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It seems so. If the input were the full 10v, it would give a current dependant on the output capability of the 10v source. The input terminal has to have some protection against too much current. It is also filtered to get rid of any induced voltage spikes to give a more stable control of the speed.
You see that's where it is confusing... looking at the diagramQuote:
Originally Posted by cropwell
Attachment 28860
The circuits (to me) at the bottom reopresent the circuit a user has to add. This is because the switch from 5 - 7 is not on/in the board, the user adds that.
Does that make sense?
I'm in agreement with Cropwell's interpretation. There's any engineering "wet thumb" rule that you design the input impedance at least 10 times the source impedance in order to not substantially influence the input. So, 10k input pot, 100k input impedance - rule is met. (ignoring balanced input design etc).
Don't get hung up on impedance over resistance.
And Nemo - Because I'm in a good mood I'm going to agree with you too - at least re. the circuit outside of the box being that that the user has to add.
I'm still lost... "Set speed Term 3: Potentiometer minimum or 0 to +10V, impedence 100K ohm filtered"Quote:
Originally Posted by Doddy
Do I need the 10k pot if using 0 - 10v from a spible card?
It's written to confuse. My interpretation that I'm very comfortable with:
"Potentiometer 10k minimum" : There are other on-board controls to set the minimum speed and I guess they don't want to influence that circuitry by the installer using a silly-small potentiometer. I would expect the on-board pot to be a lowish value to provide an offset above zero when placed in series with the speed control pot. If the speed control pot was too small then the on-board pot would become more dominant as part of the potential divider, and the effective speed control would be limited to a higher minimum speed. Or, potentially disk damaging the on-board control. However, this is belts-and-braces stuff.
"Impedance 100k ohm filtered" - as Cropwell said - it characterised the input equivalent circuit, more or less. Nothing to do with the external control.
"0 to +10V" - as an alternative to using the recommended minimum 10k potentiometer, you are able to drive a speed command in the range 0-10V, wrt 0V/pin 4 (note: NOT wrt pin-2 used for the potentiometer).
So, yes, you can drive the input with 0-10V from the spindle card (your words), provided that you can share a common 0V reference from the spindle-card output to the 0V on this SM 16 / pin 4.
Firstly - thank you for taking time to reply to my mind-numbingly simple posts.Quote:
Originally Posted by Doddy
So 0v to pin 4 (what does wrt denote please?). Do I connect +10v to pin 1 or pin 3?
Pic below is MB3 0-10v analogue
Attachment 28864
thanks again
Connect A0 to pin 3, AG to pin 4. Leave pin unconnected.
Wrt = with respect to. A voltage is the potential difference between two points, not necessarily always to ground or 0V, you could look at the wiring for the pot and decide to measure the analogue demand between pins 2&3 for example. Pedantic perhaps but I tend to reference where a voltage is measured between, and being lazy I say what point, with respect to another point.
If you have to ask the question, then I’ve failed in providing an answer that isn’t clear.
Actually that is what I thought, I was just double checking. I really didn't want to blow it up for the sake of asking again.Quote:
Originally Posted by Doddy
News: Tried it - started first time, brilliant.
Thank you - your time and effort is most appreciated.