# Simple Op Amp & Zener Diode Questions

Discussion in 'Homework Help' started by Student01, May 11, 2009.

1. ### Student01 Thread Starter Active Member

Apr 15, 2009
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The following questions is from a class where the tutor solved the question and I copied down the procedure for later reference.

Can someone tell me why I've copied from the board Vo2 = 7V? I get a different answer now that I look at it again:

10-5 = 5V

V0.5k/(V1.5k+V0.5k) x V = 0.5/(0.5+1.5) x 5 = 1.25 V ie, Vo2 = 6.25V

Where am I going wrong?

2. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
I'm also having trouble with part a of the following:

In my notes on how to solve this question, is the following:

KVL At Zener Junction
Ismax= Izmax + Iload = 100mA
Ismin = Izmin[ + Iload = 50mA

Shouldn't Izmax and Izmin be equal to 100mA and 50mA, respectively, not Ismax and Ismin?

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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For question 4, Vo2 = 7V is correct.

The method works as follows:

1. The 1kΩ to ground has 5V and the current in the 1kΩ is 5mA.

2. The 5kΩ has a voltage drop of 5V as shown and the current in the 5kΩ is 1mA

3. The current in the 0.5kΩ is 5mA-1mA = 4mA.

4. The voltage across the 0.5kΩ is then 2V [0.5kΩ x 4mA]

5. So Vo2 = 5V + 2V = 7V

See how you go from there......

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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For question 1, part (a) proceed as follows:-

1. Work out the Thevenin resistance, excluding the diode but including the RL= 200Ω looking back towards the source. You basically have a voltage divider comprising 100Ω and 200Ω fed by the source, Vs.

2. You would get Rth = 100Ω // 200Ω and Vth = (2/3)xVs - you don't know Vs - but that's OK.

3. Given the Thevenin source may drive the 10V zener diode from 0 to 50mA what would the range of Vth that would achieve this? Obviously at 0mA the Vth=10V (min value). At 50mA the Vth = 10V + Rthx50mA (max value).

4. So you then have the range of Vth from which you can then find the range of Vs, given Vs = (3/2)xVth

5. I'll leave the rest for you to consider.

5. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
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Couple of questions:

Question 4
Why can't you look at the 1.5k and 0.5k resistors, see there is a 5V drop and use a voltage divider? It gets an incorrect answer, but it seems as if it should work to me.

Question 1
In step two, why do you treat the two 50 ohm resistors and the 200 ohm like they're in parallel? How do you know they're in parallel?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The lower amplifier is driving the node at the junction of the two resistors - it's effectively a voltage source feeding the node and the simple voltage divider rule does not apply.

Redraw the circuit without the zener diode. Denote the node where the diode top (cathode) terminal was connected as node A. You have the source Vs connected with two 50Ω resistors [100Ω] to node A and 200Ω from node A to the common node. The Thevenin equivalent would be that which you would "see" between node A and the common node. The node A Thevenin voltage Vth, in series with Rth would then be connected to the zener diode for the purposes of analysis.

7. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
I follow what you did here, but I don't really understand why you did it. I don't suppose it's possible to explain further?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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May I suggest you attempt a solution [using a method you are comfortable with] and post your working. Then it's possible for someone to see where you may need guidance.

In response to your initial question .....

In my notes on how to solve this question, is the following:

KVL At Zener Junction
Ismax= Izmax + Iload = 100mA
Ismin = Izmin[ + Iload = 50mA

Shouldn't Izmax and Izmin be equal to 100mA and 50mA, respectively, not Ismax and Ismin?

The question explicitly states that you assume Izmin = 0mA. So it follows that it can't be 50mA.

Here's a hint :-

If Iz=0mA, then all the current flowing from the source is going through the load. So under this condition, if the Load voltage is 10V, what is the Load current? Hence what is the (minimum) source current?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In fact, solving using the Thevenin equivalent method isn't the most efficient one anyway.

10. ### mik3 Senior Member

Feb 4, 2008
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70
First, you can't use the voltage divider rule because you don't know the current drawn by the lower op amp.

Just make the analysis step by step.

The 1K resistor has a 1mA current through it because the node above it is at 5V.

The node below the 0.8K is at 10V and thus the current through the 5K is 1mA.

Thus the current through the 0.5K is 5mA-1mA=4mA

Thus the Vo2=4mA*0.5K+5V=7V

Therefore, the current through the 1.5K is (10-7)V/1.5K=2mA

Thus the current through the 0.8K is 1mA+2mA=3mA

Thus Vo1=10V+3mA*0.8K=12.4V

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Suppose the opamps aren't ideal, but each have a gain of exactly 10.

What are the node voltages then?

12. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
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I didn't know you had to know the current the op amp draws. I thought if you knew the total potential difference across a branch, you could calculate the potential difference across each resistor in that branch using a voltage divider.

But now I think about it, I seem to remember the voltage divider being derived from KVL, so that would require I know all the voltage drops around a closed loop before I was able to use a voltage divider. Is this correct?

13. ### mik3 Senior Member

Feb 4, 2008
4,846
70
The voltage divider rule works fine when there is no load on the voltage divider network. If there is a load then you need to include the resistance of the load into your calculations.

In your case with the op amp you don't know the resistance seen when looking into the output of the op amp and thus you can't use the voltage divider rule.