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gracy
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That's what I was thinking.But take a look at this videogneill said:. That's not how parallel components behave.
What are the needed corrections?gneill said:They are taking certain liberties with circuit theory, omitting complicating details in order present a concept
The major omission is a resistance between the source voltage and the zener that would host the potential drop between them.gracy said:What are the needed corrections?
It's just what it says: variation in the line voltage. "Line voltage" typically refers to a primary voltage supply, such as as that of the power lines entering your house. It can also refer to the supply voltage presented to a circuit from some unspecified source (maybe a separate power supply).gracy said:What does line voltage variation means?I have googled it but didn't find anything.
resistance between the source voltage and the zener causes potential difference between these two ,so why load resistance RL not causes potential drop?gneill said:resistance between the source voltage and the zener that would host the potential drop between them.
It does. But that drop is the same as that across the zener (they are in parallel). It is the desired operating condition of the load.gracy said:I don't understand one thing.
resistance between the source voltage and the zener causes potential difference between these two ,so why load resistance RL not causes potential drop?
The resistor Rs in your diagram is in series with the zener//load combination. Its voltage drop depends upon the current through it. The current through it is Is in the diagram, and is comprised of Iz + IL.
The load is connected in parallel with the zener diode, so the voltage across RL is always the same as the zener voltage,how connecting resistance in series and parallel differs?
Yes, to protect the components you want to remain within their safe operating parameters. It is also good engineering practice not to waste power unnecessarily. Batteries are expensive.gracy said:In my textbook it is written
With no load connected to the circuit, the load current will be zero, ( IL = 0 ), and all the circuit current passes through the zener diode which in turn dissipates its maximum power. Also a small value of the series resistor RS will result in a greater diode current which in turn dissipates its maximum power.That's why we connect load and greater value of RS so that it's power is not dissipated.But Why we want to save zener power?
I think it is because greater current(greater than max current of zener will cause it to damage.Right?
gneill said:The resistor Rs in your diagram is in series with the zener//load combination. Its voltage drop depends upon the current through it.
In parallel ,there is a point where current divides .Current divided with a specific pattern,where there is a larger resistance current will be less,where resistance is less current will be more so that voltage across them remains same.But in series ,there is no such point where current divides so whole current flows through ,so voltage depends on current through resistance and resistance itself.Right?gneill said:But that drop is the same as that across the zener (they are in parallel).
Right. That is in essence Ohm's Law.gracy said:But in series... ...so voltage depends on current through resistance and resistance itself.Right?
Yes it is correct.gracy said:Is my post 13 completely correct?
gneill said:the "excess" voltage is dropped across Rs.
The value of "excess" voltage which is dropped across Rs is same as that tiny increase in voltage,right?gneill said:At and past the breakdown voltage very tiny increases in voltage lead to relatively massive increases in the current through the device
Yes. When the diode is conducting in breakdown mode there is some resistance; the V-I curve for a real device is not perfectly vertical there as it would be for an ideal one. It is generally a fairly small value of a few Ohms to a few tens of Ohms. Values in the teens are typical.gracy said:is there any resistance of zener diode?
It is the difference between the supply voltage and the voltage across the zener. If the current through the zener changes (say due to a change in the supply voltage), then the bulk of the change will be dropped across Rs. Rs and the zener's small resistance form a voltage divider, and Rs being the larger resistance expresses the bulk of the change.gracy said:The value of "excess" voltage which is dropped across Rs is same as that tiny increase in voltage,right?
Small changes in the voltage across the zener due to the zener's own resistance can't be avoided, but can be minimized by choosing a modest operating current for the zener.
Is this voltage (voltage of primary source when massive increase in current takes place) obtained by the formulagneill said:. At and past the breakdown voltage very tiny increases in voltage lead to relatively massive increases in the current through the device.
You should define what "total resistance" is in this case.gracy said:Is this voltage (voltage of primary source when massive increase in current takes place) obtained by the formula
high zener current (Iz)+IL(load current)*total resistance
A useful equivalent model for purposes of design and analysis is the series connection of an ideal diode, resistance, and fixed voltage supply to set the zener voltage):lightgrav said:best to not treat a zener like a resistor, because its IV curve is so non-linear.
Vsource - Vzener = (Izener + Iload) Rs
Vsource - Vzener it can be small as well as very largelightgrav said:Vsource - Vzener =
Not quite. If Iload*Rs drops the voltage below the zener voltage then the zener is cut off (stops conducting) and effectively disappears from the circuit. The load voltage becomes unregulated at that point.gracy said:When Vsource = Vzener or Vzener> Vsource
Vsource= Iload* (Rs+Rzener diode)
Right?
So,If Iload*Rs drops the voltage below the zener voltage Vsource= Iload*( Rs+RL).Right?gneill said:If Iload*Rs drops the voltage below the zener voltage then the zener is cut off (stops conducting) and effectively disappears from the circuit.
Why?When Iload*Rs drops the voltage below the zener voltage ,does zener offer a large amount of resistance so that at current dividing point no current goes in zener?Right?gneill said:If Iload*Rs drops the voltage below the zener voltage then the zener is cut off (stops conducting)
gracy said:when it is small how this equation holds?
Vsource - Vzener= (Izener + Iload) Rs
So,when Vsource - Vzener is small Iz is minimum.If we maintain This v source so that this small Vsource - Vzener remain constant,will there be constant minimum Iz?gneill said:Refer to the discussion in your video about the minimum zener current.