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08-01-2018 #1
Again, we are interpreting a Chinglish manual (or diagram, in this case) so there is a bit of guesswork here! The given diagram shows a pot supplied with 12V. If the idea is that the pot is uncalibrated and you use it in conjunction with the VFD display, then this is fine - in effect, it just reduces the rotation of the pot spindle to go from zero to full. Ditto, I guess, if the pot has a calibrated scale associated but again you won't use the full pot travel.
Slightly different situation where the analogue input is driven from an external source, although it's a pity that the BOB or whatever doesn't regulate the DC input to ensure that the output range is correct.
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09-01-2018 #2
Guys,
With the right pot the source voltage is close to irrelevant so long as you can't blow the VFD input with the output.
Instead of arguing semantics why not posit a functional solution to the problem as it stands.
Offsetting voltage with diodes chops off the bottom of the range BTW ;-)
- NickYou think that's too expensive? You're not a Model Engineer are you? :D
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09-01-2018 #3
No it doesn't. Read my post - it's reducing the Vref voltage from 12V to 10V. The controller's charge-pump will then generate a voltage proportional to the M-S ratio of the demand vs Vref - from 0v to 10v.
It is a functional solution.
I didn't think the OP had a "pot" - if he did, then the solution would be to introduce a positive-side static resistance, e.g. 2k on top of a 10k pot, which would give a sufficient offset from the 12V reference voltage. But that's not the design case here.
Re. the chop-off. Nope, what you have interpreted is the insertion of the diode chain into the Ain1 input, that's not what I posted.Last edited by Doddy; 09-01-2018 at 12:19 AM.
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09-01-2018 #4
So the BOB will function properly with 10V input rather than 12V?
Bonus, you've fixed it!
Nice one!You think that's too expensive? You're not a Model Engineer are you? :D
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09-01-2018 #5
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09-01-2018 #6
I know it would be a bit more faff but if it really needs 10V in from a 12V supply I'd put a regulator in ;-)
You think that's too expensive? You're not a Model Engineer are you? :D
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