Im having a hard time finding the spects for this piece I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?

enter image description here

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  • 8
    Since the reference designator is L1, I suspect that it is an inductor, not a resistor. – Peter Bennett Nov 20 at 21:01
  • 7
    +1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;) – Wouter van Ooijen Nov 20 at 22:07

So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.

The hard part really is

where can I purchase a ton of this

because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.

That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.

So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.

I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.

A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is $$E=\frac12 LI^2\text,$$ leading to a total maximum energy stored in your ton of inductors of \begin{align} E_{tot}&=N\cdot\frac12 LI^2\\ &=\left\lceil\frac{1\,\text{Mg}}{185\,\text{mg}} \right\rceil\cdot \frac12 47\,\mathrm\mu\text{H}\,\left(205\,\text{mA}\right)^2\\ &\approx 5.4\cdot 10^6\,\cdot\, 2.4\cdot10^{-6} \,\text{kg}\,\text{m}^2\,\text{s}^{−2}\,\text{A}^{−2}\,\cdot \,4.2\cdot10^{-3}\,\text{A}^2\\ &=5.4\cdot2.4\cdot4.2\cdot10^{-3}\,\text{N·m}\\ &\approx 54 \,\text{J.} \end{align}

That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.

  • 14
    Again with the dry comedy! – winny Nov 20 at 22:09
  • 2
    Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel. – Johnny Nov 20 at 22:55
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    I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;) – Marcus Müller Nov 20 at 22:59
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    @Johnny - Even better, the one where they're all wired in series. – WhatRoughBeast Nov 21 at 1:12
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    +1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K. – Spehro Pefhany Nov 21 at 2:00

NO, it is an inductor -- you can see on PCB they define it as L1

the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%

  • 2
    I think you mean 47uH, right? – Spehro Pefhany Nov 20 at 21:08
  • @SpehroPefhany: Hmm, do you think it black not brown? – Rev1.0 Nov 20 at 21:09
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    It looks black on my monitor but if it's brown then 470uH. – Spehro Pefhany Nov 20 at 21:10
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    @SpehroPefhany, its brown actually – Satish Singupuram Nov 20 at 21:11
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    Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check. – pipe Nov 21 at 10:56

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