(Topic ID: 228133)

In line resistor not reducing voltage

By Cserold

5 years ago


Topic Heartbeat

Topic Stats

You

Linked Games

No games have been linked to this topic.

    Topic Gallery

    View topic image gallery

    Terminal_Strip (resized).jpg
    66148FBA-F63E-41C3-A89F-6D2CF5C6BFF7 (resized).jpeg
    pasted_image (resized).png
    image (resized).jpg
    image (resized).jpg
    image (resized).jpg
    pasted_image (resized).png
    DE7F3210-94B8-4CEA-9915-C5226B2163BE (resized).jpeg
    #1 5 years ago

    Why am I such an idiot? Trying to get the 12V DC coming out of MAIDEN Le Down to 5V, 0.125A. Attached a resistor in-line to the 12v line and I can’t get any voltage drop when using my DMM on the non connected side of the resistor. What the hell am I doing wrong? I’m sure I’m misunderstanding something basic. Appreciate any help you can provide.

    DE7F3210-94B8-4CEA-9915-C5226B2163BE (resized).jpegDE7F3210-94B8-4CEA-9915-C5226B2163BE (resized).jpeg
    #2 5 years ago

    The obvious first step is connect it to the other lead, then test again.

    Electrons have a negative charge. Electricity is electrons flowing. They flow from negative to positive.

    #3 5 years ago

    Where are you putting the two DMM leads? One on the end of the resistor, the other on the black wire?

    Also, hard to tell the scale, but that looks like a standard 1/4 W resistor. To drop 12 V to 5V at the 0.125A you specified, you need minimum of a 1 W resistor to be safe ...

    #4 5 years ago
    Quoted from setzkor:

    Where are you putting the two DMM leads? One on the end of the resistor, the other on the black wire?
    Also, hard to tell the scale, but that looks like a standard 1/4 W resistor. To drop 12 V to 5V at the 0.125A you specified, you need minimum of a 1 W resistor to be safe ...

    Was putting the red lead on free end of resistor and black lead on the black wire.

    I’m attaching a pic of the resistors I’m using. How do I know wattage?

    image (resized).jpgimage (resized).jpg
    #5 5 years ago

    No voltage drop. Testing with resistor in pic

    image (resized).jpgimage (resized).jpgimage (resized).jpgimage (resized).jpg
    #6 5 years ago

    You won't see a voltage drop doing that. Look up "voltage divider circuit".

    #7 5 years ago

    There is no current flowing through the resistor, so there will be no voltage drop. V=IR. It's the law!

    #8 5 years ago

    Thanks guys. So, is this the complete wrong approach to dropping voltage from 12v to supply an LED needing 5v, 0.125A? How can I do that?

    #9 5 years ago
    Quoted from Cserold:

    Thanks guys. So, is this the complete wrong approach to dropping voltage from 12v to supply an LED needing 5v, 0.125A? How can I do that?

    pasted_image (resized).pngpasted_image (resized).pngpasted_image (resized).pngpasted_image (resized).png

    #10 5 years ago

    As mentioned, the reason you are not seeing a drop, is you haven't yet completed the circuit...you would need to connect a load to the end of the resistor and the black wire. I would not do this with your LED, it could be damaged if the resistor is not correct. Also, if that is the standard kit from Amazon, those are 1/4W, you would have to put 4 resistors in parallel. In your case, 4 of the 220 ohm resistors in parallel is equivalent to 55 ohms which should drop close to 7V. Here, the 4 resistors would split the near 1W of power. But...

    To answer your last question: There are several approaches to do this...yours is one way, not terribly efficient, but can work with the right components.

    https://electronics.stackexchange.com/questions/127525/reducing-voltage-with-resistors

    The first main answer lists 5, it is a good list. Depending on what you what to power, you may be able to get away with the series resistor. I'm sure other posters here have examples of how they did this for their mods.

    FYI, here are examples of the Buck Converter:

    https://www.amazon.com/SMAKN-Converter-Power-Supply-Module/dp/B00CXKBJI2

    https://www.amazon.com/Converter-DROK-Regulator-Step-Down-Transformer/dp/B0758ZTS61

    #11 5 years ago
    Quoted from barakandl:

    [quoted image][quoted image]

    Appreciate this reply but this is like looking at heiroglyhiscs to me. Lol. I need more “hey dummy, take this resistor solder this end to this wire...”.

    #12 5 years ago
    Quoted from setzkor:

    As mentioned, the reason you are not seeing a drop, is you haven't yet completed the circuit...you would need to connect a load to the end of the resistor and the black wire. I would not do this with your LED, it could be damaged if the resistor is not correct. Also, if that is the standard kit from Amazon, those are 1/4W, you would have to put 4 resistors in parallel. In your case, 4 of the 220 ohm resistors in parallel is equivalent to 55 ohms which should drop close to 7V. Here, the 4 resistors would split the near 1W of power. But...
    To answer your last question: There are several approaches to do this...yours is one way, not terribly efficient, but can work with the right components.
    https://electronics.stackexchange.com/questions/127525/reducing-voltage-with-resistors
    The first main answer lists 5, it is a good list. Depending on what you what to power, you may be able to get away with the series resistor. I'm sure other posters here have examples of how they did this for their mods.
    FYI, here are examples of the Buck Converter:
    amazon.com link »
    amazon.com link »

    Thanks very much. This makes sense. Really appreciate all the replies guys!

    #13 5 years ago

    I have one all made up for you if you like. Sent PM.

    66148FBA-F63E-41C3-A89F-6D2CF5C6BFF7 (resized).jpeg66148FBA-F63E-41C3-A89F-6D2CF5C6BFF7 (resized).jpeg
    #14 5 years ago
    Quoted from Cserold:

    Thanks guys. So, is this the complete wrong approach to dropping voltage from 12v to supply an LED needing 5v, 0.125A? How can I do that?

    As others mentioned - the method was wrong. You need a complete voltage divider circuit. An accurate and consistent voltage divider is difficult to built for most applications as the load current tends to vary... which changes the voltage drop across the resistor.

    Another downside to this method. You state that you want to go from 12V to 5V with a load current of 0.125A. If going with resistors configured as a voltage divider - the resistor needs to drop 7 volts across it. If you pick the right resistance values then you have a second problem. A resistor dropping 7 volts across it and conducting 0.125 amps means the resistor is dissipating 7 * 0.125 = 0.875 watts. The resistors you show appear to be 0.25W resistors - these would pretty much turn into little space heaters for a brief period of time.

    The easiest and highly reliable method would be as BarakandI listed above - just use a simple 7805 regulator. A small heatsink should be used such as a Aavid 507302. For this simple of a circuit, you can easily just mount it on a terminal strip such as the one shown below. Or you could go with a cool running pre-built switcher such as suggested by Rock914 or setzkor.

    Terminal_Strip (resized).jpgTerminal_Strip (resized).jpg
    #15 5 years ago

    I think a thing called a buck converter will do this also for a few bucks.

    #16 5 years ago

    USB supplies 5v and phone adaptors are readily available. Plug one into your service outlet. Use an adapter with output appropriate for your use case. Most will handle a one amp load, many up to two. Splice the USB power wires to your mod.

    Reply

    Wanna join the discussion? Please sign in to reply to this topic.

    Hey there! Welcome to Pinside!

    Donate to Pinside

    Great to see you're enjoying Pinside! Did you know Pinside is able to run without any 3rd-party banners or ads, thanks to the support from our visitors? Please consider a donation to Pinside and get anext to your username to show for it! Or better yet, subscribe to Pinside+!


    This page was printed from https://pinside.com/pinball/forum/topic/in-line-resistor-not-reducing-voltage and we tried optimising it for printing. Some page elements may have been deliberately hidden.

    Scan the QR code on the left to jump to the URL this document was printed from.