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The input signal has a range of 1.15 - 0.45 = 0.70 volts. The corresponding
output needs to be 10 - 0.3 = 9.7 volts. The gain must then be
9.7 / 0.7 = 13.86
An additive offset must now be calculated. If an input of 0.45 V gives
an output of 0.3 V and the gain is 13.86, then what is the output when
the input is zero?
Vo = Offset + 13.86 * Vin
0.3 = Offset + 13.86 * 0.45
Offset = 0.3 - 13.86 * .45 = 0.3 - 6.237 = - 5.937 V

The standard non-inverting amplifier uses two resistors: R1 from the
output of the amplifier to the inverting input, and R2 from the inverting
input to ground. The gain determines the ratio between the two resistors:
Gain = (R1 + R2) / R2 = 1 + (R1/R2) = 13.86

Either R1 or R2 can be chosen arbitrarily and the other can be found
from the above equation. Of course, the initial value is only arbitrary
in a theoretical sense; practical values will depend on amplifier
bias current and overall power consumption.

Now that the gain is taken care of, the offset must be added in. The
usual place to add this is between the bottom of R2 and ground. A
voltage source must be placed here such that with the input (the
non-inverting input of the amp) grounded, the output will be - 5.937V.
Since the supply is positive only, you will never see a negative
output, but that doesn't matter. The output will saturate if the input
is zero. It only matters that the output will be 0.3 V when the input
is 0.45V.

The amplifier seen from the low end of R2 looks like an inverting
amplifier with input resistor R2, feedback resistor R1, and the
amp + input grounded. The voltage that is needed to be applied
to the low end of R2 is then

- 5.937 = Vin * (- R1 / R2)
But (R1 / R2) = 13.86 - 1 = 12.86, so
- 5.937 = -12.86 * Vin, or
Vin = (- 5.937 / -12.86) = 0.461V

A voltage divider across +10V that gives 0.461V output will
work. This assumes that R2 does not load down the output
of the divider. This issue can be accomodated by seeing that
the divider and R2 (consisting of 3 resistors total) can be
replaced by its Thevenin equivalent circuit. In other words,
a divider from +10V to ground can have its output connected
to the inverting input directly, eliminating R2, if the resistance
looking into the divider (value of its 2 resistors in parallel) is R2
and the open circuit output is 0.461V. This completely
determines both resistors in the divider with no ambiguity. If
this is too much trouble, just make sure that the lower
resistor in the offset divider is much smaller than R2.

John Power

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John,

Thank you so much for the explanation. If I understand your email
correctly the circuit should look something like the attached .bmp.

R1=9K, R2=700
(9000/700)+1=13.86

Steve

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part 2 12163 bytes content-type:image/bmp; name="op_amp.bmp" (decode)   At 07:04 AM 4/8/2004 -0700, you wrote:
>John,
>
>Thank you so much for the explanation. If I understand your email
>correctly the circuit should look something like the attached .bmp.
>
>R1=9K, R2=700
>(9000/700)+1=13.86
>
>
>Steve

The offset voltage is wrong and the resistor values are rather
low.

Try this instead (view in fixed-width font such as Courier):

Rs     +12
___    |\|
o--|___|---|+\
|  >------+---o
+-|-/       |
| |/|       |
|  ===      |
R1   |  GND      |
___  |  ___      |
+12 -|___|-+-|___|-----+
|    R3
.-.
| |
| |
R2 '-'
|
|
===
GND

Rs = 6K49
R1 = 182K + 1K00
R2 = 7K32
R3 = 88K7 + 1K78

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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