burning out a low power resistor

Discussion in 'General Electronics Chat' started by opeets, Apr 22, 2015.

  1. opeets

    Thread Starter Member

    Mar 16, 2015
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    In an earlier thread I came to the conclusion that I had to revise the plan for helping my son with his 5th grade science project. The objective is to demonstrate how Ohm's law plays a role in circuit design by preventing damage to sensitive components with the use of current limiting resistance.

    Our original plan was to calculate resistor values to drive an LED with nominal current, extremely low current, and extremely high current. After a few days of experimenting and posting results on this forum it became apparent that due to the non-linear nature of LEDs we needed to change course.

    Instead we are now going to try burning out a low power 1/8 watt resistor which will play the role of the "sensitive" load. The plan is to use decreasing R values of high power current limiting resistors connected in series to the low power load resistor until it burns out.

    A few hours ago I went out and purchased some high powered resistors that will act as current limiters and some low power load resistors.

    To start with simple matters first, I measured the actual resistance of the 5.1Ω resistor with DMM1. It was exactly 5.1Ω.

    I then measured the voltages for 5 of my 6 power supply presets by directly hooking up the leads to DMM1. Those values were as follows (rounded to two decimal points) :

    3V preset - 3.00V
    4.5V preset - 4.47V
    6V preset - 5.94V
    7.5V preset - 7.45V
    9V preset - 8.98V

    I then connected just the 5.1Ω resistor into a breadboard and added two jumper wires for the power supply grabber clips. I also connected the leads of DMM1 across the resistors leads so I can measure the voltage drop for the resistor at the different power supply presets.

    The following are measurements of voltage drop across the 5.1 Ω resistor (rounded to two decimal points) for those 5 presets. These were taken with no ammeter connected to the circuit.

    2.87V
    4.29V
    5.69V
    7.16V
    8.62V

    1) Considering nothing else was connected to this circuit is this "difference" between voltage drop and voltage supply expected at such a low R value?


    Continuing on I then connected DMM2 as an ammeter between the resistor and the negative terminal of the power supply and repeated the voltage drop measurements with DMM1 and included the corresponding current measurements with DMM2. Note: it was necessary for me to move my lead on DMM2 to the 10A scale (and the dial as well).

    DMM1 = 2.805V DMM2 = 0.55 A
    DMM1 = 4.18V DMM2 = 0.82 A
    DMM1 = 5.56V DMM2 = 1.10 A
    DMM1 = 7.0V DMM2 = 1.38 A
    DMM1 = 8.43V DMM2 = 1.66 A

    2) Is this change in voltage drop across the same resistor due to the ammeter?


    I then replaced the 5.1Ω 10W resistor with a 100Ω 10W resistor and the voltage drops across the resistor matched up perfectly with the measured power supply preset values.

    3) Does this make sense? Why did I not get the same voltage drops with the 5.1Ω resistor?
     
    Last edited: Apr 22, 2015
  2. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Yes, to confirm measure the voltage across the ammeter.
    It would appear your meter has a .118 ohm sense resistor.
     
  3. opeets

    Thread Starter Member

    Mar 16, 2015
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    For the 3V preset the voltage drop measured across the ammeter is 7.38mV.
    Based on the voltage drop readings it should be (2.87V - 2.805) = 6.50mV

    Close enough, right?

    And when I replaced the 5.1Ω resistor with the 100Ω that difference became a non-factor which is why my voltage drops were right on the money, right?
     
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I think so. There may also be some resistance in your wires and breadboard. My guess is the resistor in your meter is 0.1 ohms and all the other resistances make up the other .018 ohms.

    Edit:
    Did you mean 65 mv?
     
    Last edited: Apr 22, 2015
  5. WBahn

    Moderator

    Mar 31, 2012
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    As was described several times by several people in the other thread (but you are definitely making progress), your power supply has an internal resistance that acts, more or less, like a resistor in series with an ideal supply. Let's call this resistance Rs.

    So when you connect up a load resistor, let's call it Ro, to the supply, the total resistance seen by the ideal supply, let's call that Vs, is (Rs+Ro). The total current in both resistors, since they are in series, is

    Is = Io = Vs/(Rs+Ro)

    The voltage across the load resistor is the resistance multiplied by the current. Let's call that Vo:

    Vo = Io·Ro = [Vs/(Rs+Ro)]·Ro = Vs·[Ro/(Rs+Ro)]

    When we remove the load resistor, it is like making it infinite and then

    Vo=Vs.

    If we measure Vo for known values of Ro, we can use the measured (open circuit) value of Vs to find Rs:

    Rs = Ro·[(Vs-Vo)/Vo]

    Run the numbers for all of your first set of measurements and you will see that they something right in the 0.215Ω range give or take 10%.

    So the supply output impedance is about 4% to 5% of the load impedance and so you expect to drop about 5% of the supply voltage across the internal resistance with that load.

    With a 100Ω load the supply impedance is only about 0.2% of the load impedance and so you expect to drop about 0.2% of the supply voltage across the load. Note that this is considerably less than the basic accuracy of your meter. At 3V that's only 7mV and at 9V it's less than 20mV. So you do not expect to see much difference at all.

    When you have the ammeter in the circuit, you know have three resistances in series -- the supply internal resistance (which is now known), the load resistor (which is known) and the meter resistance (which is unknown).

    Follow the same procedure as before to find the meter resistance.
     
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  6. opeets

    Thread Starter Member

    Mar 16, 2015
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    Success!

    At 6V and a 5.1Ω current limiting resistor, I was not able to get the 3.9Ω resistor to smoke but I did smell burning.

    The approximate power being dissipated was:
    I = V/R = 6/(5.1 + 3.9) = .67A
    P = I*I*R = (.67)(.67)(3.9) = 1.75 watts

    At 9V and a 5.1Ω current limiting resistor, instant smoke.

    The approximate power being dissipated (before it went poof) was:
    I = V/R = 9/(5.1 + 3.9) = 1A
    P = I*I*R = (1)(1)(3.9) = 3.9 watts

    Interesting observation. I disconnected the 3.9 Ω resistor after meltdown and measured its resistance.
    It came in at 1.5kΩ. Is that a valid reading?
     
  7. WBahn

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    Mar 31, 2012
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    No, particularly since (2.87V - 2.805V) is 65 mV.

    You are comparing apples and oranges.

    You introduced a meter into the circuit and so the additional resistance is going to reduce the current in both the internal resistance and the load resistance.

    Do .... the .... math.

    Your total resistance is no (Rs+Ro+Rm), where Rm is the meter resistance.

    The total current is Vs/(Rs+Ro+Rm).

    The voltage across the load resistor is Vo = Ro·[Vs/(Rs+Ro+Rm)]

    So the meter resistance is (Rs+Ro+Rm) = Ro·[(Vs-Vo)/Vo)] - Rs

    When I run those numbers I get a meter resistance of 0.30 Ω, give or take about 3%.

    So let's look at the following model circuit for the middle case, Vs=6V nominally:

    Vs = 5.94V; Rs = 220 mΩ; Rm = 125 mΩ; R0 = 5.1Ω

    Without the meter, you expect a voltage across the load of:

    Vo = Vs·[Ro/(Rs+Ro] = 5.94V·[5.1Ω/(5.1Ω+0.22Ω)] = 5.69 V

    With the meter, you expect a voltage across the load of:

    Vo = Vs·[Ro/(Rs+Ro+Rm] = 5.94V·[5.1Ω/(5.1Ω+0.22Ω+0.125Ω)] = 5.56 V

    and a current of

    I0 = Vs·/(Rs+Ro+Rm = 5.94V/(5.1Ω+0.22Ω+0.125Ω) = 1.091 A
     
    Last edited: Apr 23, 2015
  8. WBahn

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    Probably. There is still some material there, but much of it has been burned away completely or has change to a different, higher resistivity substance.
     
  9. opeets

    Thread Starter Member

    Mar 16, 2015
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    This will have to wait until morning. My brain needs to recharge.
     
  10. opeets

    Thread Starter Member

    Mar 16, 2015
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    One last thought came to mind before I turn in for the night....what rationale (and I mean a very simplified version) can we provide in the writeup as to why the load resistor did not burn up at 1/4 watt, 1/2 watt, 1 watt, 2 watts, upwards to somewhere in the vicinity of 4 watts. Sure it started to feel warm and a certain power level, and it certainly smelled like it was burning up as the power increased even more, but it wasn't until we hit a whopping 4 watts (relatively large to the 0.125 watt rating) that it went poof.
     
  11. WBahn

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    Mar 31, 2012
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    If you buy a chair that has a label saying that the maximum weight is 200 lb, do you expect, or would you want, the chair to collapse at 201 lb?

    Consider this -- and this might be a test you want to include -- if you take that 5.1Ω resistor and dissipate 1W in it for a couple of days, how close it is to 5.1Ω both while it is in use and after it has cooled back down? If it is not within the specified tolerance for the resistor, then that resistor has failed. If it's a 5% resistor, then if it becomes a 5.4Ω resistor at room temp, it has failed. You don't have to have smoke and flame for it to be considered a failed part.

    There are a host of factors that go into designing a resistor for a certain power rating. You want the temperature in no-air-flow enclosed conditions to remain below a certain level. You want the maximum resistance change due to temperature to remain below a certain level. You want the permanent change in resistance over the life of the resistor to remain below a certain level. There are certainly several more. Once you design a resistor that can dissipate a certain amount of power continuously (not just for a few seconds, but for many years) and stay within all of those limits, you then build in a safety factor (actually, you build in the safety factor earlier in the process, but conceptually it's the same thing). You also take into account the statistical nature of both the manufacturing process and the environmental conditions and you design it not so that the average resistor meets those specs, but that only a tiny fraction of resistors in the tail of the distributions fail to meet them. The end result is that the typical resistor can handle significantly more power than what it is rated at without failing catastrophically.
     
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  12. opeets

    Thread Starter Member

    Mar 16, 2015
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    Okay..... I'd REALLY like to figure this whole missing voltage drop thing so to make life simpler for me I've used round numbers in two simple experiments below.

    Experiment 1: 100Ω 10 watt resistor connected to the 6V preset on my power supply. Ammeter connected in series to measure current.

    Measured Voltage (connected to circuit) across power supply: 5.9V
    Measured Voltage drop across resistor: 5.31V
    Measured Current: 53mA

    Experiment 2: 100Ω 10 watt resistor connected in series with another identical resistor. Rest is same as above.

    Measured Voltage (connected to circuit) across power supply: 5.9V
    Measured Voltage drop across each (and both) resistors: R1 = 2.799V, R2 = 2.811V, Rtotal = 5.61V (the math checks out)
    Measured Current: 28.11mA

    When I doubled my resistance, the missing voltage drop went from ~600mV to ~300mV.
    Will this tell me what the resistance of my ammeter is or do I need to perform additional tests?

    I sorta feel like an idiot now not knowing how to figure this out. But I was never taught these sort of things in my EE classes.
     
  13. opeets

    Thread Starter Member

    Mar 16, 2015
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    Experiment 3: Increased voltage to 9V. Rest is same as Experiment 2.

    Measured Voltage (connected to circuit) across power supply: 8.93V
    Measured Voltage drop across both resistors: Rtotal = 8.48V

    Experiment 4: Increased voltage to 12V. Rest is same as Experiment 2

    Measured Voltage (connected to circuit) across power supply: 11.87V
    Measured Voltage drop across both resistors: Rtotal = 11.28V

    When I doubled the voltage (Experiment 4 vs. Experiment 2), the missing voltage drop went from ~300mV back to ~600mV.
    This comes as no surprise, I just don't know how to put into simple words "why".
     
  14. opeets

    Thread Starter Member

    Mar 16, 2015
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    After measuring the voltage drop across the ammeter and using Ohm's Law I believe I determined the value of the my meter's internal resistance to be in the ballpark 10.67Ω value.

    Does that sound correct?

    Is there any other way to measure the current with a DMM so that the measured value is the actual value that should be flowing through the circuit, rather than slightly less because of the added resistance of the DMM?
     
  15. WBahn

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    Your ammeter has a resistance and when you run current through it you develop a voltage drop across it that you are not accounting for.

    When you doubled the resistance in the overall circuit (and the ammeter resistance is tiny compared to the 100Ω), the current was cut in half. What happens to the voltage that is developed across a resistor, such as the one in your ammeter, when the current is cut in half?
    I walked through in detail how to calculate the internal resistance of your ammeter.

    In your first case, you have a total of 5.9V across TWO resistors that are in series. One of them you know, it is 100Ω. The second one is unknown, namely the one in the meter.

    You know the current in both resistors. You know the voltage across the 100Ω resistor, namely 5.31V, so you know that the voltage across the unknown resistor is 590mV. If you know the voltage across a resistor and you know the current through the resistor, you can use Ohm's Law to find the resistance.

    Rm = 590mV/53mA = 11.1Ω

    Note that this is ignoring the internal resistance of the supply, but we know from prior data that it is about 220mΩ and so that isn't a factor for this.

    So, with that in mind, let's see what we would expect with a second 100Ω resistor. We have 211Ω across 5.9V so we expect a current of 28.0 mA with 2.80V dropped across each 100Ω resistor and 0.31V dropped across the meter.
     
  16. opeets

    Thread Starter Member

    Mar 16, 2015
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    Understood.

    So for these simple examples (which are the first basic experiments that my son will document) what do you suggest we write when the ammeter reads something like 53 mA but the calculated current value per Ohm's Law should be 59.1 mA. I understand the two additional resistances exist (that of the meter and that of the supply) but how would you explain to a 5th grade class?

    Are there any DMMs that contain negligible resistances?
     
  17. WBahn

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    When you step on a scale without clothes on you get a different reading then when you step on a scale with clothes, right?

    Can a fifth grader understand that concept?

    If your doctor wants to estimate your weight without clothes but the scale is sitting out in a publicly viewable area, what would they do? They'd take the scale indication and subtract an estimate for the weight of the clothes.

    As for what you would write down, that depends on what you want the number written down to mean. If you want it to mean the amount of current that was actually flowing in the circuit, you write down the number that the ammeter reads. But your calculations for how much the ammeter should read need to take into account the impact of the meter and/or the power supply resistance. Just as when you walk into the doctor's office you would adjust your morning weight by the expected effect of your clothes and whatever you've consumed/eliminated in order to determine what you expect the doctor's scale to read. If you want it to mean the amount of current that would flow in the circuit if/when the meter isn't there, then you use the information about what the circuit is doing with the meter and calculated what it should do with the meter removed.

    This is, admittedly, harder to understand since it is more abstract. But you can make it less so by demonstrating the effect of putting in multiple resistors and what happens if you try to ignore one of them.
     
  18. opeets

    Thread Starter Member

    Mar 16, 2015
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    Understood. So I guess there is no way to get an exact current reading unless you go with really high resistances (relative to the DMMs resitance). Something like 1kΩ resistors.

    Since we're working with current-limiting resistors that are 100Ω and less to connect to our load resistor, for each example we'll just have to explicitly account for the internal resistance of the meter when measuring the current. The voltage drop measurement will be done with DMM1 without DMM2 connected as the ammeter.
     
  19. opeets

    Thread Starter Member

    Mar 16, 2015
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    So my plan at this point is to set up a series of structured well-document experiments for my son to make it easier for him to understand what we're doing.
    Otherwise he will just get frustrated and do it for the sake of doing it. I really want him to grasp this information and turn it into an exciting experience.
    With so many distractions these days it is getting really hard for a parent to get a child excited about learning something, especially at this crucial age just before entering middle school.

    So....starting with the most basic circuit (and taking into account what I've learned about internal DMM resistances), refer to the image below for Experiment #1.
    Does anything in the procedure look out of place?

    experiment1.JPG
     
  20. WBahn

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    Basically, yes. That is why I keep coming back to NOT using a DMM as an ammeter. Your meter has a basic accuracy rating of 1.5% of reading plus 3 digits. Let's roughly call that 2%. That means that if the resistance in series with the ammeter is more than 50x the ammeter resistance, then your accuracy is limited by the basic meter accuracy and not the perturbing effect of the ammeter resistance.

    Any time you make any measurement, you are going to perturb the system being measured. The question is whether it can be perturbed to a small enough degree so that the perturbation can be ignored. If not, then can it be perturbed in a way that can be taken into account. If not, then is the perturbation so large and unpredictable that it renders the measurement meaningless to begin with.

    Be careful about mixing measurements taken with and without the ammeter in place. If you take the voltage drop across the resistor without the ammeter in place then you are measuring the voltage across a resistor that has a different current flowing in it than when the ammeter IS in place.
     
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