diodes

I used a multimeter to identify the anode and cathode of the diode and the Led
I measured the voltage in forward biased direction
Led:1.77v
diode:0.56v
measured voltage in reverse direction
Led and diode were open because n type meterial is more negative and when connected to positive side of battery it attracts increasing the size of the completion zone,same prosec happens with p type meterial on positive side of the battery.depolition zone increases to such a size that it stops the flow of electricity
the cathode can be identified without a multimeter.On diodes there is a stripe on the cathode side and on Leds the flat edge is the cathode.It is realy important to be able to identify the anode and cathode of a diode
I built a circuit on a breadboard the circuit has a 5v power supply and a 1kOhm resistor with a diode of 1n4007.After I made my circuit i calculated the value of current 5-0.7/1000=4.3mA 5is because of the voltage suply 0.7 is the diode and 1000 is the Ohms.I measured this with a multimeter and got 4.8mA
I expected this result because my voltage was a bit high wich means my curent would also measure a bit high.The picture shown here is a picture i took of my breadbord.

I calculated the voltage drop across the diode and got a result of 0.6-0.7v and then i mesured it with my multimeter and got a result of 0.64v so the result i calculated was within range of measured with multimeter.
After i made the circuit on the breadboard with a diode I replaced the diode with a Led.Icalculated the curent and got a result of 3.6mA I worked out by using ohms law A=V/R 3.6/1000=.0036A= 3.6mA 3.6 is the volts i got and 1000 is the resistance. i mesaured the current with a multimeter and got a result of 2mA the big diferance in measured and calculated can be cause by alot of thing but in this case i think it has to do with the multimeter because it had a blown fuse on mA so i measured it with amps.
I created a circuit with 2 100Ohmresistors and a 400mWzener diode with a 12v power supply.I mesaured the Vz and got a result of 4.9V then put a 15V supply on it and got a result of 5.15V because the more voltage you ad the more gets pushed through the zener diode.this circuit could be used as a voltage regulator because of the voltage amount it carries.After this i put the zener diode in reverse polarity and got a reading of 0.8V because the negative was on the negative side of the diode and the positive on the positive side of the diode.
The 4th circuit i built on the breadbord was a 1kresistor+400mWzener diode and a 1n4007 diode and mesured it once on 10V supply other on 15V supply  got result of:
                   10V                                                15V
zenner diode:4.65V                                            4.7V
diode:0.65V                                                       0.67V
both:5.32V                                                         5.46V
resistor:5.7V                                                      9.58V
current:5.7mA                                                    9.5mA
shows that the higher the voltage supply the more voltage flows throug the diodes and resistor.