Discussion in 'Homework Help' started by aliustun14, Nov 4, 2017.

1. ### aliustun14 Thread Starter New Member

Nov 4, 2017
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Hi,
I've got confused on this question and i couldn't find any sources to study on this subjects. From the beginning, thanks for your suggestions and helps.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But what part of this question confused you the most? Also try to show us your work.

3. ### aliustun14 Thread Starter New Member

Nov 4, 2017
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I tried to analysis the nodes. I put the voltages to every node and tried to understand how does this circuit will act but i couldn't figure it out. Seems like there will be no current on D4 or D2 but actually there must be to complete the circuit. I think i need to study more on the diodes but i'm trying to figure out this circuit.

4. ### dl324 AAC Fanatic!

Mar 30, 2015
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Under what conditions would this be the case?

I'd start by analyzing the circuit with the input voltage at 20V. Since you need to do this for multiple values of input voltage, choose some method that will facilitate that.

Show your work and members will try to guide you past where you're having difficulty.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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All I can say is that for a positive voltage at Vo. Current want to flow through the diodes in this path:

D1 and D2 diode will start conduct only if the anode voltage is 0.7V larger than the cathode voltage. And for the Zener diode (D5) the cathode voltage needs to be larger than the Zener voltage.

Nov 4, 2017
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7. ### dl324 AAC Fanatic!

Mar 30, 2015
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Think again.

How did you get the equation for I?

You should get into the habit of including units in your work.

How are you supposed to calculate Vz when Iz isn't 10mA?

How are you supposed to do the calculations when the diode drop isn't 0.7V and their impedance isn't 30 ohms?

8. ### aliustun14 Thread Starter New Member

Nov 4, 2017
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Opps This time i change my solution. Since zener diode's voltage is constant. And actually i couldn't understand what you meant with when voltage drop isn't 0.7 V and impedance because usually we neglect change of voltage drop and resistance of diode. Am i wrong ?

9. ### dl324 AAC Fanatic!

Mar 30, 2015
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We usually assume a forward voltage for the diodes and ignore their resistance. After initial calculations with the assumption(s), we iterate as necessary to obtain the solution. In some cases, not many, we can ignore the voltage drop of diodes.

You were asked to plot the output voltage for -20V < Vi < 20V. How are you going to determine the voltage drops of the diodes and their resistance when they're barely conducting? How are you going to determine Vz when the current is less than 10mA? You weren't given an IV graph for any of the diodes.

Not knowing what you've studied and are studying, or the objective of the problem, this seems like a poorly constructed question to me...

10. ### dl324 AAC Fanatic!

Mar 30, 2015
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As an example. If I had this circuit and wanted to calculate Vout for +V=10V.

I would assume that the diode voltage was 0.7V. That would give 9.3V across 10.5k ohms, or 0.89mA. Using the graph below, I'd determine that the typical forward voltage of the diode would be 0.61V.

Now I need to decide if the error is significant enough to iterate the calculation. If I did another iteration, I'd have Vd = 0.61V which would give I = 9.39V/10.5k ohms = 0.89mA.

I'd likely decide that it would be sufficient to change the diode voltage and say Vout would be 0.61V + 0.9mA * 500 ohms = 1.06V versus the 1.15V the first iteration would have given. If 10% accuracy was sufficient, I could have stopped at the first iteration.

Note that I used two significant digits to highlight the difference between iterations.

Whoever wrote the problem you're trying to solve didn't give you sufficient information to solve it for cases where Vd wasn't 0.7V and Iz wasn't 10mA. Therefore, taken in this context, it's impossible to give an accurate solution.

Last edited: Nov 4, 2017
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11. ### WBahn Moderator

Mar 31, 2012
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About the best you can do is assume ideal diode characteristics except where explicitly given otherwise (namely the forward resistive component of the four regular diodes).

You might plot the characteristic for that assumption, and then look at a data sheet for a 7.5 V zener and see what reasonable assumption you might make for it that is comparable (i.e., a forward resistive component) and redo the analysis and overlay the plot.

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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13. ### WBahn Moderator

Mar 31, 2012
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The problem is that the problem doesn't give the information needed to model the zener this way. It gives the the voltage and current at ONE point and no information about which to establish either the voltage and current at another point on that operating segment or the slope of that segment.

The information on the regular diodes is also a bit ambiguous since the current at which Vd = 0.7 V isn't specified. But here you can (and pretty much have to) make the assumption that the 0.7 V occurs at the zero current breakpoint between conducting and nonconducting. The only problem with this is that it's pretty unrealistic. With a 30 Ω resistance, it would result in a diode forward voltage of 1.0 V at a current of just 10 mA.

14. ### dl324 AAC Fanatic!

Mar 30, 2015
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Another seemingly poorly crafted question when taken without context. If we knew what the OP had already been taught and what they're supposed to be learning now, we might understand what assumptions are being made and it would make more sense.

15. ### WBahn Moderator

Mar 31, 2012
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Agreed.

Sometimes the question is just poorly crafted, period. But so often more is the case where the question is quite reasonable within the context of what is happening in that class at that time, but we have no way of knowing what that context is and, to be fair, the student generally lacks the experience to reasonably be expected to recognize what context is missing so that they can provide it. It's frustrating, but as long as the TS is willing to answer questions and show work, we can usually slug through it (and the TS often walks away with more gained knowledge than they were hoping for).