Re: 4-20mA scaling ???????
From: CFoley1064 (cfoley1064_at_aol.com)
Date: 08/19/04
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Date: 19 Aug 2004 18:10:45 GMT
>Subject: Re: 4-20mA scaling ???????
>From: "Pablo" paul@notsoclose.com
>Date: 8/19/2004 10:32 AM Central Daylight Time
>Message-id: <QP2Vc.82$Mk4.1811@news20.bellglobal.com>
>
>I have now two options (that is bought type options) and I am waiting for
>salesman on the one. First is a KB product which does exactly what I need it
>to, however at a cost of 175.00 canadian, add that with a good ten turn or
>pushbutton pot which cost about $50-75, thus overall cost would be about
>250.00 per machine. We are looking at 8 machines. The other option which is
>getting me firm is to use a operator interface which Control Techniques
>sells and use it to scale the current output.
>
(Originally posted at s.e.b.):
Hi, Paul. I kind of wish you'd given more information, specifically the
manufacturer and model number of the VFD (variable frequency drive, used to
vary the speed of AC motors by varying their frequency). Nonetheless, there's
still some general advice you can safely get on your problem.
Quite a few modern VFDs have a load resistor at the input, typically 500 ohms
or so, and use an analog-to-digital converter and software to interpret the
signal for 0-10V, 0-5V, 0-20mA, or 4-20 mA. If you're set up for 4-20 mA, you
typically also have the ability to program minimum current (the current reading
below which you'll get 0 Hz output) and maximum current (above which you'll get
100% of programmed speed). If you use a potentiometer set up as a shunt
resistor, you'll cut the slope of your input current, but you'll also bring
your minimum current well below 4 mA. That may cause a fault lockout on your
VFD (which you also might be able to program around).
If you're going to do the shunt pot route, you need to figure out the minimum
current necessary to keep a fault lockout from occurring, and also how much
attenuation you can deal with when the pot is turned all the way up. Here's an
example based on the assumption of a 500 ohm load resistance at the VFD input
(view in fixed font or M$ Notepad):
VFD Current Divider
.----------o------.
| | |
| V D |
| - | .-------------.
o+ | | +| VFD Input |
4-20mA V D '--------o--. |
signal - | | |
o- | | | |
| V D | | |
=== - | .-.500 ohm |
GND | | | |internal |
| | | |load |
.-. | '-'resistor |
10K| |<---. | | |
| | | | | |
'-' | | | |
| | | | |
o-----' .--o--' |
| | -| |
.-. | '-------------'
220 ohm | | |
| | |
'-' |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
Use a 1N4001 or Si diode of your choice. This circuit has a couple of
advantages:
* Because of the diodes, it won't start attenuating until there's about 1.8V
across the 500 ohm resistor (which will reduce your problem with error lockout)
* When the 10K pot is set for maximum resistance, you'll have minimal
attenuation of your signal. Of the 20mA, 19mA will go into the VFD. Actually,
you should be able to program around this with your maximum current setting.
* When the pot is at minimum, less than 1/3 of the current will go into the
VFD input. That would mean that 20mA output from your signal source will give
less than 33% full speed.
The only real negative of this setup is that almost all the control will be in
the first 1K of resistance. If you want to invest in a 10-turn pot with a
calibrated dial, that'll help a lot. You can also tell the machine operators
where to set the dial (that's assuming anyone on your factory floor can set a
verneir dial).
On the other hand, this is a simple and relatively bulletproof setup that
doesn't require external power, and that a maintenance person will have a hard
time messing up.
If you need/want a more complicated solution, you'll need an external power
supply and a couple of op amps. Have fun, and
Good luck
Chris
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