I've built a DC signal mixer that doesn't quite work as expected. The supply voltage is 9V, virtual ground at 4.5V and the mixer uses a TL074 quad op amp. While the input voltage is 0-9V, the output voltage is 1.3-9V. How can I remove that 1.3V offset? I tried biasing the input and the output with no success so far.
Any ideas why? I need my signal to 'sit'at 0V at minimum.
If your supply is 9V and you have created a midrail at 4.5V that is a good start however, to expect any op-amp to produce an output voltage that drops down to the 0V rail is naive. OK that's being a little harsh for newbies in electronics.
If you look at page 7 of this document, about halfway down is the spec of the peak voltage swing and typically this is +/-13.5V. Note the power supply voltage is +/-15V and what it's telling you is that the op-amp cannot swing to within 1.5V of the power rails.
You can get better op-amps - these are usually referred to as "rail to rail" devices because they can swing their outputs (and usually their inputs) to very close to both supply rails. Make sure you pick one that can operate from your 9V supply - a lot of them are intended for 5V applications.
The LT1677 is a good example - it can swing its outputs to about 200mV of either supply but there are probably better ones that you will find that achieve maybe 50mV.
If you really need 0.0mV swing on the output then you should consider haveing a negative supply.
The issue is that TL07x is not a rail-to-rail opamp... From the documentation from Texas instruments:
As you can see, the single 9V supply is +-4.5V here, translates roughly to +-3V output, which just about describes the situation you're in.
Andy aka beat me on this... I blame IrfaneView :)
