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I have an AC signal and want to give it a new DC offset. There are some questions about my experiment results.

(The original signal bias and terminal output bias are the same because I want to know if there is the attenuation or phase shift or not.)

(yellow-TP,blue-V2)

  1. Can I apply these two circuits to provide DC bias? Why are the waveforms changed in each circuit with different capacitors?

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  1. Similar question, but this time is with different resistor. enter image description here enter image description here
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You're creating a high pass filter between your capacitor and whatever resistance you're connecting to Vref with. In the first circuit, I can't see the impedance to Vref, but it's apparently not zero. In the second circuit, it's the output impedance of U_2; again, it's unknown to me. I might be able to estimate them if I knew the frequency of your test signal.

In the 3rd circuit, the 10K with the 0.1uF gives you a corner of about 159Hz; using the 330K brings the corner down to about 5Hz. This leads me to suspect your signal is in the 50-80Hz range.

The 4th circuit pits your 0.1uF against a 50K load, giving you a corner of about 32Hz. Changing to 680K resistors brings the load up to 340K (they're effectively in parallel from a loading standpoint) and again close to 5Hz corner.

You need to pick your R and C so that the lowest frequency of interest is higher than the 3dB point of the filter; you might want to make it about 3-5 times higher if you're trying to minimize distortion.

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You are passing what appears to be 60 Hz through a 100 nF capacitor. That capacitor will have an impedance of: -

$$|X_C| = \dfrac{1}{2\pi f C}$$

Plug in the numbers and you get 26.5 kohm so, if you need very little attenuation of the AC signal the resistor value needs to be many, many times the value of 26.5 kohm.

In your last scenario the equivalent resistance is 680/2 kohm and this is much bigger than 26.5 kohm hence, it works (somewhat). In your early experiments you didn't have a resistor at all i.e. the capacitor output fed straight to Vref and there is no guarantee that Vref will have anything other than a few ohms to a few hundred ohms impedance effective series impedance.

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