I have a 2.5V DC output signal ( its range is 0.5V to 4.5V), it needs to be scaled from 25% of supply to 75% of supply voltage. The supply voltage is a variable.Due to constraints i cannot use microprocessors or other fancy signal processing parts, need to stick with basic parts like opamps , resistors bridges ect. 0.5V output needs to be scaled to 25% of supply voltage and 4.5V output output needs to be scaled to 75% of supply voltage. Essentially the output needs to be ratiometric to supply voltage. Is there a way to do it with basic circuit like opamp etc?. If easier to explain please draw it if needed.Thank you.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
scaling and offsetting analog output 1
- Thread starter E2005
- Start date
I think I've seen something like this in a load cell resistive bridge. Though in my case I took great pains to make the supply voltage as close to a steady 5.0 Volts as possible. Allowing the supply voltage to fluctuate makes the instrument amp and the analog-digital conversion more difficult to accomplish.
Would you be able to install a high-accuracy voltage regulator in the sensor, that allows it to accept input voltages up to 20V, while regulating the voltage across the sensor's element a constant 5V? It would look a little like the attachment below...
As for the other separate requirement, it took me a while to tease the two of them apart, given how they are related to one another. Does the sensor have to be referenced to the same ground as the supply voltage? Could you create a false ground for the purpose of this sensor? The reason I ask is that the upper and lower limits you stated for output (0.5v and 4.5v) aren't starting out a 25% and 75% of the 5V base supply. So the slope of the Y=Mx+B lines going from the supply voltage's 25% and 75% ratios aren't going to be the same, so you add a lot of complexity. Maybe you can fool the sensor by making it think that ground is some other voltage, causing the output to be the sum of the "high ground" voltage plus the lower scale sensor output voltage. And you choose that to equal 25% of the supply voltage.
Then all you need for the scaling with respect to the upper limit of the supply voltage is a resistive voltage divider with points at 25% and 75% to pick from. I think I'm over my head now, but maybe you can see the other side of the river from here.
STF
Would you be able to install a high-accuracy voltage regulator in the sensor, that allows it to accept input voltages up to 20V, while regulating the voltage across the sensor's element a constant 5V? It would look a little like the attachment below...
As for the other separate requirement, it took me a while to tease the two of them apart, given how they are related to one another. Does the sensor have to be referenced to the same ground as the supply voltage? Could you create a false ground for the purpose of this sensor? The reason I ask is that the upper and lower limits you stated for output (0.5v and 4.5v) aren't starting out a 25% and 75% of the 5V base supply. So the slope of the Y=Mx+B lines going from the supply voltage's 25% and 75% ratios aren't going to be the same, so you add a lot of complexity. Maybe you can fool the sensor by making it think that ground is some other voltage, causing the output to be the sum of the "high ground" voltage plus the lower scale sensor output voltage. And you choose that to equal 25% of the supply voltage.
Then all you need for the scaling with respect to the upper limit of the supply voltage is a resistive voltage divider with points at 25% and 75% to pick from. I think I'm over my head now, but maybe you can see the other side of the river from here.
STF
Assuming that the sensor's output ALWAYS varies between 10% and 90% of the supply voltage, a basic resistance divider can be built like this:
[tt]
o -- +Vss (in)
|
[] R1
|
o-----------------o Sensor Vss
|................ |
[] R2 `````````` [ ]o----------------o Sensor output
|```````````````` |
o-----------------o Sensor Gnd
|
[] R3
|
o -- Gnd (input)
[/tt]
(ASCII art is such fun, especially when you can't use more than one space at a time).
So pick R1, R2 and R3. R1 has to offset the Vss down so that the sensor's Vo can't be higher than 75% and R3 has to offset the GND up so that the sensor's Vo can't be lower than 25%. Since the natural range of Vo is between 10% and 90% of Vss, then R1 needs to offset Vss down enough that the Vo output ends up at 75% of Vss, and it seems like R3 needs the same value to do the reverse to Gnd. R2 needs a value to drop the intervening fraction of the Vss between them.
Example:
Vss = 15.0 Volts
Vo (max) = 75% * 15V = 11.25V
Vo (min) = 25% * 15V = 3.75V
Full swing = Vo (max) - Vo (min) = 11.25v - 3.75v = 7.5v
R1 = 187 ohm (Voltage drop is 2.8125 volts)
R2 = 750 ohm (Voltage drop is 9.375 volts)
R3 = 187 ohm (Voltage drop is 2.8125 volts)
total 1000 ohm (current through divider is 15 mA)
Sensor supply voltage is 9.375V
90% of 9.375V = 8.4375V
10% of 9.375V = 0.9375V
Full swing = 8.4375v - 0.9375v = 7.500v (confirmed)
I doubt that the sensor can be driven with only 15mA, I was just starting with even numbers. You must scale the resistance values down as necessary to allow a suitable power supply. Then a further adjustment will be needed to account for the sensor's OWN power drain from R1 to R2, which will throw off the scale calculations above if it is a significant fraction of the current through the divider.
STF
[tt]
o -- +Vss (in)
|
[] R1
|
o-----------------o Sensor Vss
|................ |
[] R2 `````````` [ ]o----------------o Sensor output
|```````````````` |
o-----------------o Sensor Gnd
|
[] R3
|
o -- Gnd (input)
[/tt]
(ASCII art is such fun, especially when you can't use more than one space at a time).
So pick R1, R2 and R3. R1 has to offset the Vss down so that the sensor's Vo can't be higher than 75% and R3 has to offset the GND up so that the sensor's Vo can't be lower than 25%. Since the natural range of Vo is between 10% and 90% of Vss, then R1 needs to offset Vss down enough that the Vo output ends up at 75% of Vss, and it seems like R3 needs the same value to do the reverse to Gnd. R2 needs a value to drop the intervening fraction of the Vss between them.
Example:
Vss = 15.0 Volts
Vo (max) = 75% * 15V = 11.25V
Vo (min) = 25% * 15V = 3.75V
Full swing = Vo (max) - Vo (min) = 11.25v - 3.75v = 7.5v
R1 = 187 ohm (Voltage drop is 2.8125 volts)
R2 = 750 ohm (Voltage drop is 9.375 volts)
R3 = 187 ohm (Voltage drop is 2.8125 volts)
total 1000 ohm (current through divider is 15 mA)
Sensor supply voltage is 9.375V
90% of 9.375V = 8.4375V
10% of 9.375V = 0.9375V
Full swing = 8.4375v - 0.9375v = 7.500v (confirmed)
I doubt that the sensor can be driven with only 15mA, I was just starting with even numbers. You must scale the resistance values down as necessary to allow a suitable power supply. Then a further adjustment will be needed to account for the sensor's OWN power drain from R1 to R2, which will throw off the scale calculations above if it is a significant fraction of the current through the divider.
STF
OperaHouse
Electrical
Maybe I'm missing something, but the only viable solution is a microcontroller. It is cheap simple and easy. It may not even need a buffer if the load is light. With the multiple AD inputs you can also monitor the output and correct. An additional PWM output can add an extra bit and be used to adjust the output for load. No calibration needed. It is a foregone conclusion that an analog multiplier is not realistic from a cost and accuracy standpoint.
- Status
Similar threads
- Locked
- Question