Zenor diode. What happens to the voltage?

[SimonTHK]

Trying to figure out the zenor diode. It lets current run through when it reach a certain voltage. The certain voltage will be used over the zener diode or will it be able to send all voltage further into the circuit.

Like 12v to the zener diode and it breakdown at 5,6v what will be after the zener diode? 12v or 12-5,6="the answear"

 

[audioguru]

A zener diode conducts when the voltage to it is at or is more than its rated voltage.
If you do not limit the current then connecting 12V to a 5.6V zener diode will burn the zener diode and maybe also burn the 12V power supply.

 

[SimonTHK]

I knew that but thanks Do you know the answear for my quistion? When it conduct, does it mean open for all 12v or only 12+5,6= 6,4v (the rest)

 

[audioguru]

When a zener diode conducts then it regulates the voltage across it to its voltage rating (5.6V) if it has some current in it.
Any extra voltage (12V - 5.6V= 6.4V) appears across the current-limiting resistor that feeds it.
If there is no load then the zener diode conducts a fairly high current which makes it and the current-limiting resistor hot and wastes power.
If the load current is too high then there is too much voltage dropped across the current-limiting resistor so the zener diode does not conduct anymore and there is no voltage regulation.

Your question does not make sense. The load is connected directly to the zener diode so of course the voltage across the load is exactly the same voltage that is across the zener diode.

 

[SimonTHK]

so if we assume that we have a resistor in series with the zener diode and we knew that 0,03 amps would run through and 6,4v over the resistor, then the resistor should be 6,4/0,03 = 213,3 ohm (well known that this resistor does not exist).

 

[SimonTHK]

or drop in a 50 ohm resistor then it would be 6,4v/50ohm = 0,128ampere

then assuming that this zener diode is limited to 0,5 W (BZX79C5V6)

then 6,4v * 0,128A = 0,81w would burn the zener diode off?

thanks in advance and yes I am studying and trying to grap the most basic things

 

[audioguru]

assuming that this zener diode is limited to 0,5 W (BZX79C5V6)

then 6,4v * 0,128A = 0,81w would burn the zener diode off?

Yes, the zener diode will fry if that circuit has no load.
If there is a load that draws 50mA then the zener diode will dissipate only (128mA - 50mA) x 6.4V= 499.2mW and the zener diode will be at its max allowed dissipation and will be extremely hot but it will survive.

 

[SimonTHK]

Yes, the zener diode will fry if that circuit has no load.
If there is a load that draws 50mA then the zener diode will dissipate only (128mA - 50mA) x 6.4V= 499.2mW and the zener diode will be at its max allowed dissipation and will be extremely hot but it will survive.

thank you very much, helps me alot