Diodes LED and Zener diodes

Diodes and LED
Diode works like a one way valve, let the current flow in one direction and LED is a diode that can emit light when it works.

To identify the cathode without a multimeter, there is a brand marked at cathode for diode.

(ref: http://www.markallen.com/teaching/images/electronics/diode.jpg)

Longer wire is anode and shorter wire is cathode for LED.

(ref: http://www.societyofrobots.com/images/electronics_led_diagram.png)

Measure the voltage drop of diode and LED
    LED:   voltage drop in forward biased direction: 1.774V  voltage drop in reverse biased direction: N/A
    Diode: voltage drop in forward biased direction: 0.565V  voltage drop in reverse biased direction: N/A


Exercise: For Vs=5V, R=1KOhm, D=1N4007 build in series circuit on a breadboard.
      The circuit show as blew:

The1000Ohm resistance, one 1N4007 diode in series circuit with 5V supplier.

Calculate the first value of current flowing through the diode:
 Id=Ir=Vd/R=(Vs-Vd)/R=(5V-0.7)/1000Ohm=4.3mA.
Measure the current flowing through the diode, put multimeter in series in to the circuit.
The reading is 4.8mA







The reading is close to my calculation, but not 100% matched, because I took the 0.7V as the knee voltage of the diode, maybe a little different to the fact one.








I measured the voltage drop across the diode.
The reading is 0.648V. It is not 0.7V.






Here is the datasheet of 1N4007 diode

The average value of the current that can flow through the diode is 1.0A at 75degree.


For the safety, the current through the diode should not be over 1A. If we have 1000ohm resistor in circuit, the maximum voltage drop across the resistor is V=IR=1A x 1000ohm=1000V.
Vs=Vr+Vd  The maximum Vs=1000V+0.7V=1000.7V fro safety.

Replace the diode by an LED and measure the current, the reading is 3.5mA.


Calculate the current:
I=Vr/R=(5v-1.8v)/1000ohm=3.2mA.


The voltage drop across the LED is not but close to 1.8V, it cause the calculation is different from the measurement.






Zener Diodes
1. Built the circuit show blew. two 100ohm resistor, Vs=12v, one 5v1 400mW zener diode

    I build the circuit on breadboad show as below:

    Measure the voltage drop across the zener diode and the reading is 4.96V when Vs=12V.

    Vary Vs from 10V to 15V, the value of Vz is vary but still close to 5V

     Vz is vary in small range and always close to 5V, that is because the zener voltage of zener diode is 5.1V,       the circuit just let 5.1V to make the current through the zener diode in reverse direction, no matter what Vs is.So, zener diodes can be used to protect the components from the high voltage damage if build in parallel circuit.

Reverse the polarity of the zener diodes and measure the Vz.

   The reading of Vz is 0.84V, that is because the knee voltage of the zener diode is 0.84V.

2.Built the circuit show as below: one 1Kohms resistor, one 5v1 400mW zener diode, one 1N4007 diode, 

   Vs=10&15V

   I build the circuit show as below:

   

   Set the Vs= 10V and measure the V1, V2, V3, V4

V1=4.63V

V2=0.64V

V3=5.28V

V4=4.94V

   The current in the circuit is I=I4=V4/R=4.94V/1000ohm=4.94mA.

   Set Vs=15V and measure the V1 V2 V3 and V4

V1=4.79V

V2=0.67V

V3=5.47V

V4=9.54V

   The current in the circuit is I=V/R=9.54V/1000ohm=9.54mA

   In this circuit, resistor zener diode and diode in series circuit. V3=V1+v2, Vs=V3+V4. The current through any component is even. If Vs vary, V1 always is zener voltage, V2 always is the knee voltage, so just V4 changes.