26.6 ohm Resistor 2 --- 329 ohm
Calculated value 1&2 in series R1=27 ohm, R2=330 ohm______total R1+R2=357 ohm
Mesured value 1&2 in series 356 ohm
Calculated value 1&2 in parallel 25 ohm 1/R1+1/R2=1/Rt
Mesured value 1&2 in parallel 24.6 ohm
Explanation:
In series circuits total resistance adds together, but in parallel circuits total resistance lower then lowest resistor.