Discussion in 'Homework Help' started by kimfung100, Oct 9, 2013.

1. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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I do not know how to find the value of the series resistor(Rs) because there are many variable. It is hard to decide which current(knee/maximum) or voltage should I select . Can some one teach me??

Question:
Given that the zener voltage Vz = 8.2V, with the condition that the zener current 75mA ≤ Iz ≤ 1A and the load resistance RL = 9Ω. Calculate resistance Rs so that load voltage VL is regulated to 8.2V while the input voltage Vin varies by ±10% from its nominal value of 12V. Also check that the zener current Iz is larger than 75mA at lowest input voltage Vin .

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2. ### studiot AAC Fanatic!

Nov 9, 2007
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What have you worked out so far?

For instance what is the load current?

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3. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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I only know the load current = 8.2/9 = 0.9111111A and the highest/ lowest input
voltage(13.2V/10.8V) of the source.

4. ### studiot AAC Fanatic!

Nov 9, 2007
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OK so it is obviously wasteful to draw more current than is necessary and that extra current is dissipated as heat in the circuitry so is undesirable.

The best current draw is therefore the sum of the load current plus the minimum current that the zener needs to stay on.

This gives you a current through the supply resistor at max and min supply voltages.

Can you determine the ideal resistors for these conditions?

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5. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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Rs(max)=(13.2-8.2)/(0.9111+0.075)=5.070
Rs(min)=(10.8-8.2)/(0.9111+1)=1.360

Also, how to find the
Iz at min supply voltage??

6. ### studiot AAC Fanatic!

Nov 9, 2007
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why 0.9111+1?

Yes this one is correct.

Last edited: Oct 9, 2013
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7. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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Oh, I know what's wrong.
Rs(min)=(10.8-8.2)/0.9111=2.854Ω

Then, how to find Iz at min supply voltage??

8. ### studiot AAC Fanatic!

Nov 9, 2007
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Don't rush ahead, I am asking you to work out slightly more than you need to help fully understand what is going on.

You should still have the minimum zener current.

Rs(min)=(10.8-8.2)/(0.9111+.075)= ?Ω

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9. ### studiot AAC Fanatic!

Nov 9, 2007
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This is surely a simple calculation?

I make it 2.64Ω

Don't worry the purpose of all this is to show that only half of the calculations are necessary.

10. ### studiot AAC Fanatic!

Nov 9, 2007
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What's the problem?

11. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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In finding Rs(max), I know that I should select which current.
However, I don't understand select which current in finding Rs(min)

12. ### studiot AAC Fanatic!

Nov 9, 2007
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So why didn't you do the calculation?
The quicker you respond the quicker we can finish this.

If we choose Rs as the calculated value for max efficiency at 13.2 volts supply - 5.070Ω - and the supply falls to 10.8 volts then the zener diode will turn off. There will not be enough current available to keep it on.
It will, in fact draw a small amount of current but it is beyond the scope of this question to say how much, certainly microamps, not milliamps.

So we just regard it as off.

This means that the zener is not maintaining 8.2 volts across the load.

The supply now sees Rs (5.07Ω) in series with the 9Ω load so I make that a current of 0.768amps.

The point about that is that it is a safe current.

But we do not need to worry about this since I hope you can now see that any resistance above Rs = 2.64Ω will lead to this situation (zener off) at minimum supply voltage.

So what we now need to do is use Rs = 2.64 and check that the current at maximum supply voltage does not exceed the design parameters. Your questions asked this at the end.

Can you now see what I meant by only doing half the calculations?

13. ### WBahn Moderator

Mar 31, 2012
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For now, keep going with the path the Studiot is leading you down. When you are done, we can come back and look at another way to approach the problem, which is to directly apply the constraints given.

Vz = 8.2V
Rl = 9Ω
75mA ≤ Iz ≤ 1A
12V-10% ≤ Vin ≤ 12V+10%

Now, assuming that Vz is 8.2V (which is the whole point of the above constraint), then what is Iz in terms of the source voltage, Vin, the source resistor, Rs, and the load resistor, Rl?

Plug that into the above relation and then solve for Rs in the middle. That will give you something of the form:

Rmin(Vin,Rl) ≤ Rs ≤ Rmax(Vin, Rl)

Now you just have to deal with the variability of Vin. The key here is to note that Rs has to be larger than ANY possible value of Rmin and that it must be smaller than ANY possible value of Rmax. So you set up those relations accordingly.

14. ### kimfung100 Thread Starter New Member

Oct 9, 2013
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Thank you.
I think it is too hard to understand because I am studying in Chinese about whole topic.

15. ### studiot AAC Fanatic!

Nov 9, 2007
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513
That's OK, we'd like to help.

I will post some sketches (pictures) that should help. Look again in about half an hour after the post time of this.

16. ### studiot AAC Fanatic!

Nov 9, 2007
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fig1
OK , I have started with a potential divider of two resistors supplied by a voltage V.
The voltage at A is shown.

fig2
If I now connect a second resistor in parallel with the lower resistor as shown.
The parallel combination of RL and Rp will be lower than RL and thus lower the potential (voltage) at point A.

fig3

I now make the parallel resistor variable and employ a little man to vary it. I can make it vary so that it exactly compensates for variation in the supply V, since if V goes up (increases) a suitable resistor in parallel with RL will bring the voltage at point A back down to what it was.

Note that I cannot allow for V falling by this method so V must always be taken as the lowest value it will ever have.

I have shown that this is how a zener diode works in the second part of fig3.

fig4

I have shown the values from your problem with the lowest value of V used.

fig 5

Shows that this is OK since without the zener to hold point A at 8.2 volts it would be higher at 8.35.

figs 6 and 7

Show that this is not the case for the higher value minimum resistor we calculated when the supply drops from 13.2 volts to 10.8.