1
\$\begingroup\$

I tried to scale the signal by 3 times with Av=1+R2/R1. but how do I shift it up by 1v? for example: a 0.2/2V VIN into 0.6/6V and shift it up by 1v to 1.6/7v? I managed to scale it but I don't know how to shift the Vout signal.

\$\endgroup\$
0
0
\$\begingroup\$

Vin+ controls offset. Vin- controls offset and gain.

If you split Rin=R1 into two resistors such that Thevenin Req is the same value but Veq = Voutput offset/Av then you achieve this with one Op Amp.

Do for Av-=-2 you need a Vin- offset of -0.5V or you could do the same with Vin+ .

Attenuate with pullup to attenuate and add offset so get overall ( 1+|Av-|) Vin * R3/(R3+R4) = 3Vin+1V.

That assumes you can use Vcc to get an accurate offset then amplify that + input signal to get desired transfer function.

enter image description here

Tap the switch, observe Vpp gain of 3

\$\endgroup\$
2
  • \$\begingroup\$ This is the simplest way. Any questions? The switch is just for proof . \$\endgroup\$ May 17 at 20:23
  • \$\begingroup\$ thank you, I got it. And thanks again for that website that was a lot of help \$\endgroup\$ May 17 at 21:42
4
\$\begingroup\$

Do you have a negative voltage available? If so, shift R1 input by -0.5V. This will translate to the output as:

  • Vout = (inverting input * -R2/R1) + (non-inverting input * (1 + R2/R1) )

Example:

enter image description here

More generally, if you treat your amplifier as a differential you have options to inject voltage on the (+) or (-) side to achieve the offset you want.

More about differential amplifiers here: https://www.electronics-tutorials.ws/opamp/opamp_5.html

This version works on the (+) side so doesn't need any negative voltage. enter image description here

Notice the inverting gain is adjusted to be 5, for a total non-inverting gain of 6. This is because the input is attenuated by 50% (Why? Look into how a voltage summer works), so we have a net non-inverting gain of 3.

Finally, 0.3333V (1/3 of 1V) is a weird voltage. You can work out a Thevenin equivalent for it using a voltage divider and achieve the same thing (that is, 1/3V in series with 1k.)

\$\endgroup\$
3
  • \$\begingroup\$ Duplicated...... \$\endgroup\$ May 17 at 19:34
  • \$\begingroup\$ I already worked it out \$\endgroup\$ May 17 at 20:22
  • \$\begingroup\$ Thank you i got it \$\endgroup\$ May 17 at 21:44
-1
\$\begingroup\$

You can use a second op-amp, a negative voltage reference or require the input impedance to be low and capable of sinking some current.

\$\endgroup\$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .