# AC vs DC on an Oscilloscope

I am looking at an oscilloscope (LG 3040D) that is connected to the output a half-wave rectifier. The input of the half wave rectifier is a 20Vpp at 60Hz, a stepped down voltage from the wall. I know that the signal that am I looking at is, at the output, only a DC signal - that's the whole point of a half-wave rectifier!

Since the oscilloscope has two modes (AC and DC) changing in between the modes, from what I understand, introduces a capacitor into the circuit which cuts out any constant DC voltage from the signal that is being viewed on the oscilloscope. But, since our signal is not a constant DC signal, it can still pass through the capacitor and is displayed on the screen even in AC mode.

Here comes the crux of my question: on this oscilloscope, there is a GRD feature which allows you to ground the circuit and use it as a point of reference. In DC mode, when you ground the circuit and then view the signal again, you see the signal on top or "above" the GRD reference. This makes sense to me, the half-wave rectifier is only taking the positive portion of the sine wave since the diode can only conduct in one direction, leading to only half of your input signal being "used".

But when you then put the oscilloscope into AC mode, ground the circuit and then observe the signal, it is not "above" the reference or entirely below, it is nearly centered on the signal. Why does this happen and how can it be explained? Does it matter where the signal is in AC mode? Is it supposed to be anywhere in particular with respect to the reference?

I understand this is a convoluted question that ultimately doesn't need to be discussed and comes from a lack of full understanding of how things work, but I am totally confused at the moment.

Attached is a crummy drawing in paint.

Thanks! EDIT: I have an additional question:

As a follow up question: for a half-wave rectifier, what is the "AC" voltage and how do you measure it? Additionally, What is the DC voltage and how do you measure it?

From another answer, the DC voltage is the "offset" between the AC and DC signals when viewed in AC and then DC mode. How do I then measure AC signal? Go into AC mode and measure from the peak to GND?

If I do it that way, the those values are: VDC= 2.9V and VAC=6.0V (from reference to peak). Does that seem correct? Finally, is it incorrect to talk about the voltage in this way?

## 3 Answers

A full wave signal looks like this: -

http://electriciantraining.tpub.com/14178/img/14178_117_2.jpg

It has an average value somewhere around the middle of the signal (not unexpectedly)

When you view it on the scope using "AC", that average value seen above aligns itself with the zero-volts trace position for GRD

• With the have wave rectifier, my peak value is ~ 9.2V. The reference (treating the bottom of the rectified wave as 0V) is at ~ 3.2V. So the majority of the signal is above it – metropolis Jun 3 '15 at 16:47
• ... time averaged value, 2*less signal = lower average, almost exactly half. My bad, thanks for the straightforward answer! – metropolis Jun 3 '15 at 16:55
• Average of half wave is peak/pi. If peak is 9.2 volts then average is 2.93 volts so on AC you should see 2.93 volts below 0V line. – Andy aka Jun 3 '15 at 17:00
• Yes, I took the lower portion of the rectified wave (the flat portions of the signal are not symmetrical, one is slightly higher than the other) – metropolis Jun 3 '15 at 17:28
• AC - That's more complex - you have to calculate the RMS of the waveform then subtract the DC value using Pythagoras difference of squares. The DC voltage is the average above. You need an RMS measuring voltmeter to measure the RMS AC signal. You can't take pin-points and construct an accurate evaluation. Maybe raise a new question. It's too complex to answer in comments. – Andy aka Jun 4 '15 at 7:17

In AC mode the oscilloscope displays the signal with average voltage of 0V.

It does this by using a series capacitor and a resistor to ground. The RC time constant of the resistor and capacitor is long enough that most signals won't be distorted much and fast enough that you (as a human) don't have to wait 'too long' for the display to stabilize. Below diagrams from UofW lab notes.  If you try to measure a signal such as a very low frequency square wave using AC coupling you'll see distortion (in that case, a slope towards zero after each zero crossing).

The main purpose of AC coupling is to allow you to see something like a small ripple or signal on top of a large DC voltage. For example, if you have an amplifier biased at 4V out and there is a 800uV signal present at that point, you will not be able to see it on a 1V per division setting. In fact, the signal will not even be captured by the oscilloscope ADC.

By using AC coupling, the scope can be set to 200uV per division and the signal can be displayed in all its glory. If you need to know the offset, you can switch the input to DC coupling and you'll see a flat line at 4V.

Just consider your signal as consisting of: Vdc + Vac where Vdc is the (long time) average value of your signal and Vac is the time varying part. The AC option just filters out the constant Vdc part allowing you to view the Vac separately. There are also some explanatory videos on Youtube, search for "AC DC oscilloscope".