Ok. Circuit without load:
I=V/R=24/470=51mA
Or do you mean figure out what Iknee and Imax are? Cause I'm lost with that too.
Vmax from what I got from the lecture is always 0.7V.
The only info I have is 470ohm resistor and 24v supply. I have been reading a number of texts today but no joy on understanding how to figure thing out.

Is any of this on the right track?

 

Without load I=(Vsource-Vz)\470 ohms. Zener diodes can drop about .7 volts when forward biased, but they work in the reverse bias condition when used as regulators.

 

Without a load, the 24V supply is dropped across the 470 ohm resistor and the zener diode (Vz = 12V). In the forward direction the diode would achieve a voltage drop of 0.7V, however, in the reverse direction it will have a voltage drop determined by the Vz.

You calculated the current as though the zener diode is not there... keep the zener diode in place. Assume the zener is regulating well. What is the voltage drop across the zener? What is the voltage drop across the resistor?

Is the calculated current through the resistor, which all goes through the diode if there is no load, enough to keep the diode regulated? Hint: Look at the graph and the previous I definitions from before...

 

ahh! That may have led me to the correct info from the book.
Iz=Vin-Vz/R=24-12/470=12/470=25.5mA
When Iload=0A, Izis maximum and equal to Icct. Therefore,
Iload(max)=Icct-Iz(min)=25.5-3=22.5mA
Given the max value of Iload, the minimum value of Rload(min)=Vz/Iz(max)
=12v/22.5mA=533.3ohms.

So, if this is correct. I have now:
Iload(min)=25.5mA
Iload(max)=22.5mA
Rload(min)=533.3ohms
I just need Iload(max) - Is that found from Vz/Iz(min)?

And why is my Iload.max smaller than Iload.min?

 

25.5 mA is not really Iload(min). It is the current drawn from the supply through the resistor and zener when Iload = 0mA.

 

ok. I need to calculate the voltage drop across the diode and the resistor. What is a voltage drop and how do I calculate it?

 

I am not introducing anything new here. You know that the supply voltage is 24V and the voltage across the zener and load is 12V. To this point you have done things correctly.

I am only reminding you that the calculation that resulted in 25.5 mA was done, and correctly so, without a load resistor. That means the load current was 0 mA. And, since the calculated current of 25.5 mA is between the 3 mA knee current and the 90 mA maximum current the diode can handle, the zener will still be rectifying correctly.

The circuit behaves correctly with very high load resistances, i.e. Iload down to 0 mA.

 

look im sorry. I'm obviously not going to be able to understand what is needed by tomorrow. Nothing is making sense for me. Can you direct me to some very basic reading on diodes and loads - things we have been discussing please? Perhaps with some applets or similar.

 

I am sure there is information about zener diodes in the AllAboutCircuit book from one of the tabs above, probably in Volume 3. I do not know that it will have any such load type calculations.

Let me also say that your calculations that you did are correct to find the lowest resistance that would allow the zener to still regulate...

I(Supply-24V)...I(R-470).....I(Zener).....I(load).....R(load)
---------------------------------------------------------------
25.5 mA...........25.5 mA.....25.5 mA.....0.............infinite (very large)
25.5 mA...........25.5 mA.....3 mA.........22.5 mA....533 ohms

If I(load) > 22.5 mA then I(Zener) < 3 mA and therefore does not regulate correctly. That is the same as saying: R(load) < 533 ohms cause the zener to stop regulating correctly.

And, there is no load that could cause this circuit to make I(zener) > 90 mA = I(zm).

 

Hi StayatHomeElectronics:
How does that tell me what the maximum Rl that can be used is?

 

It should work with any R(load) > 533 ohms.

 

I found this reading on Forward Voltage Drop

Electricity uses up a little energy pushing its way through the diode, rather like a person pushing through a door with a spring. This means that there is a small voltage across a conducting diode, it is called the forward voltage drop and is about 0.7V for all normal diodes which are made from silicon. The forward voltage drop of a diode is almost constant whatever the current passing through the diode so they have a very steep characteristic (current-voltage graph).

So, does this mean that we consider the voltage as energy? And that it takes 0.7v of 'energy' to push a current through a diode. So the voltage supply (battery) now only has its original-0.7V left for the rest of the circuit components? And this is why they call it a drop, because the original voltage has dropped in value?

 

Stayathome:
You put this chart on: How did you acquire the value of 533ohms?

I(Supply-24V)...I(R-470).....I(Zener).....I(load).....R(load)
---------------------------------------------------------------
25.5 mA...........25.5 mA.....25.5 mA.....0.............infinite (very large)
25.5 mA...........25.5 mA.....3 mA.........22.5 mA....533 ohms


If I(load) > 22.5 mA then I(Zener) < 3 mA and therefore does not regulate correctly. That is the same as saying: R(load) < 533 ohms cause the zener to stop regulating correctly.

 

I just used the number you calculated in post #24 of this topic.

 

ok. I'm gonna try to summarize everything so far.
The zener is equivalent to a variable diode which adjusts itself to find its 'zener' voltage(set by the doping). Below this there is no conduction. Below Izk (the zener's minimum current value), there is also no conduction. The limiting and load resistors in series form a voltage drop, which this minimum current must overcome across the zener voltage.
The load restricts the current through the zener, avoiding it overheating because of a too high wattage.
The current drawn from the supply, through the Resistor and zener is found by I=Vsource-Vz/R. Where Iload=0.
To find the minimum Rload, Vz/Iz. Vz and Iz being obtained when no load was attached. If the current through the zener is less than Izk, as Rload is in parallel with the zener it will receive the current the zener should have and is therefore not regulating properly.
Not sure is this mean that the zener is faulty?
This is the same as saying, if Rload is less than Rload(min), the zener is not regulating correctly. The Rload steals current from the zener but if Rload is below Rload(min) then not enough current will go through the zener to pass Iknee. Therefore, Iload needs to be above Iload(min).

Well. Hopefully I have come to some sense on this. It seems zener diodes are trickier than they first appear.
Please let me know how my conclusion sums up.

 

Here is the characteristic curve of the zener diode, it comes from Microsemi's website at http://www.microsemi.com/micnotes/200ser/201.htm

For the reverse voltage across the zener to be Vz (12V) the current must be greater than or equal to Izk (3mA). Currents larger than Izm (90mA) will physically damage the diode.

The best way to analyse the circuit is to assume the zener is regulating correctly, i.e. it has 12V across it, and find the currents through the limiting and load resistors.

I(limiting) = (24V(supply) - 12V(zener))/470ohms = 25.5 mA
I(load) = 12V(zener) / R(load)

I(limiting) is the maximum current that can be supplied by the circuit if the diode is regulating. Now, we know that the zener needs at least 3mA to maintain its voltage. Any additional current can be supplied safely to the load without disturbing the regulating zener diode. So, a maximum current of I(limiting) - 3mA = 22.5mA = I(load) can be supplied to the load without disturbing the regulating zener diode. You can think of it as the load is stealing this current from the diode.

Substitute I(load)max = 22.5 mA into the I(load) equation and we get R(load) = 533 ohms. That is the lowest resistance that the circuit can have to regulate correctly.

Without a load, 25.5mA would all flow through the diode. This is fine for the diode since the Izm was given to be 90 mA.

*** Next part not used in you example ***

If the calculated I(limiting) had been above 90mA, you would have needed to add a load to steal some of the current away just so the diode would not get damaged. This was not the case in your example though. If it had been, you would have used I(excess) = I(limiting) - 90ma to find the current needed to be drawn away by the load.

The load resistance needed to absorb the excess current would be determined by

I(excess) = 12V(zener) / R(load)

This would have given you another limit on the resistance range...

 

Wow - thanks StayatHomeElectronics.
That should all seriously help my understanding.