True for a series regulator but not for a shunt regulator.

now im completly confused

 

I don't think I've heard a rule of thumb that the zener current should be 10x the load current, either.

Your right. I had my rules of thumb mixed up.

The zener current just needs to keep it well past the knee.

 

now im completly confused

I am not surprised. That previous comment was not meant for you. Just ignore it.

 

now im completly confused

In a shunt regulator the current through the resistor to the zener always carries the same current since the voltage across that resistor is constant, being equal to the source voltage minus the zener voltage.
The current is just divided between the zener and the load.
Thus the more load current, the less zener current, and the better the efficiency.
At low load currents, the zener is carrying most of the current and the efficiency is low, since the supply current hasn't changed.

Make sense?

 

In a shunt regulator the current through the resistor to the zener always carries the same current since the voltage across that resistor is constant, being equal to the source voltage minus the zener voltage.
The current is just divided between the zener and the load.
Thus the more load current, the less zener current, and the better the efficiency.
At low load currents, the zener is carrying most of the current and the efficiency is low, since the supply current hasn't changed.

Make sense?

yes make sense
lets say the load resistor is 500ohm and Vz=6v R=200ohm Vin=12v what is the efficiency ?

 

yes make sense
lets say the load resistor is 500ohm and Vz=6v R=200ohm Vin=12v what is the efficiency ?

First thing to consider is whether or not this is even operating as a regulator.

Remove the zener and figure out the voltage across the load without it. If that voltage is less than Vz, then the diode might as well not be there.

If it IS in regulation, then consider the following:

What is the current from the supply?

How much voltage is across that series 200 Ω resistor?

How much current is in the load?

What is the power being delivered to the load?

What is the power being delivered by the supply?

What is the efficiency?

 

Current drawn from battery is 0.03A @ 12V
Current drawn from load is 0.012A @ 6V

Efficiency = ( 6V x 0.012A ) / (12V x 0.03A) x 100 = 20%

 

Current drawn from battery is 0.03A @ 12V
Current drawn from load is 0.012A @ 6V
Efficiency = ( 6V x 0.012A ) / (12V x 0.03A) x 100 = 20%

when i change the resistor of the load to 20 ohm IL=0.3A that more than the input current
why the current of the load is more than the input ? can you explain to me please

 

why the current of the load is more than the input ? can you explain to me please

Look at the numbers carefully.

Current from battery is 0.03A = 30mA
Current through the load is 0.012A = 12mA

 

Look at the numbers carefully.

Current from battery is 0.03A = 30mA
Current through the load is 0.012A = 12mA

when i changed the resistor load to 20ohm the current is 0,3A

 

when i changed the resistor load to 20ohm the current is 0,3A

That is what you think. But the voltage across the load is no longer 6V.

The max current from the battery is now 12V/(200Ω + 20Ω) = 0.055A = 55mA

You cannot get 0,3A with the 200Ω resistor in series.

 

That is what you think. But the voltage across the load is no longer 6V.

The max current from the battery is now 12V/(200Ω + 20Ω) = 0.055A = 55mA

You cannot get 0,3A with the 200Ω resistor in series.

oh ok thx

 

now i did understand how diode zener work as voltage regulator
i want to know linear voltage regulator how it work

 

now i did understand how diode zener work as voltage regulator
i want to know linear voltage regulator how it work

Look what I found when I googled for "linear voltage regulator block diagram":

 

Look what I found when I googled for "linear voltage regulator block diagram":

Diode zener still used in linear regulator that mean there will be current loss like in my first circuit i posted

 

Diode zener still used in linear regulator that mean there will be current loss like in my first circuit i posted

Yeah, but it's only going to be on the order of 10mA.

 

If r2=500ohm and r1=200ohm vin=12 vz=6v does this circuit have the same efficiency as the simple diode zener circuit ?

 

If r2=500ohm and r1=200ohm vin=12 vz=6v does this circuit have the same efficiency as the simple diode zener circuit ?

I believe we've given you enough information so you should be able to figure that out yourself.

 

when i changed the resistor load to 20ohm the current is 0,3A

What was the first thing that I said?

First thing to consider is whether or not this is even operating as a regulator.

Remove the zener and figure out the voltage across the load without it. If that voltage is less than Vz, then the diode might as well not be there.

With a 200 Ω series resistor and 20 Ω load resistor, the voltage across the load is (12 V)(20 Ω / 220 Ω) = 1.09 V. Since this is less than the Vz of 6 V, the circuit is NOT in regulation and the zener diode might as well not be there. So the current through the load is 12 V / 220 Ω = 54.5 mA.

What is the largest value that the series pass resistor can be and have the circuit in regulation? Hint: This means no load (zero load current) and just the 100 mA minimum zener current I stipulated in the problem?