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Are the computed values in Load regulation correct?

  1. Sep 19, 2016 #1
    1. The problem statement, all variables and given/known data
    The whole image is at here http://imgur.com/a/Dcbvv , but I will try to elaborate more. I have a 12V fixed source with an RS of 220 Ohms with a 1N4733 zener diode model connected and a varying load parallel to the diode. Find the Source Current, Z current, Voltage at R1, and Load Current.

    We haven't done the experiment yet so we're only doing the computed values for now. However I have doubts on what my groupmates have computed here.Their results are written in the table on the imgur link. While my computations are written on the left side of it


    2. Relevant equations
    IS = IZ + IL Where the Source is fixed,

    IS= (VS - Zener voltage) / RS

    3. The attempt at a solution

    I did the solution on the left side of the imgur image but I will also write it down again.

    Using equation 2, (12V-5V)/220Ohms = 31.82mA. Which should be the constant IS.

    Using IS = IZ + IL, I have 31.82 = IZ + 5 mA for 1K resistor which should yield IZ = 26.82mA. The rest have the same logic but I also have a problem with the VR1, should I assume it to be equal to 5V of the Zener Diode?
     
    Last edited by a moderator: Sep 19, 2016
  2. jcsd
  3. Sep 19, 2016 #2

    gneill

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    Staff: Mentor

    The 1N4733 zener is actually rated at 5.1 V. If you want to get more accurate in your predictions you might want to look up the datasheet for this diode and see what the zener resistance is and account for it (usually a fairly small value, typically between 5 and 20 Ohms. It varies from zener to zener).

    Your predictions may be inaccurate when your load draws a lot of current and the drop across the 220 Ohm resistor drops the output voltage lower than the zener's 5.1 V (It can't supply more voltage to make up the difference if too much is lost across the dropping resistor; it can only limit what's available!). So check to see what the output voltage would be for a given load without the zener in place first! If you remove the zener you're left with a simple two-resistor potential divider.
     
  4. Sep 19, 2016 #3

    I see, so should I simply change the value to 5.1V? And are my equations correct?
     
  5. Sep 19, 2016 #4

    gneill

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    Your equations will do for a reasonable approximation, unless as I said, the source draws so much current that the drop across the 220 Ω resistor causes the voltage available to the zener to fall below the zener voltage. Then the zener will stop regulating.

    Take a careful look at what happens when the smallest load resistance is used. Ignore the zener (pretend it's not in the circuit) and calculate the output of the potential divider.
     
  6. Sep 19, 2016 #5
    A bit confused on your advice, if I remove the zener diode, my output would be 8.33 V from looking at 500Ohms as an RL on the formula Output = (Voltage*RL)/(RL+RS). Is this right?
     
  7. Sep 20, 2016 #6

    gneill

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    Yes, that's right. So in this case the zener would be able to function to limit the voltage to zener voltage as there's (8.33 - 5.1) = 3.2 V of "headroom" for it to work with.

    If the voltage divider produced less than the zener voltage, the zener would not be able to conduct and the output would be unregulated. The output would be the same with or without the zener in place. Check what voltage your voltage divider equation gives you when the load resistor is 100 Ω.
     
  8. Sep 20, 2016 #7
    I would have a 3.75V output? And it yields a negative when subtracted with 5.1V so the output would be unregulated?
     
  9. Sep 20, 2016 #8

    gneill

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    Right. If the load draws so much current that the drop across the 220 Ohm resistor pulls the output below 5.1 V, the zener cannot stop it from happening. The zener will turn off and the output will just follow the dictates of the voltage divider.
     
  10. Oct 3, 2016 #9

    Sorry for the late reply, another question though. How does the output voltage affect the Voltage of R1 and the current of the source, load and zener?
     
  11. Oct 3, 2016 #10

    gneill

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    Isn't R1 the load resistor? As such the output voltage is the voltage across R1.

    upload_2016-10-3_6-38-54.png

    The most important thing, if the circuit is to work as a regulator, is that the current drawn by the load does not cause a potential drop across the 220 Ω resistor of more than 12 - 5.1 = 6.9 V, because then the zener won't have any "room to work" so to speak. A zener can't add voltage to a circuit, it's not a battery. It can only draw current so that the potential drop across the dropping resistor (220 Ω resistor) keeps the output at 5.1 V.
     
  12. Oct 4, 2016 #11
    Strange, my groupmates used a VOM on the same circuit but the values for the Output Voltage were all 5.19V. Did we do something wrong or is it the polarity of the probing?
     
  13. Oct 4, 2016 #12

    gneill

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    They measured 5.19V even for a load resistance of 100 Ω ? Did you confirm with the VOM resistance scale the values of the resistors used?
     
  14. Oct 4, 2016 #13
    According to them, the potentiometer did measure 100Ohms up to 1000Ohms respectively by using a Digital Multimeter. But I'm seriously doubting them because they used a somewhat defective multimeter due to the fact it reads the negative polarity when measuring the power supply voltage and it was quite shaky at times. Also, they did measure the red probe on one side of the zener diode where the black line was and the black probe at the other side.

    I'm guessing something went awry?
     
  15. Oct 4, 2016 #14

    gneill

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    That would be my guess, yes.

    Can you post the data set collected?
     
  16. Oct 4, 2016 #15
    My groupmates are holding the data but the only thing they wrote at the measured value was the output voltage having a constant 5.19V across 100 Ohms to 1K Ohms.

    It's going to be hard for me to convince them the data is erroneous

    Edit: My hypothesis on this erroneous data is probably the probing? Or the multimeter?
     
    Last edited: Oct 4, 2016
  17. Oct 4, 2016 #16

    gneill

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    Just show them the calculation for the output voltage for the unregulated voltage divider for each of the load resistance values.

    upload_2016-10-4_9-13-26.png

    In particular note the output voltage when the load is 100 Ω. If that output is less than VZ, then the zener can't hold regulation. It just can't.

    One possibility that occurs to me is the the dropping resistor (the 220 Ω resistor) wasn't in fact 220 Ω. If it was smaller, say closer to 100 Ω, then the output would remain above the zener threshold. The fairly high measured values for your output voltages supports this premise: I'd expect the regulated values to be closer to 5.1 V than 5.2 V, since that zener is a commercial part with a spec of 5.1 V . If the zener were being forced to handle larger currents than normal in order to maintain regulation, then the potential drop across the inherent resistance of the device would nudge that value upwards.
     
  18. Oct 6, 2016 #17
    Since I got kicked out from my group (long story) , I'm doing the experiment again alone. It's both a relief and kind of stressful.

    Based on the attached circuit above, I would have a computed output of 9.836V when having an RL of 1000Ohms?
     
  19. Oct 6, 2016 #18

    gneill

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    Okay. At least you'll have a chance to check things that might have been missed before.
    Yes. And since 9.8 V is greater than VZ at 5.1 V, when the zener is in place it can pull the output down to 5.1 V.

    When you re-do the lab, be sure to verify that the 220 Ω resistor is in fact 220 Ω, and check that the power supply is really 12 V.
     
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