I have been asked to convert some +/- 10V sensor signals to 0 - +5
volt for use with a Basicx 24. Is there a way to do this without using
a 3rd party signal conditioning board? I've been shown a simple
schematic that uses a resistor network to do this but I don't see how
it works.
Any help will be greatly appreciated.
TIA,
Paul Dubinsky
+/- 10V to +5V Conversion
Started by ●October 3, 2005
Reply by ●October 3, 20052005-10-03
At 07:41 PM 10/3/2005 -0000, you wrote:
>>>>
I have been asked to convert some +/- 10V sensor signals to 0 - +5
volt for use with a Basicx 24. Is there a way to do this without using
a 3rd party signal conditioning board? I've been shown a simple
schematic that uses a resistor network to do this but I don't see how
it works.
<---Sounds like RS232 to TTL conversion to me. It isn't analog, is it?
Ken
------
President and CTO - Arcom Communications
Makers of state-of-the-art repeater controllers and accessories.
http://www.ah6le.net/arcom/index.html
We offer complete Kenwood TKR repeater packages!
We are now an authorized Telewave Dealer!
AH6LE/R - IRLP Node 3000
http://www.irlp.net
>>>>
I have been asked to convert some +/- 10V sensor signals to 0 - +5
volt for use with a Basicx 24. Is there a way to do this without using
a 3rd party signal conditioning board? I've been shown a simple
schematic that uses a resistor network to do this but I don't see how
it works.
<---Sounds like RS232 to TTL conversion to me. It isn't analog, is it?
Ken
------
President and CTO - Arcom Communications
Makers of state-of-the-art repeater controllers and accessories.
http://www.ah6le.net/arcom/index.html
We offer complete Kenwood TKR repeater packages!
We are now an authorized Telewave Dealer!
AH6LE/R - IRLP Node 3000
http://www.irlp.net
Reply by ●October 3, 20052005-10-03
At 03:45 PM 10/3/2005, you wrote:
><...>
>
><---Sounds like RS232 to TTL conversion to me. It isn't analog, is it?
That's exactly what is - an analog -10 to + 10 voltage voltage from
an analog sensor.
Paul "I've seen the future, and you're not going to like it."
-- Vincent van Gui
><...>
>
><---Sounds like RS232 to TTL conversion to me. It isn't analog, is it?
That's exactly what is - an analog -10 to + 10 voltage voltage from
an analog sensor.
Paul "I've seen the future, and you're not going to like it."
-- Vincent van Gui
Reply by ●October 3, 20052005-10-03
--- 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.
After you effect the divide-by-four, you'll have a signal with a range
of -2.5 to +2.5. The remaining task, then, is to level-shift the
signal so that its range is 0 to +5. This is a simple matter of
adding +2.5 volts to the divided signal. The 2.5 volts is easily
obtainable from an existing 5V supply by dividing by 2. Depending on
other factors in your system, you'll probably want to/need to filter
the 5 volts or use a separate supply.
You can "add" the two voltages together using a resistor summer.
The "resistor network" is that you mentioned probably comprises the
two dividers and the summer. The biggest problem in designing such a
circuit is being able to select suitable values for each stage so that
they don't interfere with each other.
It's easier to achieve your objective if you include an op amp in the
circuit because its high input impedance and low output impedance
simplifies the choice of component values in the other portions. If
you want to use an op-amp based solution, you might benefit from
getting "Op Amps for Everyone" available at
http://focus.ti.com/lit/an/slod006b/slod006b.pdf
Don
> [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.
After you effect the divide-by-four, you'll have a signal with a range
of -2.5 to +2.5. The remaining task, then, is to level-shift the
signal so that its range is 0 to +5. This is a simple matter of
adding +2.5 volts to the divided signal. The 2.5 volts is easily
obtainable from an existing 5V supply by dividing by 2. Depending on
other factors in your system, you'll probably want to/need to filter
the 5 volts or use a separate supply.
You can "add" the two voltages together using a resistor summer.
The "resistor network" is that you mentioned probably comprises the
two dividers and the summer. The biggest problem in designing such a
circuit is being able to select suitable values for each stage so that
they don't interfere with each other.
It's easier to achieve your objective if you include an op amp in the
circuit because its high input impedance and low output impedance
simplifies the choice of component values in the other portions. If
you want to use an op-amp based solution, you might benefit from
getting "Op Amps for Everyone" available at
http://focus.ti.com/lit/an/slod006b/slod006b.pdf
Don
Reply by ●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
>--- 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
Reply by ●October 3, 20052005-10-03
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
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 ●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,
Something more like:
+/-10V Input
|
301K
|
|--------------------> to BX24 Analog Input
100K
|
|
+5V |
| |
10K |
| |
|-----|
|
10K
|
Gnd
Cheers,
lauren
>--- 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,
Something more like:
+/-10V Input
|
301K
|
|--------------------> to BX24 Analog Input
100K
|
|
+5V |
| |
10K |
| |
|-----|
|
10K
|
Gnd
Cheers,
lauren
Reply by ●October 3, 20052005-10-03
Of course. Thanks for helping an old dummy.
Thanks,
Paul
At 06:18 PM 10/3/2005, you wrote:
>At 05:17 PM 10/3/2005, you wrote:<...>
>Paul,
>
>Something more like:
>
> +/-10V Input
> |
> 301K
> |
> |--------------------> to BX24 Analog Input
> 100K
> |
> |
> +5V |
> | |
> 10K |
> | |
> |-----|
> |
> 10K
> |
> Gnd
>
>Cheers,
>lauren
>
"I've seen the future, and you're not going to like it."
-- Vincent van Gui
Thanks,
Paul
At 06:18 PM 10/3/2005, you wrote:
>At 05:17 PM 10/3/2005, you wrote:<...>
>Paul,
>
>Something more like:
>
> +/-10V Input
> |
> 301K
> |
> |--------------------> to BX24 Analog Input
> 100K
> |
> |
> +5V |
> | |
> 10K |
> | |
> |-----|
> |
> 10K
> |
> Gnd
>
>Cheers,
>lauren
>
"I've seen the future, and you're not going to like it."
-- Vincent van Gui
Reply by ●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.
Don
> 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.
Don
Reply by ●October 4, 20052005-10-04
--- 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
> 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
