Potential Difference Across Sensitive Instrument: 7.5V

[gracy]

Homework Statement

:What will be potential difference across the sensitive instrument?[/B]

Homework Equations

:

[/B]

The Attempt at a Solution

:I think the potential difference across the sensitive instrument would be 7.5v.[/B]

 

[gneill]

Is the diagram correct and complete? It shows that all the components are connected in parallel, yet states that the potential difference across the zener diode differs from that of the supply voltage. That's not how parallel components behave.

 

[gracy]

. That's not how parallel components behave.

That's what I was thinking.But take a look at this video

from time 1:22 to 1:39

 

[gneill]

They are taking certain liberties with circuit theory, omitting complicating details in order present a concept. In the context of way the material is presented your answer is fine.

 

[gracy]

They are taking certain liberties with circuit theory, omitting complicating details in order present a concept

What are the needed corrections?

 

[gneill]

What are the needed corrections?

The major omission is a resistance between the source voltage and the zener that would host the potential drop between them.

 

[gracy]

What does line voltage variation means?I have googled it but didn't find anything.

 

[gneill]

What does line voltage variation means?I have googled it but didn't find anything.

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]

I don't understand one thing.

resistance between the source voltage and the zener that would host the potential drop between them.

resistance between the source voltage and the zener causes potential difference between these two ,so why load resistance RL not causes potential drop?

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?

 

[gracy]

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]

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?

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.

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?

The resistor R_s 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 I_z + I_L.

 

[gneill]

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?

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]

The resistor Rs in your diagram is in series with the zener//load combination. Its voltage drop depends upon the current through it.

But that drop is the same as that across the zener (they are in parallel).

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?
I tend to express my views about concepts and then say Right?because I think it is useful and wise to take corrections from experts on your concepts than growing up with wrong concepts.

 

[gneill]

But in series... ...so voltage depends on current through resistance and resistance itself.Right?

Right. That is in essence Ohm's Law.

 

[gracy]

Is my post 13 completely correct?

 

[gneill]

Is my post 13 completely correct?

Yes it is correct.

 

[gracy]

Can you please explain how zener maintain constant voltage?I have gone through many sites but no one explains this.

 

[gneill]

The video you linked to showed the I-V characteristic curve of the zener. It is the heart of the explanation. The voltage across the zener simply cannot go (much) beyond its reverse breakdown voltage. At and past the breakdown voltage very tiny increases in voltage lead to relatively massive increases in the current through the device. That increased current passes through the Rs of your previous diagram and causes a potential drop there. So the voltage across the zener remains steady and the "excess" voltage is dropped across Rs.

 

[gracy]

is there any resistance of zener diode?

 

[gracy]

the "excess" voltage is dropped across Rs.

At and past the breakdown voltage very tiny increases in voltage lead to relatively massive increases in the current through the device

The value of "excess" voltage which is dropped across Rs is same as that tiny increase in voltage,right?

 

[gneill]

is there any resistance of zener diode?

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.

The value of "excess" voltage which is dropped across Rs is same as that tiny increase in voltage,right?

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.

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.

 

[gracy]

. At and past the breakdown voltage very tiny increases in voltage lead to relatively massive increases in the current through the device.

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[/QUOTE]

 

[lightgrav]

best to not treat a zener like a resistor, because its IV curve is so non-linear.
V_source - V_zener = (I_zener + I_load) R_s

 

[gneill]

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

You should define what "total resistance" is in this case.

Note that the "massive increase in *zener* current" I referred to takes place if a voltage greater than the zener breakdown voltage is impressed across the zener. This behavior can be read right off the V-I characteristic curve.

The resistor Rs is in place to moderate such currents when the source voltage rises. Remember I said that Rs and the zener resistance form a voltage divider that greatly reduces the magnitude of the voltage change that appears across the zener compared to a change in the source voltage.

 

[gneill]

best to not treat a zener like a resistor, because its IV curve is so non-linear.
V_source - V_zener = (I_zener + I_load) R_s

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):

 

[gracy]

Vsource - Vzener =

Vsource - Vzener it can be small as well as very large
when it is small how this equation holds?
Vsource - Vzener= (Izener + Iload) Rs[/QUOTE]

 

[gneill]

Does something make you believe it wouldn't hold?

The formula will hold so long as Izener is sufficient to hold the zener past its breakdown "knee". Refer to the discussion in your video about the minimum zener current.

 

[gracy]

When Vsource = Vzener or Vzener> Vsource
Vsource= Iload* (Rs+Rzener diode)
Right?

 

[gneill]

When Vsource = Vzener or Vzener> Vsource
Vsource= Iload* (Rs+Rzener diode)
Right?

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.

Vsource being equal to Vzener is not practical as there will be some voltage drop across Rs due to the load drawing current through it, even if the zener is cut off. Again the zener would be ineffective as a regulator.

A correct operating point for the zener is to maintain a current through it that exceeds its "knee" current by a comfortable margin when the source voltage is at its lowest anticipated value and the load drawing its maximum current. This is why you'll often see source voltages specified as a nominal value with a +/- percentage associated with it. It tells you the minimum and maximum values to expect during operation.

 

[gracy]

If Iload*Rs drops the voltage below the zener voltage then the zener is cut off (stops conducting) and effectively disappears from the circuit.

So,If Iload*Rs drops the voltage below the zener voltage Vsource= Iload*( Rs+RL).Right?

 

[Bystander]

Yes.

 

[gracy]

If Iload*Rs drops the voltage below the zener voltage then the zener is cut off (stops conducting)

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?

 

[Bystander]

Yes

 

[gracy]

One more question
I had asked you a question

when it is small how this equation holds?
Vsource - Vzener= (Izener + Iload) Rs

Refer to the discussion in your video about the minimum zener current.

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?

 

[Bystander]

Yes, it looks as if you're now the forum zener expert.

 

[gracy]

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?

Yes

So ,where do all resistance go when breakdown voltage is achieved?

 

[Bystander]

It doesn't "go anywhere." This is a characteristic of this type of component, a very non-linear current-voltage curve, and at least two different conduction mechanisms that produce/explain it, or result in zener behavior, neither of which I can explain to myself right at the moment, let alone to someone learning the topic.

 

[gracy]

Take a look at this video from time 1:32 to 2:50

in forward bias why diode behaves as short circuit?And in reverse bias why it behaves as open circuit?

 

[Bystander]

The circled "~" at the left side of the schematic indicates that an AC signal is being applied to the circuit; for the first half of the cycle, the voltage is positive, and for the second half it is negative.

 

[gracy]

The circled "~" at the left side of the schematic indicates that an AC signal is being applied to the circuit; for the first half of the cycle, the voltage is positive, and for the second half it is negative.

Yes,but in forward bias why diode behaves as short circuit?And in reverse bias why it behaves as open circuit?

 

[Bystander]

It's a regular diode, and when forward biased it passes current with very little resistance. I can't get the video to reset to show me what the load is, but there is a complete circuit, and the "lower" conductor in the circuit diagram is arbitrarily assigned a potential of zero volts. The upper conductor is alternating between positive and negative potentials/emfs.

 

[gracy]

In normal diode forward bias there is a lot more current than in reverse bias.That's why in video forward bias is regarded s short circuit and reverse bias as open circuit.

 

[Bystander]

Yes.

 

[gracy]

What is special about zener diode?I mean if we will connect any other load in place of zener, the current will not drop the voltage across Rs by same amount by which the voltage has increased so that the voltage across that load remain constant?

 

[gneill]

The special thing about the zener diode is its I-V characteristic curve. This curve results from exploiting certain features of semiconductor physics, and in particular, how a semiconductor junction that is reverse biased can break down by way of certain modes of "failure". Careful engineering of the chemistry of the materials and structure of the junction encourage and "tune" these features to particular "knee" voltages and make the "failure" survivable for the component.

In theory you could replace a zener diode with a handful of other parts that mimic its behavior (an equivalent circuit model). But having a single inexpensive component that delivers this functionality is very practical.

 

[gracy]

I got really good marks in my class test of zener diode.Thanks gneill ,bystander and lightgrav.Can you please explain what is current gain in transistors?

 

[gneill]

I got really good marks in my class test of zener diode.Thanks gneill ,bystander and lightgrav.

You're welcome! Glad to hear you did well on your test.

Can you please explain what is current gain in transistors?

That's a new topic, so you need to start a new thread . Be sure to include the current understanding you have from your class and reading your text.

 

[gracy]

That's a new topic, so you need to start a new thread

Ok.But please help me in the new thread also.