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I have a signal coming out of a basic DAC (resistor and capacitor) as follows:

1.4V min 4.8V max ~2.8V pp

I have a bipolar supply that is -12V - 0 - +12V

What is the best way to shift this signal to +-5V (10V pp)

As I understand this can be done with a single non-inverting opamp. However being a real electronics newbie I have no idea where to start figuring this out.

Alternately maybe it would be easier (and more beneficial in terms of learning) to shift the signal to ground (-1.4V - +1.4V) and then amplify.

Not looking for an exact answer would just like to be pointed in the right direction. As in what type of non inverting opamp circuit is used to shift the voltage?

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Not looking for an exact answer would just like to be pointed in the right direction.

If it's an analogue signal that naturally settles at 3.1 volts DC when there is no signal present AND the low frequency components (as in audio below 20 Hz) are unimportant then use a simlpe RC high pass filter to remove the 3.1 volts thus leaving you with 2.8 volts p-p centred at 0 volts. Amplify-up and you're done.

If the 3.1 volt DC level is important then use a couple of resistors to re-centre the signal about 0 volts. One resistor in series with the orignal signal meeting another resistor biasing the output from the - 12 volt rail. The easiest way to implement this is with a potentiometer if you are not up-to the fairly straightforward math (or use a sim tool). Amplify-up to finish.

what type of non inverting opamp circuit is used to shift the voltage?

Pretty much the standard type with a feedback resistor and resistor to 0 volts. Don't forget 100 nF op-amp power supply rail decouplers.

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  • \$\begingroup\$ Thank you, that put me on the right track. Used a buffered voltage divider and then a simple opamp amplifier and it worked nicely! \$\endgroup\$ – someuser Mar 11 '17 at 15:48
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  • INPUT = 1.4V min 4.8V max ~2.8V pp
  • OUTPUT = +/- 5V (10V pp)

thus design adds;

  • OFFSET = -½(1.4+4.8) = -3.1V
  • GAIN = 10/2.8 = 3.6

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I"ll add it depends how accurate he want's this. If this circuit is for control purposes it might be more prudent to use a biased ZENER or other voltage reference instead of the POT. That would also give you isolation from any noise from that supply you chose Tony. If you want the system to be really accurate I'd feed the output back to the Micro, via a suitable circuit, so you can self calibrate too. \$\endgroup\$ – Trevor_G Mar 11 '17 at 14:59
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    \$\begingroup\$ Yes V1 was intended to mean any stable voltage reference. A pot is not necessary if it is accurate as long as it is Vref=1.2V or Vref*(Gain ratio)=input OFFSET for DC coupling, if needed. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 11 '17 at 15:44

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