I've been playing around with getting both gain and offset on a single non-
inverting op amp and I'm not quite getting how to calculate the offset
input voltage needed to get 1.27 volts out. I'm hoping someone might know
what I'm missing.

What I attempted was to get 2.5x gain with a +1.25 volt offset - so a zero
volt input on the non-inverting input would give 1.25 volts out while a 1
volt input would give 3.75 volts out.

At first, I thought that since there was to be a 2.5x gain, I needed to add
-1.25/2.5= -0.5V to the inverting input. I used a voltage divider to create
the -0.5V. from the -5V supply.

Next I went to tackle the gain. I calculated the thevenin equivilent
resistance of the voltage divider then calculated the size of the feedback
resistor and put it all together on a breadboard.

I got the gain right - but with 0V in I get 0.750V out - which is only 1.5
times the offset voltage I supplied. Ok, so my assumption about the affect
of gain on the supplied offset voltage was wrong - I went looking on google
to see if I could find out how to calculate the voltage required: two days
ago!

To sum up: I'm using an LMC6482 CMOS amp with 10 teraohm input impedance.
Input signal is a 10k resistor to either ground or a voltage divider
putting out 1V. I can calculate the gain with Thevenin's but I cannot
figure out how to calculate the offset voltage required.

Anyone able to steer me in the right direction?

P.S. I have this working with two opamps (one chip). Its just bugging me
that I cannot figure out how to do it with a single stage.

Z wrote:
I've been playing around with getting both gain and offset on a
single non- inverting op amp and I'm not quite getting how to
calculate the offset input voltage needed to get 1.27 volts out. I'm
hoping someone might know what I'm missing.

What I attempted was to get 2.5x gain with a +1.25 volt offset - so a
zero volt input on the non-inverting input would give 1.25 volts out
while a 1 volt input would give 3.75 volts out.


You have to use a non-inverting configuration. Vref is a negative reference,
which can be derived from the negative rail as well. You can use trimmers in
series with R2 and R3 to adjust gain and offset. The equations are the sum
of currents in the inv.input node. This reminds me of homework though. :-(

|\ 1.25/R1-Vref/R3=0
o-------|+\ 3.75/R1-Vref/R3+1/R2=1
Vin | --+-----o
+--|-/ | Vout
| |/ .-.
| | |R1
| | |
| '-'
| |
+--------+
| |
.-. .-.
| |R2 | |R3
| | | |
'-' '-'
| |
=== o
GND -Vref
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

--
ciao Ban
Apricale, Italy

On 2006-08-02, Z wrote:
Z wrote in :

No, the circuit I was asking about is a buffer/gain/offset for a pH
sensor. The pH sensor is a battery that actually puts out a positive
voltage when its below neutral (acidic) and a negative voltage when it
is above neutral (base). My ADC is single supply so it cannot handle the
negative voltage.

I'm using a 2.5V reference for the ADC and the pH sensor will never put
out more than +/-450MV so I went with 2.5 times gain (+/- 1.125) and an
offset that would put neutral (zero from sensor) smack dab in the middle
of the ADC range. with the offset that would give an ADC input between
.125V and 2.375V.

The range I'm actually interested in is pH7 to pH8 and I need two
decimal places, so I'm using a 12 bit ADC. I would use a higher gain to
have 6-9 PH span the entire ADC range, but I cannot figure out how to
clamp the ADC input at zero volts. The ADC can withstand -0.5V on an
input and I considered using a small signal diode from the ADC input to
ground, but the voltage drop is larger than the -0.5V limit on the
input.

use a germanium small-signal diode: 0.2V,
or a schottky diode: 0.3V

The three amp and two amp versions work perfect but I want to learn how
to do it with one amp!

what's the output impedance of the pH sensor.
is the output ground referenced, isolated, or differential?

Bye.
Jasen

jasen wrote in :


what's the output impedance of the pH sensor.
is the output ground referenced, isolated, or differential?

Bye.
Jasen


Oops, the ADC input tolerance is -0.3V, not -0.5. I'll look for germanium
diodes though! -0.2V would work.

The output impedence is somewhere around 500K-1Meg, which is why I'm using
a 10 teraohm input impedenace op-amp. The sensor ground is simple connected
to circuit ground.

On 2006-08-03, Z wrote:
jasen wrote in :

Oops, the ADC input tolerance is -0.3V, not -0.5. I'll look for germanium
diodes though! -0.2V would work.

The output impedence is somewhere around 500K-1Meg, which is why I'm using
a 10 teraohm input impedenace op-amp. The sensor ground is simple connected
to circuit ground.

could you connect the PH sensor ground to a voltage divider instead ?

Bye.
Jasen

jasen wrote in
:

On 2006-08-03, Z wrote:
jasen wrote in
:

Oops, the ADC input tolerance is -0.3V, not -0.5. I'll look for
germanium diodes though! -0.2V would work.

The output impedence is somewhere around 500K-1Meg, which is why I'm
using a 10 teraohm input impedenace op-amp. The sensor ground is
simple connected to circuit ground.

could you connect the PH sensor ground to a voltage divider instead ?

Bye.
Jasen


I've considered that, but I don't know enough about batteries to know if
that would damage the sensor or not.

On 2006-08-04, Z wrote:
jasen wrote in
:

On 2006-08-03, Z wrote:
jasen wrote in
:

Oops, the ADC input tolerance is -0.3V, not -0.5. I'll look for
germanium diodes though! -0.2V would work.

The output impedence is somewhere around 500K-1Meg, which is why I'm
using a 10 teraohm input impedenace op-amp. The sensor ground is
simple connected to circuit ground.

could you connect the PH sensor ground to a voltage divider instead ?


I've considered that, but I don't know enough about batteries to know if
that would damage the sensor or not.

if the only electrical connection from the ph sensor to the ADC if through
those two wires it should me OK. assuming there's no electrical connection
from the ADC to the liquid being tested...

Bye.
Jasen