Zener diode help (self-learning)

[pokaymahn]

sorry just found out this is the wrong thread

 

[yungman]

What is your question?

 

[pokaymahn]

Homework Statement

Hello, I'm trying to teach myself the basics of EE to kind of get an idea of what i'd do in college. I'm finding this question particularly troubling. Is this due to my low intelligence, or is it normal?

I've attached the photo of the problem. It is asking to find the power rating that the zener diode should have. Also, I included the answer that the book provides; the most confusing this is that it says the assume a zener current of .5 amps which doesn't make sense to me because I don't know how they came up with .5 amps. If any more info is needed let me know, thanks!

Homework Equations

P=IV V=IR

The Attempt at a Solution

I know that the zener diode must get the same voltage as the lamp because it is in parallel so it has 15V. For the resistor on top, I figured that it must have a voltage drop of 9 to 45 volts. However, even with this I am unable to find the current or resistance through the resistor. This is where I get stuck. The solution in gray says to assume a zener current of .5 amps, but I have no idea where this number comes from. Please help!

 

[Studiot]

Don't know where you got this question but it is ridiculous.

The 15 volt zener needs to draw about 0.02A to be considered 'on'.

Additionally it needs to be able to cope with the lamp current, when the lamp is off or broken.

This is stated to be 0.075 amps

So the zener needs a standing current of 0.02 + 0.075 = 0.1 amps.

Power = volts x amps = 0.1 x 15 =1.5 watts say 2 watts for safety.

The idea of a zener is that it is (as you say) connected in parallel with the load (lamp) and it adjusts its own current draw so that the result is always 15 volts across it.

 

[pokaymahn]

Thanks for the reply studiot. The answer says 42 watts so I am confused, why do you say it is ridiculous?

 

[Studiot]

It's wrong.

It's wrong to choose 0.5 amps, that's way too much for the job.

But let us say that we did choose this standing current.

Then the wattage dissipated by the zener = 0.5 x 15 = 7.5 watts, not 42.

 

[pokaymahn]

I am incredibly confused, do you have any external resources I could use to learn more about zener diodes?

 

[Studiot]

Do you understand what a potential (voltage) divider is?

 

[pokaymahn]

Not quite sure... I'm fairly new to all this.

 

[berkeman]

I am incredibly confused, do you have any external resources I could use to learn more about zener diodes?

http://en.wikipedia.org/wiki/Zener_diode

.

 

[Studiot]

OK this is very important to learn as it is the basis of many circuit arrangements.

A voltage divider divides the input voltage in a defined ratio.

It is very simple - at its simplest it is just two resistors in series with the input voltage connected across them.

In my sketch I have shown this with an input of 1 volt the drop across the larger resistor is 90% of the voltage ie 0.9 volts.

In your circuit that is what is used.
Your input resistor is my R1
The parallel combination of the zener and the lamp is my R2

The object is to waste as little power as possible and the only component we really want to supply is the lamp.

Are you with me so far?

 

[pokaymahn]

It is very simple - at its simplest it is just two resistors in series with the input voltage connected across them.

Why are the resistors in series? How is that going to divide voltage?

 

[Studiot]

What happens to to the voltage if you place it across one single resistor?

 

[pokaymahn]

It drops while current stays the same.

 

[Studiot]

Not really. I think you need to revise the basics.

Remember this is for circuit theory, not physics. The physics is confusing at first until you have got hold of simple circuit theory.

The voltage is what I measure with a voltmeter.

Let us say we have a 12 volt auto battery.

If I connect it across a 12 ohm resistor 1 amp flows

If I connect it across a 24 ohm resistor 0.5 amps flow

If I connect it across a 48 ohm resistor 0.25 amps flow

If I connect it across a 1 ohm resistor 12 amps flow

But the voltage is always 12 volts.

Now what happens if my 12 ohm resistor is actually two 6 ohms resistors in series?

I have 12 volts across the series combination, but 6 volts across each resistor.

Are we doing any better?

 

[pokaymahn]

Wait what? How is that wrong? "The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage."http://www.tpub.com/neets/book1/chapter3/1-12.htm

 

[Studiot]

"The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage."

Don't you think that is a bit of a mouthful?

It's not actually wrong, but what does it mean?

What is the 'total voltage across the circuit' and how would you recognise it walking down the street?

Seriously - don't go there - that way makes a simple thing very complicated.

 

[berkeman]

Wait what? How is that wrong? "The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage."


http://www.tpub.com/neets/book1/chapter3/1-12.htm

Your link says exactly what Studiot was saying. What is the confusion?

 

[Studiot]

A voltage is what I measure with a voltmeter.

A voltmeter has two leads: you connect one to each of two different points in your circuit.

Here is a (slightly) more complicated arrangement to show what you would read on a voltmeter, connected at various locations.

 

[pokaymahn]

I thought the link was easier to understand... :(

 

[Studiot]

So how did your link lead you to think that the current remains the same but the voltage drops in your post 14?

 

[pokaymahn]

Well in a series circuit the same current is used by all the components. And all the components use a certain voltage which leaves less for the rest. Am I wrong?

 

[Studiot]

Yes that is correct.

And all the components use a certain voltage which leaves less for the rest.

This bit is another way of saying they divide the available total voltage.
Or another way of saying that they form a potential divider.

Which brings us back to post12

 

[pokaymahn]

Go on. Do you have a EE degree? How do you know this stuff?

 

[Studiot]

My first degree was applied maths, but I have been doing this for quite a few years.

Since you are self learning you might like to investigate this site

I have placed the link in vol 3 of their online textbook.
Look at chap 3 for zener diodes

http://www.allaboutcircuits.com/vol_3/index.html

I need to go to bed now but keep asking questions

 

[medwatt]

Looks like you need a good understanding of the basics first before you jump right away solving zener diode problems.
Here is a good book which is lucid and explains all the basic stuff well:

http://books.google.com/books?id=91lgEno4-H8C&printsec=frontcover&dq=electronics+bakshi&source=bl&ots=6MmTEyGq5v&sig=BuY0lKVX66RwwECd9sGD4VAdsWs&hl=en&sa=X&ei=u8ctUJPwK6nq2AXPy4CQCQ&ved=0CEsQ6AEwBA

This is from google books and you will be able to view some 200 pages before an restrictions.