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| Find highest value of 4 inputs without microcontroller? |
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| spec:
--- Quote from: GeorgeOfTheJungle on November 19, 2018, 12:49:46 pm ---I've heard there's a 3 cents^W 7 cents µC with twelve twelve bits ADCs: pms132. --- End quote --- Wow 12 bits: that is useful. Thanks for posting |
| spec:
--- Quote from: T3sl4co1l on November 19, 2018, 06:35:01 pm ---Yabbut... you need imagination to do design. I interpreted this thread as a design thread. Can words not generate an image in your mind? ??? Tim --- End quote --- I think we have seen the difference between imagination and an actual circuit. Would you kindly desist. I am only interested in technical matters. |
| spec:
--- Quote from: SiliconWizard on November 18, 2018, 08:44:32 pm --- --- Quote from: spec on November 18, 2018, 08:40:13 pm ---It's a tantalizing problem- you just feel that there ought to be a simpler way, and of course there is: a tiny MCU Sorry I mentioned that word :palm: --- End quote --- In terms of part count, obviously ;D Although, depending on your input voltage sources, you may need additional buffering anyway - so wouldn't be that clear then! Also, you'd be limited to the MCU's Vdd, whereas the opamp based design gives you a lot more freedom. --- End quote --- Ah yes! I hadn't thought about voltage range :palm: It's always the simple things that get you. |
| T3sl4co1l:
Voltage range is easy, put a divider in front of everything. ;D (Matched resistors, of course.) Visual aid: https://www.seventransistorlabs.com/Calc/ResDiv.html#three a calculator I wrote just for this kind of purpose, actually. :) Tim |
| spec:
Nice one Tim. So the parts count is MCU + 8 resistors: two resistors per channel. I don't know about you all, but that circuit would get my vote. :) It has the advantages of simplicity, low cost, and flexibility. To add another input just requires 2 more resistors. Can you please post a schematic of this attenuator approach that you speak of- sorry, couldn't resist it :-DD |
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