# How can I convert 0-5V to +/-2V using a dual supply op amp? [duplicate]

I have a 0-5 volt variable frequency sine wave signal source I am trying to convert to +/-2V.

I would like to preserve the frequencies 20Hz-20KHz.

I have a +5 and -5 volt signals I would like to use as power supplies.

How can I level shift this signal down 2.5 volts so that it is centered at 0, and amplify the signal by 0.8 using a dual supply op-amp? • What kind of OA arrangement (schematics) did you attempt to solve this problem? – Ale..chenski Feb 15 '18 at 2:20
• You already posted the same question two weeks ago, electronics.stackexchange.com/q/352787/117785 . Why you are asking this again? – Ale..chenski Feb 15 '18 at 2:24
• @AliChen the previous question asked how to solve using a single supply op amp. I want to understand how to solve this using a dual supply op-amp and I don't know enough about op-amps to extrapolate an answer from the other question. – circuitry Feb 15 '18 at 2:37
• The DAC you were using in the original question has a built in amplifier, it still seems that you are not quite clear on what the correct question is. For reference, the question I think you really want to answer is "How do I drive a line-level audio signal from my DAC". This would prompt better answers that could offer best practice when it comes to audio outputs. – loudnoises Feb 15 '18 at 15:26

The following schematic gets rid of DC offset with a capacitor. The gain is set for 80%. You may need a 1K trim pot inbetween R1 and R2 if accuracy is important. Don't forget to add bypass capacitors to the power supply pins. A .1 uF with a 10uF in parallel should be good enough. I picked a TL072 because of its low noise.

NOTE: There are also .1% tolerance resistors on the market. Cheaper and much more stable than a trim pot. simulate this circuit – Schematic created using CircuitLab

• this is great, but it doesn't utilize my dual supply. Do you know how I would do this using +/-5 volts rail to rail? Also, here the capacitor alone is what's responsible for shifting the signal down -2.5 volts? – circuitry Feb 15 '18 at 13:54
• Pins 8 and 4 are your +/- 5 volt supply. You only need one op-amp – Sparky256 Feb 15 '18 at 14:13
• I updated the schematic to include pin numbers. – Sparky256 Feb 15 '18 at 14:38
• Thanks @Sparky256! If I used .1% resistors instead of a trim pot would pin 3 just be a 100k resistor in series with a 10 uF capacitor? Is there an equation I can look at for this type of configuration? This is what I'm having the most trouble with.. finding equations for various configurations of op-amps, and then identifying which one I need for the task. Thanks! – circuitry Feb 15 '18 at 14:40
• is the op amp in this configuration being used as a voltage follower/buffer? – circuitry Feb 15 '18 at 14:53

You can very nearly get to +/- 2V centred at 0 volts with just two resistors and the -5 volt rail: - Vout is +/- 1.6667 V.

So, if you want to avoid a capacitor (works down to DC) use the above and then amplify with a gain of 1.2 i.e. a non-iverting amplifier with 200 ohm feedback resistor and 1 kohm to ground from the inverting pin. If you can rely on your -5 volt rail being stable and fairly noise free I would consider this method.

• how would you classify this type of circuit? Dual supply resistor divider network? In your example, is +5 V2 and -5 is V1? – circuitry Feb 15 '18 at 14:56
• @circuitry V1 is the 5 volt p-p sinewave centred at +2.5 volt. V2 is 5V but with the positive terminal grounded hence the negative terminal produces -5V at the right hand side of R2. It's a potential divider. For such simple questions I might not consider my EE name to be "circuitry" lol. – Andy aka Feb 15 '18 at 15:29
• yes, that's fair – circuitry Feb 15 '18 at 15:42