Hi, i was reading up on the the zener diode on the website allaboutcircuits.com...and im confused about why is it operated in the reverse bias and not in forward bias?
(Following was given for forward bias of the diode)
Quoting from the text, "
If the power supply voltage were to be increased, the resistor's voltage drop would increase almost the same amount, and the diode's voltage drop just a little. Conversely, a decrease in power supply voltage would result in an almost equal decrease in resistor voltage drop, with just a little decrease in diode voltage drop. In a word, we could summarize this behavior by saying that the diode isregulating the voltage drop at approximately 0.7 volts."

Does this mean that suppose if 0.8 v is supplied to the circuit, then voltage drop across the zener will be 0.7v and across resistor will be0.1v? If so, then suppose 0.6 v was supplied. Then how will the voltage drop be distributed across the zener and the resistor?
Thanks in advance.

 

If you don't apply more than the break-over voltage of the diode, it won't conduct, so current through the diode becomes zero. (Assuming I guessed the circuit correctly.)

 

First off, 0.7V is the approximate forward voltage drop of a zener (and most other silicon diodes too). A zener is designed to break down in the reverse direction at a higher voltage, say 5V or 12V for example.

Give it a voltage the same of below the breakdown (or zener) voltage and it does not conduct (excepting a tiny current). Bring the voltage above breakdown and current will flow to approximately match the input voltage so about the same voltage will be across the zener.

 

This is like saying, "Why is it that when I don't put enough water in the bucket for it to overflow, it doesn't overflow."

 

Find out, look up or draw the current-voltage characteristics of a Zener diode, in both forward and reverse biased directions.

Now apply it to a circuit with a voltage supply and resistor load.
How does one solve for the operating current and voltage?
The simple graphical solution is to draw a load line and find the intersection of the I-V curve and the load line.

Now alter the supply voltage and observe how much this affects the voltage across the diode.

Which of the two configurations causes the diode voltage to change the least, forward biased or reversed biased?

 

Hi, i was reading up on the the zener diode on the website allaboutcircuits.com...and im confused about why is it operated in the reverse bias and not in forward bias?
(Following was given for forward bias of the diode)
Quoting from the text, "
If the power supply voltage were to be increased, the resistor's voltage drop would increase almost the same amount, and the diode's voltage drop just a little. Conversely, a decrease in power supply voltage would result in an almost equal decrease in resistor voltage drop, with just a little decrease in diode voltage drop. In a word, we could summarize this behavior by saying that the diode isregulating the voltage drop at approximately 0.7 volts."

Does this mean that suppose if 0.8 v is supplied to the circuit, then voltage drop across the zener will be 0.7v and across resistor will be0.1v? If so, then suppose 0.6 v was supplied. Then how will the voltage drop be distributed across the zener and the resistor?
Thanks in advance.

There is a very small range of voltage stabilizer diodes that operate in the forward conduction mode for less than about 3V, they were sometimes used for bias stabilizers in class-A/B output stages - most other voltage reference diodes operate on controlled breakdown effect when reverse biased.

I vaguely remember reading in a text book that up to 7V its called the zener effect - above 7V its called the avalanche effect, simplifying the explanation; the stray electrons that are the leakage current when reverse biased, tumble through the crystal lattice and dislodge more electrons from the structure - these in turn knock yet more electrons loose and any increase in voltage causes an avalanche of leakage electrons.

 

Hi, i was reading up on the the zener diode on the website allaboutcircuits.com...and im confused about why is it operated in the reverse bias and not in forward bias?
(Following was given for forward bias of the diode)
Quoting from the text, "
If the power supply voltage were to be increased, the resistor's voltage drop would increase almost the same amount, and the diode's voltage drop just a little. Conversely, a decrease in power supply voltage would result in an almost equal decrease in resistor voltage drop, with just a little decrease in diode voltage drop. In a word, we could summarize this behavior by saying that the diode isregulating the voltage drop at approximately 0.7 volts."

Does this mean that suppose if 0.8 v is supplied to the circuit, then voltage drop across the zener will be 0.7v and across resistor will be0.1v? If so, then suppose 0.6 v was supplied. Then how will the voltage drop be distributed across the zener and the resistor?
Thanks in advance.

If you will apply voltage lees than required by the diode to its excitation that is threshold voltage then the diode will not conduct and hence circuit will not coplete. hence there will be no flow of current.