I like the resistor divider, combined with a "summer" for its
simplicity.
I was going to offer the possibility of a divider circuit feeding a
rail-to-rail op-amp. I think you can get one which operate with
single-sided power supplies (or you can use an extra 2.5v reference
(is it called a phantom ground, or something like that?))
As a non-expert, and knowing only enough to get myself into trouble,
_I_ would try to come up with an op-amp solution.
What bandwidth do you need? Analog or Digital? As I said, I am no
expert, but I have heard that the divider networks have problems
with bandwidth. Presumably due to capacitance?
For digital purposes, you could consider a level shifter IC. I am
using one in my SD card interface project. Dividers would work,
except for possible bandwidth, in the conversion from high voltage
to low voltage. Not quite so easy in the low-to-high conversion.
Some people rely on the fact that +3v or +3.3v is considered "high"
in most 5v digital systems. Unless, of course, it doesn't work the
way it should.
-Tony
Reply by Paul Dubinsky●October 4, 20052005-10-04
At 11:10 PM 10/3/2005, you wrote:
>--- In basicx@basi..., Ken Arck <ah6le@a...>
wrote:
> > Call me dumb but how does a voltage divider protect the BX's
input
> > from a negative voltage?<...>
A great deal of thanks to Don, Lauren, Neil and others for the quick
response to my question. It never ceases to amaze how much you guys
know and how willing you are to share it.
Thanks, again,
Paul
"I've seen the future, and you're not going to like it."
-- Vincent van Gui
Reply by Don Kinzer●October 4, 20052005-10-04
--- In basicx@basi..., Jepsen Electronics Ltd <njepsen@i...>
wrote: > Insert "NOT" in the appropriate
place.
Indeed. It should have read:
... so that variations in the current flowing through the upper
divider do *not* disturb the offset voltage (very much).
Reply by Jepsen Electronics Ltd●October 4, 20052005-10-04
Insert "NOT" in the appropriate place.
neil
Don Kinzer wrote:
--- In basicx@basi..., Paul Dubinsky <pdubinsky@f...> wrote: > Of course.
Note that the circuit that Lauren suggested is designed so that the
2.5V offset is supplied with an impedance about a factor of 10 lower
than the impedance of the divider above it. That is important so that
variations in the current flowing through the upper divider do disturb
the offset voltage (very much). Ideally, you'd probably like to have
two orders of magnitude difference but if you try to do that you
either run into problems with too much current being consumed in the
offset circuit or having the impedance of the signal divider being too
high for the circuit that it is driving. The suggested circuit is
probably a good compromise. In any event it's a good starting point.
Don
Reply by Don Kinzer●October 4, 20052005-10-04
--- In basicx@basi..., Ken Arck <ah6le@a...> wrote: > Call me dumb but how does a voltage divider
protect the BX's input
> from a negative voltage?
By itself, it doesn't. Once you've limit the swing to +/-2.5 volts
and shift it up by 2.5 volts, the signal will always be 0-5V unless
there is a component failure or the output from the analog source goes
out of spec.
To fully protect the BX-24 input you'd want to add clipping diodes at
the BX pin and a limiting resistor between the pin and the level
converter. The BX-24P has the clipping diodes built into the AVR chip
so external diodes are not necessary for that part. It won't hurt
anything, though, other than the small amount of parasitic capacitance
and reverse leakage current that they will present.
To review, the clipping (or clamping) diodes are connected thusly: one
diode with cathode to +5, anode to the pin and the other one with the
cathode to the pin and the anode to ground. This will keep the pin
from going more than about 0.6V above +5 or the same below ground.
"I've seen the future, and you're not going to like it."
-- Vincent van Gui
Reply by Lauren Vanderhoof●October 3, 20052005-10-03
At 05:17 PM 10/3/2005, you wrote:
>--- In basicx@basi..., Paul Dubinsky
<pdubinsky@f...> wrote:
> > [It is] an analog -10 to + 10 voltage voltage from an analog
sensor.
>
>The conversion needs two parts. Firstly, you need to convert the 20
>volt swing (-10 to +10) into a 5 volt swing. That's easy - use a
>voltage divider. For this, you'll need two resistors that have a 3:1
>ratio. The exact values that you'll want to use depend on what else
>will be done with the scaled signal. If the circuit that follows has a
>high input impedance (i.e. it will draw very little
current from the
>voltage divider) then the only consideration is the load that the
>divider will present to the analog source.<...>
Something like this?
+/-10V Input +5V
| |
100K 10K
| |
|-------------|------> to BX24 Analog Input
33K |
| 10K
Gnd |
Gnd
At 09:17 PM 10/3/2005 -0000, you wrote:
The conversion needs two parts. Firstly, you need to convert the 20
volt swing (-10 to +10) into a 5 volt swing.
<---Call me dumb but how does a voltage divider protect the BX's input
from
a negative voltage?
Ken
Reply by Paul Dubinsky●October 3, 20052005-10-03
At 05:17 PM 10/3/2005, you wrote:
>--- In basicx@basi..., Paul Dubinsky
<pdubinsky@f...> wrote:
> > [It is] an analog -10 to + 10 voltage voltage from an analog
sensor.
>
>The conversion needs two parts. Firstly, you need to convert the 20
>volt swing (-10 to +10) into a 5 volt swing. That's easy - use a
>voltage divider. For this, you'll need two resistors that have a 3:1
>ratio. The exact values that you'll want to use depend on what else
>will be done with the scaled signal. If the circuit that follows has
>a high input impedance (i.e. it will draw very little current from the
>voltage divider) then the only consideration is the load that the
>divider will present to the analog source.<...>
Something like this?
+/-10V Input +5V
| |
100K 10K
| |
|-------------|------> to BX24 Analog Input
33K |
| 10K
Gnd |
Gnd
Paul
>
"I've seen the future, and you're not going to like it."
-- Vincent van Gui