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I made a traditional PWM signal generator using a 555 timer IC.

enter image description here

However, when I connect the output (pin 3 of the 555) to my oscilloscope and set it to AC coupling, the signal shown on the display is around -5V and 4V.

I know that I should set my oscilloscope to DC coupling but I'm wondering why there is an AC signal since I powered the circuit with a DC source. (9v battery)

enter image description here

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  • \$\begingroup\$ you probably have not balanced your probe... -0.5V offset \$\endgroup\$ – JonRB Jan 1 '18 at 17:50
  • \$\begingroup\$ as to the question... you are making an AC signal, a squarewave \$\endgroup\$ – JonRB Jan 1 '18 at 17:50
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    \$\begingroup\$ Signal duty cycle suggests your signal average seems to be right around 0 . Have you got your scope input set to AC? \$\endgroup\$ – Oldfart Jan 1 '18 at 17:59
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    \$\begingroup\$ @JonRB, or the duty cycle is not 50% (based on the screen image, it's not). \$\endgroup\$ – The Photon Jan 1 '18 at 18:00
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    \$\begingroup\$ The phrasing is so weird here that I wonder what we're missing. You explicitly tell us you have built a signal generator, but still you think it's weird that you get a signal on the scope. I don't know what you expected here or why you think there should be something else. Maybe your question isn't what we think it is. \$\endgroup\$ – pipe Jan 1 '18 at 22:38
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set the AC coupling in oscilloscope

This is what setting the oscilloscope to AC coupling does: It puts a capacitor in series with the input, so that the average voltage of the signal is removed and you see only the AC components.

If you had a 0 to 9 V square wave with 55.6% duty cycle, the average voltage would be 5 V, and the AC coupling would subtract 5 V from the signal before it reached the scope's sampling circuit. So the signal you'd see on the scope would range from -5 to +4 V.

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None of the others would be happy with this circuit if their oscilloscope displayed straight horizontal line in pin 3. You have got what people normally expect from this circuit (=pulse generator, outputs rectangular pulses). Your oscilloscope shows it right. As AC coupled it subtracts the DC component(=the average voltage) from the input and shows what's left. It does not change the signal , only the display of your scope isn't the full truth because the DC component is left undisplayed.

AC mode is useful, when the DC component is big, say one volt or more and the pulses on it are only ten millivolts. In DC mode the pulses would be invisible. In AC mode one can use low Volts/Div setting regardless of several volts DC component.

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