Applying DC offset to signal line

[SchoolKid]

I have a signal line AC waveform which goes something like -1.5 to +1.5V. Shown above as point "B". I've took it form the existing waveform switch (ramp, pulse, triangle, noise). Red color shows points where i "tapped in" my modifications. I need to DC offset it, so that it remains positive only. For that purpose i prepared 1.5V DC (shown in image above as "A").

What would be the most simple way to combine A + B?

In other words, to apply DC offset of 1.5V to AC signal. Of course without affecting the functionality of existing circuity.

Thanks!

 

[crutschow]

The question is: (1) Do you actually want 1.5V offset exactly, or (2) Do you just want the most negative part of the signal to be at 0V?

For (1), you would add DC offset with an op amp or other method depending upon the signal frequency. For (2), you would use a clamp circuit that clamps the negative part of the signal at ground.

 

[SchoolKid]

The question is: (1) Do you actually want 1.5V offset exactly, or (2) Do you just want the most negative part of the signal to be at 0V?

For (1), you would add DC offset with an op amp or other method depending upon the signal frequency. For (2), you would use a clamp circuit that clamps the negative part of the signal at ground.

Thank you sir!

I will go with the clamp circuit. It is exactly what i need.

 

[crutschow]

A simple, low offset clamp can be made with a single bipolar transistor (PNP for negative clamp). Connect the transistor collector to ground. Connect a resistor to the emitter to make an emitter follower circuit. Connect the other end of the resistor to a positive supply. Connect a capacitor in series with the transistor base and a resistor from base to ground. The other end of the capacitor is the signal input.

The diode offset of the base-collector diode, which acts as the clamp, is canceled by the emitter follower offset due to the base-emitter diode. Since both diodes are in the same device they will track well with temperature.

 

[SchoolKid]

Fantastic. I believe this is exactly what i need.

Because with classic diode clamp i got (standard) 0.5V offset which in my condition was too much as can be seen below. A significant part of the waveform was left below 0V (0V = center line), yet i need the waveform to remain almost intact, or as close as possible.

 

[SchoolKid]

Connect the transistor collector to ground. Connect a resistor to the emitter to make an emitter follower circuit. Connect the other end of the resistor to a positive supply. Connect a capacitor in series with the transistor base and a resistor from base to ground. The other end of the capacitor is the signal input.

Do you mean something like this?

In which point do i pickup the signal?

 

[crutschow]

That's it except it needs to be a PNP transistor (arrow pointing towards the base).

It's an emitter follower so the output signal is at the junction of the emitter and emitter resistor.

 

[SchoolKid]

You were right. The transistor based clamp is much better. I got additional +0.3V offset, so the less of the waveform is chopped below 0V.

yellow = transistor based clamp
blue = diode based clamp

I hope this info will be useful to others as well.

Once again, thank you crutschow!

 

[crutschow]

You might play with the input capacitor and resistor sizes to see how that affects the circuit operation.

If you wanted even a more ideal clamp for low frequencies, there are also op amp based circuits which place the clamp diode inside the feedback loop for essentially zero offset.

 

[SchoolKid]

Thank you for infos. I will try the op amp based clamp - though this is very good as well.

Cheers!