Hi,

I'm using a CMOS 555 astable timer re Ben Eater 555 circuit (that is fairly standard I assume) that generates a slow 5-10Hz clock. It looks fine on LED.

I have now purchased a second hand Tektronix TBS 1052B oscilloscope to replace the FNIRSI one I wasted money on.

I have calibrated the probe on Channel 1 (note noise!!!?)

I have set probe to x10 and coupling to DC and get a nice square wave

I now change coupling to x10 AC and get

I don't understand why I'm getting the difference. I would expect the same waveform? I'm obviously doing something stupid!

 

Hi,

I'm using a CMOS 555 astable timer re Ben Eater 555 circuit (that is fairly standard I assume) that generates a slow 5-10Hz clock. It looks fine on LED.

I have now purchased a second hand Tektronix TBS 1052B oscilloscope to replace the FNIRSI one I wasted money on.

I have calibrated the probe on Channel 1 (note noise!!!?)

I have set probe to x10 and coupling to DC and get a nice square wave

I now change coupling to x10 AC and get

I don't understand why I'm getting the difference. I would expect the same waveform? I'm obviously doing something stupid!

This comes from a misunderstanding of AC coupling. This method blocks DC from getting to the vertical amplifier by introducing a capacitor in the signal path. This will remove the DC component and depending on the frequency, show a distorted version of the waveform. Why would you even be tempted to do this?

 

hi gf,
That shift to above and below 0V is what you would expect with AC coupling capacitor.
All is OK.
E

 

The AC input is capacitively coupled, which necessarily creates a high pass filter. At your very low frequency, the waveform will be altered to have sloping lines as shown. At higher frequencies, this will not happen. It all comes down to how much the capacitor is charged/ discharged in a half cycle.

 

What is AC coupling?

AC coupling introduces a high-pass filter. This is also a differentiator. The output is the mathematical derivative of the input.



From the slope of the decay you can estimate the RC time-constant of the high-pass filter.

 

This comes from a misunderstanding of AC coupling. This method blocks DC from getting to the vertical amplifier by introducing a capacitor in the signal path. This will remove the DC component and depending on the frequency, show a distorted version of the waveform. Why would you even be tempted to do this?

Just ignorance

 

Thankyou all so much for the clear and now understandable explanation. I really appreciate your help.

Edit: When would I want to use AC coupling? I suppose at higher frequencies it saves having to move the waveform down if there is a lot of dc in the input, but is there any other reason(s)?

 

Thankyou all so much for the clear and now understandable explanation. I really appreciate your help.

Edit: When would I want to use AC coupling? I suppose at higher frequencies it saves having to move the waveform down if there is a lot of dc in the input, but is there any other reason(s)?

You need to use AC coupling when the AC signal is on top of a DC voltage much larger that the AC signal.

For example, say you are looking at a 10mV AC signal at the input if an amp that is biased at 2V DC. If you use DC coupling you will see just a small wiggle because the 2V DC offset prevents you from increasing the gain to enlarge the signal. You will probably have the scope set at about 1V / division.

Now use AC coupling and the DC part is gone so you can turn up the gain to 5mV / division and see the AC sigbal in all its detail.

 

Somewhere in the scope or probe data is the value of the AC coupling capacitor, possibly expressed as a bandwidth or corner frequency. One octave above that corner freq, the response should be pretty flat.

The coupling cap will have a voltage rating. In the past, this was routinely something like 600 V, so you could look at a tube amplifier's plate signal of a few volts of audio riding on hundreds of volts of B+. These days, I've seen low cost digital scopes with (to me) frighteningly low AC-coupling voltage ratings. Something to drill down on when shopping.

ak

 

Also, it is important to have a clear understanding of what we mean when we refer to the terms DC and AC.

In initial usage, DC stands for Direct Current, meaning that the current flows in the same direction.
AC stands for Alternating Current, i.e. the current changes directions.

The DC/AC option on an oscilloscope has a totally different interpretation.
In signal theory, all signals are DC + AC. In other words, all electrical signals can have frequency components from 0Hz to to ∞Hz.
In theory, there is no such thing as 0Hz since we would have to analyze the signal for an infinite amount of time.

From the perspective of the oscilloscope input option, DC means DC + AC input.
The AC option attenuates the DC component (and up to a defined low frequency) and only pass AC (or high frequency) components of the signal.

 

If you want to explore this further, put the scope in DC and see what happens when you put different value capacitors in series with your measurement.

 

I would even suggest that you use this circuit and try different combinations of C and R.






For further study, try the RC Low Pass Filter circuit.
Whereas the high pass filter performs mathematical differentiation, conversely, the low pass filter performs mathematical integration.

 

Hi,

I'm using a CMOS 555 astable timer re Ben Eater 555 circuit (that is fairly standard I assume) that generates a slow 5-10Hz clock. It looks fine on LED.

I have now purchased a second hand Tektronix TBS 1052B oscilloscope to replace the FNIRSI one I wasted money on.

I have calibrated the probe on Channel 1 (note noise!!!?)

I have set probe to x10 and coupling to DC and get a nice square wave

I now change coupling to x10 AC and get

I don't understand why I'm getting the difference. I would expect the same waveform? I'm obviously doing something stupid!

Consider what you would expect to see if you applied a constant DC voltage of, say, 10 V and AC coupled the scope. You would expect to see 0 V because the only signal present is a DC voltage.

Now consider what would happen if you have a a voltage of 10 V for a long time, and then you switched it to 20 V and left it there. You would expect to see 0 V up until the switching event, and you would expect to see 0 V long after the switching event. But what about AT the switching event? You would see the signal jump up 10 V, but now how is it going to get to that 0 V that you expect it to be a long time from now? It's going to slowly decay to 0 V.

That is going to be true for all signals -- they will show quick changes but then start decaying toward 0 V. If we have nothing but quick changes, we won't see that decaying behavior. It's still there, but it gets lost in changes we are seeing.

So how "quick" is a quick change and how "slow" is the decay?

That is a function of the capacitor that is introduced into the circuit when you turn on AC coupling. This capacitor interacts with the probe resistance to form an RC high-pass filter that has a certain time constant. Your waveform with the slanted tops is slow enough to let you measure an reasonable estimate of that time constant. Measure the amount of time it takes for the waveform to decay to ~37% of it's initial value after a clock edge and that gives you the time constant. If you were to slow the waveform down another factor of two or three, you could get a very good measurement of it. You can turn that time constant, τ, into a frequency by using

f = 1/(2πτ)

Any component in your signal that is at a frequency of less that three to five times that frequency is going to be significantly distorted.

So why use AC coupling at all?

Often times we have a small signal riding on top of a large DC offset. If we want to zoom in on the small signal, we quickly push the entire thing off the scope display. So we AC-couple the scope so that we can increase the sensitivity of the vertical channel while keeping the signal on the display.

 

When would I want to use AC coupling?

One application would be measuring noise/ripple on the output of a power supply.