is my DMM bad?

Discussion in 'The Projects Forum' started by coeng, Nov 4, 2009.

  1. coeng

    Thread Starter Member

    Oct 29, 2009
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    I have a Commercial Electric DMM (Model HDM350). I think I bought it at Home Depot a long time ago. I don't know a great deal about meters, in fact the only thing I've ever used it for is to make sure I have 120V coming out of an AC outlet.

    In the process of working on my garage door project, I came up with some unexpected readings. I don't have an app that draws circuit diagrams so I will just explain it because its simple:

    12V non-regulated wallwart -> positive lead connects to 1K resistor -> other side of resistor connects to anode of 1st LED -> cathode of 1st LED connects to anode of 2nd LED -> cathode of 2nd LED connects to negative lead of the wallwart.

    When I took the LEDs out of the equation and measured the voltage across the positive and negative leads of the Wallwart (with the 1K resistor soldered to the positive lead, I got 15.6V).

    When I added the LEDs back in, it read 3.45V (using the 20V DC scale).

    When I broke the circuit and added the meter in between the 1K resistor and the anode of the 1st LED to measure the current, it read 0.06 (using the 20m scale).

    Since the typical voltage of my LEDs is 2.25V (max is 2.6V), the following current should have been measured:

    (15.6-2.25-2.25)/1000 = 11.1 mA

    Is my meter really off, or am I reading the current wrong?

    Also, where does the 3.45V come from? :confused:
     
  2. CDRIVE

    Senior Member

    Jul 1, 2008
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    Something is very odd here. If you measured 15.6V at the wall wart's output with only a 1K as a load it shouldn't have dropped by inserting two Leds in series with the resistor. In fact, the load on the wall wart should have been reduced. Check your wall wart again with just the 1K load. I suspect something may have gone west!
     
  3. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    I've had quite a bit of trouble with the fuses in my DMM, so why don't you replace the fuse? See if it's burnt or looks damaged.

    Austin
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, this is about what you measured when you tried it before with no load.

    And those LEDs are supposed to have a typical Vf of 2.25v with 20mA, right?

    OK, it sounds like your "wall wart" supply is no good; you're only getting about 60 microamperes current out of it. The Vf of the LEDs at that tiny current is very low.

    Your meter is fine. The power supply is bad - OR your electrical connections to the supply are not making good contact.
    Unless the LEDs are lighting up reasonably bright - then something else is amiss.

    Your LEDs are rated for typ. Vf of 2.25@20mA. With only 0.06mA current (60uA), the Vf will be much lower. 60uA is about 0.3% of 20mA. I'm not surprised you're measuring an average of 1.725v per LED at that low current.
     
  5. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    You left out an important datum: did the LEDs light up when you were trying to measure the current? If they did not, then you probably made a mistake when measuring the current. Many multimeters require you to change the leads to different jacks to measure current. There's no way the LEDs would have been lit running on 60 μA of current, so something's fishy.

    You can easily check whether your ammeter is working: just connect a 1 kΩ resistor in series with a typical 1.5 volt battery and see that you get a current on the order of 1.5 mA. If it doesn't work, it's almost always an internal fuse that can be replaced (make sure you replace it with the correct type and amperage rating of fuse; using the wrong type or rating can ruin your meter).

    An even easier check can be done if you have a DC power supply that can be set to constant current mode. Set the supply to a convenient constant current, then connect the DMM's leads (in the proper current jacks) to the supply's output jacks and you should read the supply's set current.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Someonesdad,
    Our OP is a novice in electronics. They could try the test using the 1k resistor and a 1.5v battery with the meter set to the 10mA or 20mA range (should read ~1.5mA), but I don't think that they have access to, or know how to use a constant current source.

    coeng,
    when you try the 1k Ohm resistor in series with the DMM set to read on the 10mA/20mA scale, make sure that you are measuring current in SERIES with the resistor, not in PARALLEL with the resistor. If you accidentally connect the meter across the terminals of the battery while it is set to read current, you will immediately blow the fuse in the meter.
     
  7. coeng

    Thread Starter Member

    Oct 29, 2009
    32
    0
    Both LEDs are lit up bright red when I connect the DMM in between the resistor and anode of the 1st LED. I'll go through my measurements step by step tonight and post photos. Maybe it will be obvious then what I'm doing wrong.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    OK, so when the LEDs are lit up, what is the voltage that you are measuring across the 1k Ohm resistor?
     
  9. coeng

    Thread Starter Member

    Oct 29, 2009
    32
    0
    Is there any quicker way to do this other than to remove the shrink tubing that covers it?
     
  10. coeng

    Thread Starter Member

    Oct 29, 2009
    32
    0
    Just a thought...could the fact that I am using LONG wires have anything to do with these pecularities? From the wallwart to the block terminal is about 4 feet. From that block terminal, the wire pair traverses 100 feet or so to the my LEDs on the other side of the basement. The wire is 18 AWG solid thermostat wire.

    Like I said I will re-measure tonight and post my results.
     
  11. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    No, not with a load that light.
     
  12. coeng

    Thread Starter Member

    Oct 29, 2009
    32
    0
    OK, so here's what I did last night.

    I opened the DMM and realized my fuse was blown. How long ago this occured is unknown to me since I don't ever recall measuring current with it since I owned it. I'll pick up the fuse today and retake my current measurements tonight. Is the fuse only needed to take current measurements? If the fuse is rated for 500mA, will taking a current reading just across the wallwart (which is rated at 500mA) cause the fuse to blow?

    I then created a test setup last night using my breadboard and took some voltage readings (see steps below).

    Forgive me for the basic questions below...I'm just trying to get an understanding of my measured readings. I don't remember very much from my circuit courses I took in the early 90s because I eventually became a software engineer.

    1) The voltage measured across the wallwart (with no load) was 14.5V. The wallwart I used is the exact same model but not the same wallwart I mentioned earlier. I could not measure the current because no fuse!

    2) The 1K resistor (by itself) measured in at 977Ω

    3) I connected the 1K resistor to the wallwart and measured the voltage drop across the resistor to be 13.26V. I don't understand this number. How can it be calculated? I assume this means that 13.26/977 = 13.6mA is flowing through the resistor.

    4) I then connected one end of the 1K resistor to an LED in series. The cathode of the LED connected back to negative side of wallwart. I measured the voltage drop across the resistor to be 12.18V. I measured the voltage drop across the LED to be 1.73V. I measure the combined drop across the resistor and LED to be 13.92V which adds up. Why was the voltage drop across the LED only 1.73V? I expected something between 2.25 and 2.6V per the specification of the LED. How did adding the LED increase the total voltage drop across the load from 13.26V in step 3 to 13.92V (+0.66V)? Why was the voltage drop across the resistor 12.18V?

    5) I then added another LED in series with the first. I measured the voltage drop across the resistor to be 10.55V. I measured the voltage drop across the first LED to be 1.71V. I measured the voltage drop across the first LED to also be 1.71V. I measure the combined drop across the resistor and both LEDs to be 13.99V which adds up. Why did adding the LED only minimally affect the total voltage drop across the load from 13.92V in step 4 to 13.99V (+0.07V)? The voltage across the resistor only dropped by 1.63V when the 2nd LED was added...why not 1.71V?

    The LEDs were lit up bright red in steps 4 and 5.
     
    Last edited: Nov 5, 2009
  13. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    Typically, as you suggest, the fuse is just to protect the instrument's electronics from too much current. You were probably still able to take voltage and resistance readings using the meter.

    Do not connect the ammeter directly across the output voltage of the wall wart. If you do this, you'll be connecting a low resistance across a voltage source and it will draw lots of current -- you run the risk of blowing the fuse again. You might want to read a bit about ammeters to refresh your memory.

    You're seeing an example of non-perfect regulation. You measured the open-circuit (i.e., no load) output voltage of the wall wart, then saw it drop 1.2 V when you were drawing about 14 mA from the wall wart circuitry. If you made the same measurement with e.g. a lab DC power supply, you'd likely see better regulation.

    Unless you know the construction and parameters of the wall wart's internal electronics, there's no a priori way to calculate the voltage drop across the resistor. Most of us would just determine it empirically, as you have done.

    Perhaps the operating point is in the "knee" area of the voltage-current characteristic curve of the LED (see also here). You can measure this curve yourself by supplying a current and measuring the voltage across the device.

    Adding in the LED changed (increased) the total resistance of the load (LED plus resistor) on the wall wart. As the load resistance increases towards an indefinitely large value, the voltage drop across the load will approach the open circuit output voltage of the wall wart.
     
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