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Current question

salesm21
salesm21 over 7 years ago

Attatched I have a photo of a small circuit I set up (I know its simple but I am just starting out) and realized something weird. When I put my test leads over my resistor I got very low resistance on my resistor which I thought would take the majority of the drop since the led offers little resistance. Also odd was when I turned my MM onto amps the DC power supplies current reading went up. Why is that? image

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  • michaelkellett
    michaelkellett over 7 years ago +3
    You've got a 3.2V voltage source (your PSU) driving a resistor and LED in series. The LED behaves like a diode with quite a high forward voltage drop. I found a nice cheap blue LED a bit like yours on…
  • jw0752
    jw0752 over 7 years ago +3
    Hi Mitchell, If you turned your meter to current measurement with the leads hooked across the resistor you would have put a short across the resistor effectively taking it out of the circuit. Since the…
  • salesm21
    salesm21 over 7 years ago in reply to jw0752 +1
    John, So as I understand it, if the meter wants to know current it creates a path of little resistance so that all the current is read through the meter. However how does this work in a parallel circuit…
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  • michaelkellett
    michaelkellett over 7 years ago

    You've got a 3.2V voltage source (your PSU) driving a resistor and LED in series. The LED behaves like a diode with quite a high forward voltage drop.  I found a nice cheap blue LED a bit like yours on Farnell ( 2377467) - if you download the data sheet you'll see that they specify the forward voltage as 3.3V typical at 20mA. Diodes and LEDs are very non linear for for their forward voltage/current relationship but for your diode to drop about 2.7V at 3mA (which I can read on you psu) seems quite reasonable.

     

    Ohms law doesn't apply to diodes ! The current rises much more rapidly as the forward voltage increases than it does with a resistor.

     

    Here's a tutorial on LEDs (looks OK but their are many others):  https://www.electronics-tutorials.ws/diode/diode_8.html

     

    Hope this helps.

     

    MK

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  • michaelkellett
    michaelkellett over 7 years ago

    You've got a 3.2V voltage source (your PSU) driving a resistor and LED in series. The LED behaves like a diode with quite a high forward voltage drop.  I found a nice cheap blue LED a bit like yours on Farnell ( 2377467) - if you download the data sheet you'll see that they specify the forward voltage as 3.3V typical at 20mA. Diodes and LEDs are very non linear for for their forward voltage/current relationship but for your diode to drop about 2.7V at 3mA (which I can read on you psu) seems quite reasonable.

     

    Ohms law doesn't apply to diodes ! The current rises much more rapidly as the forward voltage increases than it does with a resistor.

     

    Here's a tutorial on LEDs (looks OK but their are many others):  https://www.electronics-tutorials.ws/diode/diode_8.html

     

    Hope this helps.

     

    MK

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