I was measuring voltage on zener diode and found unclear behavior of oscilloscope. So I have function generator which puts saw wave(or sine, doesn't matter) of 20 Vpp. It goes through 1k resistor to 5.1 zener diode. Like on the picture.
enter image description here
If I probe the diode with DC coupling set on oscilloscope channel - I get expected waveform i.e. voltage cuts on around -0.8 Volt when forward biased and around 5.1 when reverse biased.
But when I use AC coupling on the channel, the waveform looks weird and doesn't maen anything useful at all - now it swings from -2.4 to 3.8 - which is not what you expect from this circuit.
So I get that the waveform we got is not the AC, but rather DC biased AC(though we still have -0.8 volts). But what happens with AC coupled measurement on oscilloscope? Why does it take such a weird shape? Where -2.4V and 3.8V came from?

P.S. Should stop asking silly questions in the middle of the night... AC coupling just dropped the "DC bias" to zero... duh...


1 Answer 1


They come from the RC constant of the high-pass filter created by your circuit and the AC coupling of the scope itself. If you look carefully you'll notice that it still swings the same amount (3.8V-(-2.4V)=5.2V), but the waveform loses definition due to the filter.

  • \$\begingroup\$ But what portion of negative wave required for signal to be not affected by AC coupling(i.e. be genuine AC signal)? I mean if I would have signal of -2 to +5.1 volts(lets suppose zener with high forward bias) - would it be affected the same way? What about -2.5 to +5.1? -3 to 5.1? Where is the line? I guess it is different for every oscilloscope, but what is usual? \$\endgroup\$ Sep 29, 2015 at 22:11
  • \$\begingroup\$ Every signal with a frequency greater than 0Hz is an AC signal around an arbitrary DC bias. AC coupling just causes the scope to completely ignore said bias. \$\endgroup\$ Sep 29, 2015 at 22:13
  • \$\begingroup\$ If you had a sine wave signal that was evenly biased around ground (say it went to +2 volts and -2 volts), it should appear the same whether you use AC or DC coupling. A low frequency square wave would get distorted - a fast rise (or fall) on the leading edge, then droop towards 0 volts with AC coupling. \$\endgroup\$ Sep 29, 2015 at 22:21
  • \$\begingroup\$ oh, my comment was stupid, should have thought before posting it, I get it now, sorry :) Thank you. \$\endgroup\$ Sep 29, 2015 at 22:22

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