Adding input offset on a non-inverting opamp.

By the way, if the 100K input impedance of the above circuit isn't good enough, take another amplifier and use it as a voltage follower to give you the high input impedance.

Good luck Chris

The gain is (10K + 20K)/20K = 1.5

"Gain" is (10K + 20K||22K)/(20K||22K)

If 10K Resistor = R1

Then (summing currents at -Input of OpAmp):

(Vo-Vin)/R1 + (Vs-Vin)/R2 = (Vs-Vin)/R3

Solving is left as an exercise for the STUDENT ;-)

(Visually, it is obvious that the output is negative when Vin=0, so your choice of resistors is incorrect.)

...Jim Thompson

The OP DID stress input-Z was critical ;-)

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

is better, because it's much less sensitive to power supply variation. The original circuit adds noise and drift from both supplies into the output. If r2

Oh, the classic bad example of this homily is...

Thanks. I caught that a little late, I guess. Possibly the OP might prefer this...

| +5V | Vin |\| Vout = 3*Vin +100mV | o---------|+\ | | >--o---------o | .-----|-/ | | | |/| .-. | | -5V | |200K | | | | | | '-' | | | | '-----------o | | | .-. | | |100K | | | | '-' | | | -33mV o-----. | | | | .-. .-. | | |33 | |5.1K | | | | | | '-' '-' | | | | === -5V | GND created by Andy´s ASCII-Circuit v1.24.140803 Beta

This is a variation on a circuit on page 6 of AN-31.

Thanks again for the spot, Mr. Thompson.

Chris

Hello! I need to make a small-gain amplifier with extremely high impedance. Because of the latter requirement, I've chosen the TL08x opamp in a non-inverting configuration:

---------------------- + | TL08x --*------------- .-- - | | | | | | -5V | | | *---/\/\/\/---' | 10K Input | Output \ / \ 20K / \ | |

-------------------*--------------------------- Gnd

The gain of the above circuit will be x3.

I would really appreciate adding some offset, so that e.g. Output=(Input+0.1)*3, but at the same time I cannot give up on the high impedance and non-inverting configuration. Also, I'd like to do it using these TL08x parts, just because I have plenty of them.

The following circuit seemed to work well (but all the equipment I had to test it was a multimeter and a voltage source):

---------------------- + | TL08x --*------------- .-- - | | | | | | -5V | | | +5V --/\/\/\/---*---/\/\/\/---' 20K | 10K Input | Output | -5V --/\/\/\/---' 22K

----------------------------------------------- Gnd

Now my problems are:

I.e. calculate the values of the 3 resistor for a formula like:

Output=(Input+0.1)*3

(or any other values)

Thanks! TPM

This is the simple version Modified schematic:

R2 ! GND --/\/\/\----

The way you drew it the difference between the +5V and -5V voltages effects the output a lot. The change I've done. Reduces the effect of the supply voltage.

Basically the answer to both is apply Ohm's law. You know that the inputs of an op-amp are at equal voltage when the op-amp is doing its thing.

Assume that the R3 is really 10K.

You know the output voltage when the input is zero.

Assume the input voltage is 0V and figure out the currents through R3

Once you have the current apply the fact that none of this flows into the op-amp so it must go through the R1 resistor. R2 must have 0V on it and thus no current in it. Apply Ohms law to find the resistor.

Now you know the output voltage when the input is 1V (Ignore any clipping)

Use Ohm's law to find the current in R1 and R3. You now know how much current flows in R2 with 1V on it so apply Ohm's law.

How mean of you to apply the Student's t-Test...

Regards Ian ;-)