dc-offsetpulse
Below is a pulse train with a DC offset:
This will go to the circuit below as input.
The thing is this works well in simulation but in real I need to remove the offset of the input signal.
My question is:
How can I remove this DC offset without using another opAmp i.e. just by adding a component to the circuitry ect.? Is there an easy way?
simulate this circuit – Schematic created using CircuitLab
It could very easily just be this. A shunt 8V zener diode. I don't know your frequency or voltage parameters. Is there a power/voltage rating on this little box? Is it custom or is there a mfg. part number?
Edit: There is going to be a limiting resistor in there somewhere. If you measure the resistance from your IN to your OUT port, you might actually measure the current limiting if it's in series. Or, you could apply a known voltage and measure the current draw to complete your schematic.
You need to use meg instead of m for the resistor value. You have specified a 0.007\$\Omega\$ resistor, which will tend to reduce the gain to 1.
Your amplifier has a GBW product of 18MHz meaning it will only have an open-loop gain of about 500 at 40kHz, not the 14,000 you appear to be expecting. However it will have extremely high gain at DC so any offset will be amplified - so only a few uV will saturate the amplifier.
You also need to bias the amplifier so the inputs and outputs are somewhere in the operating range- right now it will sit at the negative rail- you want it to be about Vdd/2.
It's better to have a few AC-coupled stages with gain of ~100 or less rather than one stage which needs astronomical gain.
Best Answer
There has been a similar question recently: Solution for adding around 60V dc-offset on digital signal(0 and 5v) of 10MHz frequency
Except you want to remove some offset (not add), so the circuit is a little bit different, but the principles are exactly the same (and well described in this other answer - the only difference is that the capacitor is charged when the signal is at high level, not low).
Here it is:
That will lead to a signal that will swing between 0.5V and 5.5V (due to the diode voltage drop). The advantage compared to a RC filter is that the levels won't change depending on the duty cycle.