I am unable to find a proper explanation for this. When I create the passive RC circuit as shown and feed it 8 us pulses at 25KHz, The output of the RC circuit drifts for some time until it settles as shown in the waveforms below. The top signal is the input and the below signal is the output of the filter.

I assume this is caused by the RC time constant of the filter?

What is the best way to prevent this problem? Is decreasing the capacitor value an only option? Will using an active filter help?

Many thanks in advance.

enter image description here

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2 Answers 2


Your source signal is in principle an AC + DC signal. Your pulse train is your AC signal, with an amplitude at a certain value, if your plot is accurate, about 220mV I'd say (half the energy up top, half below -> the virtual middle is a little below the exact middle of the signal).

It is then offset to a DC value, around which it toggles/oscillates. Again, if your plot is correct, approximately 700mV.

Your capacitor effectively blocks low frequency signals, or this DC offset, so you remain with only the AC portion, around 0.

Either your question is:

  • How come it goes negative? Then the answer is: because AC signals swing around 0V, which means they also go negative.
  • How come it takes time to go to the steady state? Then the answer is: Yes, because of the R-C time it takes to re-balance the output after you start "exciting" the input.

When you first start toggling the input, both plates of the capacitor will want to jump up, and the resistor is quite large, so it can't prevent that. Since at the start-up the output was at 0V, it jumps a full swing up on the first flank. Then after that slowly the output plate will start to balance around 0V again, with both positive and negative peaks.

You can also see this charging effect on the low and high steady periods of the signal: Small slopes that are the start of an RC curve. If your signal gets slower, you'll see more and more of that RC-charge curve.

Want it to go to the steady state faster? Smaller capacitor, or smaller resistor.

Want it to stay above 0v? You'll need to add some semiconductor action. For such small signals it's possible you need some transistor action even, but there are Schottky diodes with 50mV forward drop at very low currents. If you connect those like this:


simulate this circuit – Schematic created using CircuitLab

You may need a (small) resistor on the input side as well, but that (and its value) would depend on more specifics you haven't given.

This way, every time the output plate swings negative, the diode pushes it back up nearly to 0V.

If you want real 0V with such a small signal that'll take transistors and/or op-amps and a supply of at least 2V (or even more complication if that 2V is not available)

  • 1
    \$\begingroup\$ Take one yourself but don't inform the mods (oops). \$\endgroup\$
    – Andy aka
    Apr 29, 2016 at 11:19

When the first pulse arrives (after a period of inactivity), the capacitor does what it must do and it transfers that rising edge from one terminal to the other terminal. This is what you initially see in your 2nd diagram.

During the "on" period of that first pulse, the capacitor witnesses a short period of DC on one terminal and this causes the small decay on the output terminal. Then the falling edge of the first pulse happens and that change in voltage transfers across the capacitor. But now, the waveform has sunk below 0V and will progressively continue to sink below 0V on repeating falling edges.

So, it's all down to what happens once the rising edge has transferred and the input pulse top is flat - this is where things are happening that ultimately cause the output pulse train to have an average DC value of zero.

If the capacitance value was infinite the output pulse train would look exactly like the input pulse train. Hopefully this diagram will help: -

enter image description here

The diagram's input square wave is centred around 0V so that the output is symmetrical (unlike your pulse train) but it's the decaying output when the pulse is constantly high or low that is important to consider.

As for recommendations I stand by my proposal in the question you raised here.

  • \$\begingroup\$ Bah, I always hate it when Andy's simultaneously answer to a quality that I'm compelled to upvote! :-P \$\endgroup\$
    – Asmyldof
    Apr 29, 2016 at 11:18
  • \$\begingroup\$ @Andy aka Thanks for that. People tell me to use an active filter instead. I don't understand. I have looked up active high-pass filter configurations and the set up is the same as I have. A passive RC filter before the positive input of an op-amp using negative feedback for the gain. Isn't this a first order active high pass filter? \$\endgroup\$ Apr 29, 2016 at 12:33
  • \$\begingroup\$ A high pass fitler is a high pass filter whether it's active or passive. It does the same job. It's a red-herring. \$\endgroup\$
    – Andy aka
    Apr 29, 2016 at 12:35

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