# Anything wrong with this level shift and amplify op amp circuit?

I posted this circuit in a question about protection clamping, but got a lot of questions around the circuit itself.

Is this not a good / the normal way to shift and amplify a low amplitude (75mv) swinging around ground signal (simulated by Vin) to Vout: 2.5V amplitude signal swinging around ca 2.5V (2.79) ?

For the sake of simplicity Vbat is a stable, regulated 12v source.

Vin is a current sensor, so Vin is practically a DC signal.

The implementation of the circuit gave me a lower than expected center (ca 1.8v) I guess that can explained by too low bias current?

simulate this circuit – Schematic created using CircuitLab

• What gain did you compute for that? Did you ground your VIN when measuring your VOUT? Have you looked at the input offset voltage for the CA3140?? – jonk Jun 9 '17 at 18:28
• If you suppress Vin and connect R4 directly to ground, Vout is ? – pasaba por aqui Jun 9 '17 at 19:13
• The gain resistors are trial and error in simulator to get 75mv = 5V(2.5), so I guess that's a gain of 33, and yes, I grounded Vin when reading the center. offset voltage should be 5 / 15 mV according to datasheet. Resistors are 1%. – ttyridal Jun 9 '17 at 19:14
• Ok. I get a gain of about 29.55, given your resistor values and assuming perfect values. $R_{TH}=\frac{R_1\cdot R_2}{R_1+R_2}, V_{TH}=V_{BAT}\cdot\frac{R_2}{R_1+R_2}, V_-=\frac{V_{OUT}\cdot \left(R_5+R_{TH}\right)+V_{TH}\cdot R_6}{R_5+R_6}, V_+=\frac{V_{IN}\cdot R_3+V_{BAT}\cdot R_4}{R_3+R_4}$ and solving $V_-=V_+$ to get $\frac{\textrm{d} V_{OUT}}{\textrm{d} V_{IN}}$. But I also now understand your question about the output voltage. It's not right. – jonk Jun 9 '17 at 19:35

I would prefer to shift the voltage with a parallel power supply. Something that resembles a zener, a shunt voltage reference. Look at something like this: http://www.ti.com/lit/ds/symlink/lm4040-n-q1.pdf

This would provide a very stable 2.5v reference. That means you don't need R1, R2 R3 and R4, eliminating errors and variance.

But anyway. If you're just trying to measure a shunt your best bet is to use an IC designed for that and you can save money and time. Take a look at this, and search for one that is better suited: http://www.ti.com/lit/ds/symlink/ina180.pdf