1
\$\begingroup\$

I want to offset an input sine wave, while being able to adjust the input voltage.

For example a sine wave of +-3 V should be offset by 1.5 V such that the sine wave never goes below 0 V. However I want the same circuit to be able to offset correctly +-2 V by 2 V, such that the lowest point of the negative cycle (-2 V) would remain exactly at 0 V once the offset is introduced. The same goes for +-1 V with an offset of 1 V etc.. I am looking to be able to offset 0 - 3 Vpp input range.

I have thought of connecting a variable resistor as a voltage divider across a well regulated voltage, and buffer the output by using a follower amplifier, however I do not wish to stay doing this manually.

What would be a good approach to go about implementing the above? Any ideas would be appreciated.

\$\endgroup\$
  • 2
    \$\begingroup\$ How precise does it need to be? en.wikipedia.org/wiki/Clamper_(electronics) shows some simple circuits that may work. \$\endgroup\$ – vofa Jan 15 '18 at 17:50
  • 1
    \$\begingroup\$ @vofa, with a little more detail, that would be an answer rather than a comment. \$\endgroup\$ – The Photon Jan 15 '18 at 17:54
  • 1
    \$\begingroup\$ If you want the RMS value you might be going about this the wrong way. This sounds like an XY problem to me. \$\endgroup\$ – Andy aka Jan 15 '18 at 18:03
  • 2
    \$\begingroup\$ Still sounds like an XY problem to me. Are you aware that a diode clamp will still generate a small negative voltage circa -0.5 volts. If you are trying to calculate RMS then it's better to take the AC through a fixed dc offseting circuit. Then you know the DC offset and the peak is what your MCU measures minus the offset. I am trying to guide you to make the best choice. \$\endgroup\$ – Andy aka Jan 15 '18 at 18:10
  • 1
    \$\begingroup\$ @Andyaka, I appreciate your helpful comments. That is surely simpler and may even provide a better solution. I will implement and test both. Regarding the original question (forgetting the RMS part), before asking this question I searched rather extensively for a solution to the above question and the answer below is what I was looking for. \$\endgroup\$ – rrz0 Jan 15 '18 at 18:27
0
\$\begingroup\$

Thanks to vofa's comments I found the below Positive Unbiased Clamp.

enter image description here

As per the referenced web page:

In the negative cycle of the input AC signal, the diode is forward biased and conducts, charging the capacitor to the peak negative value of VIN. During the positive cycle, the diode is reverse biased and thus does not conduct. The output voltage is therefore equal to the voltage stored in the capacitor plus the input voltage, so VOUT = VIN + VINpeak.

This circuit will do the job.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ The bottom of the signal will be about 0.5 volts negative using a capacitor-diode clamp. I'm not saying it isn't good enough; I'm just pointing it out. \$\endgroup\$ – Andy aka Jan 15 '18 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.