A DIY Current Source Demonstrating Ohm's Law - DC To Daylight

Поделиться
HTML-код
  • Опубликовано: 7 июл 2024
  • In this episode, we take a look at the shortcomings of measuring low resistances with a standard digital multimeter. We demonstrate measuring a small resistance using both the two-wire and four-wire kelvin methods. Then we build a standalone current source that will be used in conjunction with a standard DMM, allowing us to make four-wire measurements, without an expensive benchtop meter: bit.ly/30t9dKj
    Engage with the element14 presents team on the element14 Community - suggest builds, find project files and behind the scenes video: bit.ly/2MFMG0v
    Visit the element14 Community for more great activities and free hardware:
    Tech spotlights: bit.ly/2KLz0TS
    Roadtest and Reviews: bit.ly/2KH4fj3
    Project14: bit.ly/2wPnajx
    #0:00 Welcome to DC to Daylight
    #4:33 The Circuit
    #7:51 Build It
    #10:39 Test It
    #12:00 Give Your Feedback
    #ohmslaw #currentsource #DIY #4-wiremeasurement
  • НаукаНаука

Комментарии • 17

  • @graemelinsdell2361
    @graemelinsdell2361 2 года назад +6

    The prototype solution is excellent , I have never seen before,.

  • @Lokalaskurar
    @Lokalaskurar Год назад +1

    The parts where the maker says "The issues I faced were" are my absolute favourite

  • @oakspines7171
    @oakspines7171 6 месяцев назад +1

    Thanks. Very helpful. Quite some work and dedication you put into this video.

  • @Mr.Beauregarde
    @Mr.Beauregarde 2 года назад +2

    More of this guy in my future watch history. Very nice

  • @Workshopshed
    @Workshopshed 2 года назад +1

    Great explaination of the 4 wire measurement, I've never dug into why that was needed before.

  • @vobbertbecking3598
    @vobbertbecking3598 2 года назад +10

    You should fit the diode before the capacitors, if that is a 22uf electorlytic it is not protected from reverse polarity.

    • @AmRadPodcast
      @AmRadPodcast 2 года назад +4

      Facepalm. Rookie mistake - you're absolutely correct. Let's just call that cap a "reverse polarity notification device"! 😉

  • @barry2718
    @barry2718 2 года назад +1

    The prototype method is a great idea. Beats veroboard or matrix board for smaller circuits.For quick small circuits I have used stiff cardboard and a pin to make the component holes. Just push component through the cardboard and bend leads at back and solder. Great video, very well put together and educational.

  • @MrAshwindersingh
    @MrAshwindersingh 8 месяцев назад

    Today I learnt new idea to prototyping thank you sir really appreciated

  • @mnoxman
    @mnoxman 2 года назад +1

    Use Enhancement mode FETs to switch in resistances. Your minimum resistance is 5000 mohms and the RdsON of 4mOhms if you were using a 2n6661 would be insignificant by comparison worst case (0.08%). It would also be consistent (wouldn't change from switch click to switch click). Using a push button, a schmitt trigger in to a toggle flip flop to a 4066 would let you do this "the old school way". You could do it all in a PIC chip for a more modern take.

  • @radman999
    @radman999 2 года назад

    Great presentation style. Excellent work!

  • @GeorgeGraves
    @GeorgeGraves 2 года назад +2

    LM317 is cool. But boy I wish someone would make a Keathly 225 knock-off with modern parts and decent accuracy. It can't be that hard right? Right?!?!? (hehehe)

  • @GeoffryGifari
    @GeoffryGifari Год назад +1

    how big are the lead, contact, and jack resistances compared to 10 Ohm resistor, in most cases?

  • @dicko195
    @dicko195 2 года назад

    @7:14 2,270 ohms? Not 2,027? TY great video and info.

  • @RaviKumar-xx7gi
    @RaviKumar-xx7gi 10 месяцев назад

    Hi I have made the circuit of LM317 on bread board. I am generating a constant current of 100mA. The input to the regulator is 3.3vdc. 100mA constant current is fed to a resistor of 100 ohms. I should measure a voltage across resistor of =10mA x 100= 1000mV=> 1V, but the voltage across the resistor I am measuring is 3.3vdc. why so? I want to use the constant current source to find the unknown resistor using current and voltage How can I do this?

  • @MrSamwien
    @MrSamwien 2 года назад

    How high is Vin?

  • @RaviKumar-xx7gi
    @RaviKumar-xx7gi 10 месяцев назад

    Hi