Technical Question - Measuring Circuitry

Kevin PeatKevin Peat Posts: 2,779Member
edited September 12 in Totally Not Guitars
My idiot's guide tells me I should measure amps in series with a component (in this case an LED.)

I did so, but the thing is this:

I first measured the LED with the multimeter needles either side of it [is this parallel ?]and the reading accorded accurately with Ohm's Law. When I put them is series (before the LED) the reading was way off my calculations.

Also how can I have such wild fluctuations in amps depending on where I measure ? The circuit had an LED, a switch and a 100 Ohm resistor. (3v DC)

Thanks in anticipation.


  • ESBlondeESBlonde Posts: 805Member
    If you think about it the multi meter in parallel will bypass the device and screw the current calculations completely. Thats why series is required for small components. With larger currents you can measure the current flow throuh a single strand of cable with a clamp on device but afaik these are not sensitive enough for milliamps.
    I write as a complete novice and would welcome an experts correction.
  • Kevin PeatKevin Peat Posts: 2,779Member

    But it got me exactly the reading I'd calculated should be there using Ohm's law. They reckon it's OK to measure voltage in parallel and this doesn't screw up readings.
  • Kevin PeatKevin Peat Posts: 2,779Member
    I'll just continue having fun building circuits and measuring them, even if they make no sense.
  • DaveBassDaveBass Posts: 3,301Member
    Okay Kevin, what you need to grasp are these points. First, an LED can be considered to be a voltage source. Eh, what? This is a technical way of saying that no matter what current is passing through it, the voltage drop across it remains constant. Ohm's Law does NOT apply to it! (It's for resistors and not much else, certainly not semiconductors.) In reality of course there's no voltage drop if the current is zero, and the voltage drop does in fact increase with the current passed through it; but not by much. So as an approximation we say that it can be modelled by a constant voltage. For the sake of argument, let's say this voltage is 2 volts. (Different LEDs will vary. Get the figure from the data sheet.)

    Secondly, an ammeter has (ideally) zero resistance. A real ammeter is a good enough approximation to this.

    So if you short-circuit the LED with an ammeter, the current you read will be pretty close to what you calculated ... because you calculated it wrong!

    I guess what you did is to take V = 3 volts and R = 100 ohms, so I = V/R (Ohm's Law) = 3/100 = 0.03 amps = 30 milliamps. And that's what you got, because the circuit comprised a 3V battery and a 100 ohm resistor; the ammeter resistance being negligible and the LED being shorted out by it.

    Now if you put the ammeter in series with the LED and the resistor (the order doesn't matter), what you have is a voltage source of 3 volts due to the battery, and a voltage drop of 2 volts due to the LED, leaving 1 volt across the resistor and ammeter. (Technically this is an application of Kirchhoff's Voltage Law or KVL: the voltages around a circuit loop add to zero. 3 - 2 - 1 = 0. Or, as we've used it, 3 - 2 = 1. There's also a Kirchhoff Current Law: the currents entering a node add to zero. In other words, what flows in must flow out. But we don't need KCL in this simple circuit, which is just a loop.)

    We'll neglect the series ammeter, assuming that its resistance is much less than 100 ohms.

    Now we know the voltage across the resistor, 1 volt, we can apply Ohm's Law to it, because Ohm's Law is for resistors: I = V/R = 1/100 amps = 10 milliamps. This is what the ammeter should read in practice.

    Circuits make perfect sense if you understand what's going on. But my experience of teaching undergrads is that most people find them very unintuitive and difficult to get to grips with. (As I did myself when just starting out.) And a.c. circuit theory adds a whole extra layer of complexity!

    As ESBlonde says, you can use a clamp-on Hall-effect probe to measure d.c. currents; and they do in fact go down to milliamps if you can afford an expensive bit of gear.

  • Kevin PeatKevin Peat Posts: 2,779Member
    Dave - Thank you most kindly for that. You've pointed me in the right direction. The calculations and figures accord exactly with what I was doing.

    I doubt I'll ever visualise what is going on in circuitry but it's fun to try.

    Once again. Thanks.
  • Kevin PeatKevin Peat Posts: 2,779Member
    edited September 19
    My readings are making a lot more sense in other circuits. I re-built this circuit and took a reading of 8.7mA in series between the negative terminal of the battery and the cathode of the LED (assuming that it didn't matter where I took the reading in the circuit.)
    Not completely out of the ball park. I assume there has been some extra resistance in the somewhere. I'll do it again tomorrow just to be sure.
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