Hello guys,
I found this valuable forum just recently and was happy to read so many intelligent and wise oppinions here. I am a student and unfortunately have troubles passing an exam in "Electronics and Measuring Technologies". I really hope that I will find some support here and answers to my probably silly questions (pleaassee )

Here comes the first one - I have the following part of an exam problem (don't lough, that's one of the easiests).

There is a zener Diode in the middle as a voltage regulator which hase a Zener Voltage of 5V. The input voltage is initially set to 11V. R1 = 150 Ohm. The Light Diode (LD) has U=2V.

a) What should be the resistance of R2, so that the LD would have a current of 20mA by U(LD)=2V. Answer: 150 Ohm

b) What current flows through R1 and what through ZD? Answer: 40mA, 20mA

c) The input voltage is now set to 9,5V. What current is flowing through the LD now? Answer: 20mA

d)The input voltage is now set to 6,5V. What current is flowing through the LD now? Answer: 15mA

This exactly is my problem - how come and why is the current of d) 15mA!?? My answer would be 20mA as before!? is it my mistake or is the given answer wrong? I do get the previous answers... but maybe I am not solving it right

Thanks very much guys

P.s. Please let me know if it is ok to post my problematic exam problem in this forum at all, really hoping to find some help 3rd attempt to pass the exam

Attached Images

ScreenShot004 (Small).jpg (20.3 KB, 25 views)

 

With a 6.5 V supply, there isn't enough current flowing through R1 to maintain 20 mA in the LED.

Therfore no current at all flows in the zener diode. The current in the LED is (6.5 - 2)/(150 + 150) = 15 mA

The voltage on the zener is 2 + (0.015 * 150) = 4.25 V

So at 6.5 V supply the zener does nothing at all.

 

Thanks very much for your reply. How do you explain the above? How do you proceed?

 

Hi Perky123,

I think Diver300's answer did not explain for you to understand. So I'll give it a try.

Supply voltage=6.5V, R1=150Ω, R2=150Ω, forward voltage drop (Uf) of the LED=2V.

My calculation looks like this: (Vs-Uf)/(R1+R2), in numbers: (6.5-2)/(150+150) or (6.5-2)/300 or 4.5/300=0.015A=15mA.

At 6.5V the 5V-zener-diode has almost no effect on the supply voltage assuming leaks.

I guess forum members generally don't like the idea helping at school work (home work).

There is a general saying in Thailand: "Thai rak Thai", meaning Thai helps Thai. To adapt it I'd say "German rak German".

Kind regards

Hans-Jürgen

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Proper Planning Prevents Piss Poor Performance

 

You're right. It isn't exactly 15mA.

Your lecturer has not told you that the LED voltage drop varies with forward current. At 15mA, the forward voltage drop might be anything say 1.95V instead of 2.0V.

Actually, the problem can only be solved using a load line on the voltage drop vs forward current characteristic graph of the LED.

__________________
L.Chung

 

Thanks guys.... but isnt't the voltage across R2 and LD always equal ot the ZD voltage (when the input voltage is greater)!? this worked for the previous questions

 

No. If one disconnects the zener diode but the junction voltage(between R1 & R2) is already lower than the 5V zener voltage, the zener "cannot" bring it up to 5V. This is exactly what happened here. The LED and the two resistors formed a network with the junction between the two resistors being less than 5V so the zener has nearly no current flow.

You can study the attached diagram where I gradually reduce the supply voltage from 12V to 3V and you can see how the voltage and current varies.

Attached Images

Zdled.gif (23.6 KB, 15 views)

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L.Chung

 

You are right. At 6.5V the zener has a breakdown current of 10mA!, hence the voltage will be stabilized to 5V, good enough for R2 and the LED to "see" just 5V.

Boncuk

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Proper Planning Prevents Piss Poor Performance

 

Thanks very much for the graph and explanation, but I still have problems understanding how to determine whether the voltage between R1 and R2 is already lower then 5V as you say without th help of a computer program. Or in other words - how do I (mathematically or theoretically) determine whether the 6,5V are enough to give 5V to the ZD and as a result 5V to the LD and R2!?

I always did it as follows:
6,5V > 5V, => U(ZD) = 5V = U(LD+R2)
U(LD)=2V => U(R2)=3V => I(R2)=I(LD)=3V/150Ohm=20mA

this apparently is wrong. what would the right solution be?

Thanks again

 

It would be 6.5V - 2.0V (Vld) = 4.5V/2 (R1 in series with R2) = 2.25V drop for R1 and R2. Add this 2.25V (R2 Vdrop) to the Vld of 2.0V and you get 4.25V at the junction of R1 and R2. Divide the 2.25V by 150Ω and you get 15mA. This is what eblc1388 was saying about the junction voltage being lower than Vz. If Boncuk is right and the zener has a breakdown current of 10mA, than by his calculation, the total current in the circuit would be 10mA. Since current splits at the parallel junction, you can't have 10mA flowing through the zener and 15mA through the LED. This would be a total current of 25mA, which does not compute. Eblc1388 is correct. He is also correct in stating that the Vfd of the LED will be less at 15mA than at 20mA. But your instructors opinion is probably "close enough". Spoken like a true technician. But, if this is an engineering course, he needs to be more precise.

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You do it in the reversed sense.

1. Assuming there is 5V across the zener, so R2+LED will takes 20mA as previously calculated.

2. Now decide what current one wants to flow through the zener so that it will do its job efficiently. In this case, one can choose any current between 5mA ~ 10mA.

3. Assume we select 5mA and 10mA as two typical case and see what happens.

4. The sum of these two currents (LED 20mA + zener 5mA or 10mA) flow through R1 and by ohm's law voltage drop across R1=25mA*150=3.75V or 4.5V respectively.

5. Now the supply for this whole network is easily worked out by adding this voltage to 5V across the zerer diode. Giving you 8.75V or 9.5V for these two particular cases.

So you'll find out 6.5V is way too low for the zener circuit to work.

__________________
L.Chung

 

Wow. That brought some brightness to the whole story! Thanks a lot eblc1388! So this is probably how I should have started with the previous question to that problem as well, isn't it? To first do the check?

So just to make sure I understand things right now:

1. Generally said, in order for a (any) ZD to function correctly, it needs 5 to 10mA current flowing through?

2. If the ZD does not get it's 5-10mA (or ZD Volage?) I can just ignore it and make the calculations as if there isn't one in the circuit?

Thanks once again!! Big Kiss!

 

There is another way to explain the circuit:

R1 and R2 make a voltage divider. Consider the LED as a dead short at a forward voltage of more than 2V, so these 2V go down the drain if the current is high enough for the LED.

So at Us of 6.5V the current across the LED calculates like this: ILED=(6.5-2)/300(R1+R2), so the LED-current is 15mA.

Hans

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Proper Planning Prevents Piss Poor Performance

 

In school we analyzed these type circuits using graphical piecewise-linear analysis. The straight lines we drew were asymptotic to the actual values of V or I.
Without this approximation tool you chase the amps and volts all around, ewig, without seeing the big picture.

You could start by modelling the Zener diode using a ideal battery, a resistor and an ideal diode.

 

Zener diodes, like transistors, have different power handling ratings from 0.1W to 10W or more. This will determine what current range it should be operating normally.

For further info, I recommend the following article by Rod Elliott: at How to Use Zener Diodes

One first do a simple circuit calculation, without the zener diode connected and determine if the node voltage is less than the zener rated voltage. If it is so, then it is an indication that the circuit design component value is wrong.

Alternatively, in your previous case, if one can change the value of R1, then it is easy to make the zener circuit works at 6.5V too. The calculation is easy:

With 6.5V supply voltage, there will be a (6.5V-5V)=1.5V across the R1. The 5V being the zener voltage. The total current in R1 we have worked out at 25mA or 30mA. Using Ohm's law will give us the value of R1 to be 60 Ohm or 50 Ohm.

See, circuit calculation is all fun, if you know how.

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L.Chung