The ABCs of Electronic Instruments (I) – Is the AC coupling of an oscilloscope used correctly?

The AC coupling of an oscilloscope is something you learn on the first day of learning oscilloscopes. I thought so at first, but it was not until a difficult customer problem was put in front of me that I began to think deeply about this topic.

Here’s a free question first: What is the lower cutoff frequency of the AC coupling of the oscilloscope?

This is easy, just check the data sheet and you will know. For example, the one I have is 5Hz.

Next question: In the case of AC coupling, how low a frequency signal can be observed?

This is also simple, just make it larger than 5Hz.

When a 6Hz 200mVrms sine wave is input, the measured amplitude is only about 145mVrms, which is 3dB attenuated. The measurement amplitude error is large, so it seems that this answer is not correct.

Uh... then 50Hz or above is enough, 10 times higher than the cutoff frequency, so it shouldn't have any impact at all.

Indeed, when observing a 60Hz 200mVrms sine wave, the measured amplitude is also about 200mVrms, with basically no attenuation, but this answer can only be given 60 points.

What's the matter with the remaining 40 points? Let's try a 60Hz square wave. No one would think this is a square wave! But the effective value is still very accurate, about 200mVrms.

So the correct answer is to analyze the specific situation (it’s the same as saying nothing, but this is the reality).

After calibrating the skew of the two channels of the oscilloscope, the phase shift of the 60Hz sine wave under DC and AC coupling is compared. The phase shift is close to 6°, and the group delay is 270us.

Continuing to increase the frequency, the phase shift is already very small at 600Hz, but the group delay is close to 1us, which is still too large. At 6kHz, both the phase shift and the group delay are negligible.

That is to say, if you want to observe a square wave using AC coupling, the frequency of the square wave must be higher than 600Hz so that no obvious deformation will be seen (600Hz vs. 60Hz and 5Hz).

This is what is called specific analysis based on specific circumstances. The key lies in what kind of waveform needs to be observed.

For a single-tone sinusoidal signal, the phase shift is not important as long as the amplitude is accurate enough (unless you are measuring the phase difference between two sinusoids, one DC-coupled and one AC-coupled).

For signals with complex frequency components, such as square waves, phase accuracy is more important. If a system has different phase shifts at the fundamental and harmonics, the observed waveform may be distorted. However, not all complex signals are sensitive to phase. For example, random noise is almost unaffected by phase.

If we are concerned about the effective value (power) of the signal and are not interested in the shape (or peak) of the signal, then it is sufficient as long as the lowest frequency component of the signal is almost not attenuated, and there is no need to worry about phase shift.

What if I want to observe a low-frequency signal superimposed on DC, but cannot use AC coupling?

This is where the DC offset of the oscilloscope comes in. For example, if AC coupling is used for the signal below, it will be severely distorted.

If DC coupling is used in conjunction with a DC offset, the signal can be observed without distortion.

However, this DC offset function has its limits. The offset is not too large, and it has a certain correlation with the gear position. In many cases, the signal cannot be moved to the center of the screen.

At this time, you need to consider adding external circuits to solve this problem, or consider whether there is a more suitable instrument to complete the task (such as a multimeter with a trend chart function and a high sampling rate).

This leads to a question: How far does the influence of a zero point (or pole) extend?

Try to simulate a first-order RC high-pass filter, set the cutoff frequency to 5Hz, the phase shift less than 1° needs to be greater than 300Hz, the group delay less than 1us needs to be higher than 800Hz, and the amplitude attenuation less than 0.1dB only needs about 30Hz.

In this case, why don't oscilloscope manufacturers make the cutoff frequency of AC coupling very low? In this way, the phase shift can be ignored at lower frequencies.

Engineering is the art of compromise. The lower the cutoff frequency of the high-pass filter, the slower the filter responds to DC signals. For a first-order RC high-pass filter,

For   , it takes about 147ms for the DC signal to decay to 1%. If   , it takes 733ms.

That is to say, if the AC coupling cutoff frequency is set to 1Hz, it takes about 1s for the AC coupling to respond to the external DC signal. When the oscilloscope changes some gear settings (the gear where the relay rings), it also takes about 1s to respond to see the waveform, and the oscilloscope will feel a bit slow. If the cutoff frequency is further reduced, it will be too slow to be used normally. However, the cutoff frequencies of 1Hz and 5Hz do not have a significant impact on the phase response, so it is not cost-effective no matter how you calculate it.