I posted relative problem a while ago here but we came to conclusion that the overshoots were just artifacts. Later, when I extended the signal lines, the overshoots increased and were about 2V of magnitude.

enter image description here

First thing I tried were unknown to me before ferrite beads. I tried several different types of bandwidths but only one had shown some positive effects. I connected one and then two FB in series making a total of 940 Ohms (MPZ1608B471A), the overshoots were smoothed out, but the undershoots remained:

enter image description here

Then I found a similar discussion about my problem here and a simple 100 Ohm series resistor fixed the whole deal:

enter image description here

Finally, questions:

  1. why a simple resistor (that kills DC as well) fixed the problem, whereas the ferrite bead that should only attenuate high frequencies was not as effective?

  2. if I run two wires (one is signal the other is GND), should the ferrite beads go on both wires or just the signal wire?

  3. do I need a pull up/pull down resistors if I connect two DI/DO interfaces?

  4. could someone, please, add "overshoot" and "undershoot" tags.

Thank you so much.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ What kind of chip are you using to drive the signal? Do you know the output impedance? Can you calculate the characteristic impedance of the PCB trace, or is it a wire? 8V 1 MHz 25% duty cycle is kinda odd. \$\endgroup\$
    – ajs410
    Mar 26, 2014 at 17:53
  • \$\begingroup\$ ajs410, I use attiny2313 and EL7156 as high frequency switch. I have wires connecting three PCBs. \$\endgroup\$
    – Nazar
    Mar 26, 2014 at 18:13

3 Answers 3


Overshoot (and undershoot as used by the TS, which is essentially negative overshoot) are created by a quick transition of the signal (arriving through a long wire) arriving at the end of the line with no place to go (read: arriving at a high impedance). Hence there are three basic ways to reduce overshoot:

  • make the wire short
  • make the edge (transition) less abrubt (a series resistor at the source will do this)
  • give the signal edge somewhere to go (a parallel resistor at the destination will do this, a series resistor will have some effect, but less).

A ferrite bead will soften the edge a little bit, but it is mainly used to prevent HF signals (like from radio stations) from entering a device, and vice versa to prevent HF signals from the device from entering the cable which would work as an antenna for these (highly unwanted) signals. HF in this sense means higher than the frequencies you are likely to look at.


A series resistor will be more effective at the source, but it will flatten the edge, which might be undesirable. A parallel resistor (to ground or power, or a combination) will be more effective at the receiving end, and will not harm the edge, but it will attenuate (lower) the received voltage, and increase power consumption. Note that to be effective the resistors must match the cable impedance. Something in the order of 30 .. 100 Ohms is a good guess.

  • \$\begingroup\$ Wouter, so you are saying that it's better to use a parallel resistor instead of series? Also, what do you think about the other questions? \$\endgroup\$
    – Nazar
    Mar 26, 2014 at 16:54
  • \$\begingroup\$ Actually, I just tried several resistor values in parallel - does not kill the overshoots. \$\endgroup\$
    – Nazar
    Mar 26, 2014 at 16:58
  • 2
    \$\begingroup\$ Technically, what you guys are calling "undershoot" is really negative overshoot. Undershoot is what you get when a system is overdamped -- excessively long transition times. \$\endgroup\$
    – Dave Tweed
    Mar 26, 2014 at 17:12

That's not a very high frequency circuit. Ferrite beads tend to be pretty high Q inductors up into the MHz. They get much more lossy at high frequencies.

As you can see from the datasheet, it doesn't cross over into looking more resistive until you hit about 5MHz.

enter image description here


It sounds like your issue is as Wouter suggests, too abrupt of a switching action. From your waveform, it would imply to me that you have a very small inductance action happening. Current starts flowing and then has nowhere to go once it hits the receiver resulting in a transient voltage spike. I'm confused as to why no one has suggested a very small capacitor on the receiver side to absorb this transient energy. If a cap is put in place it should be to ground or if it's a differential signal, between the two signals. This should also help relive your original issue of having cross-talk or transient power issues by smoothing out this signal.

And to answer your original q's: a ferrite acts on higher frequencies as others have stated. The resistor helps your issue because it's adding a load to the output of the transmitter so part of its output current gets sent to ground rather than it all getting sent only into the capacitance of the wire and reciever.

  • \$\begingroup\$ I tried 10p 100p and 1n - did not help. \$\endgroup\$
    – Nazar
    Mar 26, 2014 at 21:24
  • \$\begingroup\$ Try higher? Your resistance will likely be extremely low on the wire, so your pole will be at a very high frequency. You'll want to bring that pole lower in frequency. \$\endgroup\$
    – horta
    Mar 26, 2014 at 21:27

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