I have the following line driver and line transceiver

driver: 74VHC244

transceiver: 74VHC245

They're connected via a 0.5m long ribbon cable with an impedance of 100Ohm. The digital signals have a maximum switching frequency of 16MHz at 3.3V through the cable. The order in the ribbon cable is always [signal - ground - signal - ground]. All signals are single ended.

Do I need additional termination? If so, how would you do it? Just add 100Ohm resistors in series after the outputs of the driver?


2 Answers 2


Signal integrity headaches due to termination and reflections in transmission lines do not depend on the frequency of the signal. The crucial factor is the slew rate (or rise time), which determines the highest frequency in the bandwidth of the signal. Here is a random example from the internet:

enter image description here

Such ringing on the edges may cause one clock edge to be interpreted as several clock pulses by the receiving chip, which will corrupt your transmission. This only depends on rise time, not signal frequency: a 1Hz signal with fast ringing edges can do this too. Edges that are too slow are a problem too. Your 74VHC chip datasheet specifies a minimum input transition rise/fall rate of 20-100 ns/V depending on supply voltage. If the signal spends too much time in the transition zone, this can cause problems.

74VHC rise time is listed as 2-3 ns, which could cause problems considering the length of your cable. However, its drive current is quite low (4/8 mA for 3.3V/5V), which in your case is excellent as the chip's weak output drivers will behave more or less like a "free" source termination resistor which should be sufficient to avoid any problems.

With a 74AC or LVC chip, it would be another story, these would require a source termination resistor.

Make sure to place a decoupling cap on the cable driver chip's power pin. It needs transient current to charge the cable capacitance. Bad decoupling will cause crosstalk through the power supply, or transient supply voltage (and output voltage) sag when many outputs switch simultaneously.

You can check signal integrity at the receiver end, but make sure you use a 10x probe of adequate bandwidth and the tiny ground spring (not the alligator clip lead).

Note that you should probe at the receiver! Signal will be distorted at the driver (pic source):

enter image description here

Voltage at driver side ("A") plateaus halfway between logic levels during the whole time it takes for the signal to make a roundtrip in the transmission line. During this time, line capacitance is being charged, so the driver draws current to charge it. However, signal at the receiver end is clean, and its rise time does not depend on transmission line length and total capacitance (besides losses). This is a bit counter-intuitive, but it works. Driver supply current does depend on line length and capacitance though, as the current to charge the line capacitance comes from the supply.

  • 2
    \$\begingroup\$ They depend on the transition time which is determines the highest frequencies in the bandwidth of the signal. In other words, the transition times determine the bandwidth of the signal. The clock frequency is contained in the the bandwidth, but is, by definition, a lower frequency since it is the fundamental so does not determine the bandwidth. \$\endgroup\$
    – DKNguyen
    Aug 7, 2019 at 18:09
  • \$\begingroup\$ Yes, that's it. \$\endgroup\$
    – bobflux
    Aug 7, 2019 at 19:27

You probably don't need to worry about transmission line effects as the wavelength of 16Mhz is 9m and much higher than the 0.5m that you need to transmit. Typically transmigration line effects need to be considered more than 50Mhz (a rule of thumb) in normal PCB designs.

The thing that will affect the rise times is capacitance in the cable and the drivers, so keep that to a minimum, but that shouldn't be hard to do with a cable of a few ohms and capacitance in the 100pF to 10pF range

  • 1
    \$\begingroup\$ Should not the full wavelength for 16MHz and 70% VF Cable be \$\lambda \sim\$ 13m? For 50MHz it should be 4.2m. It should be critical for the size of the PCB? Or for the transmitted length? \$\endgroup\$
    – Brethlosze
    Aug 7, 2019 at 18:26
  • 1
    \$\begingroup\$ Actually, yeah. I calculated that wrong, for ribbon cable it would be around 9m in AWG26. If the wavelength is longer, then you don't need to worry about it, because the phase won't change much if it's like 1/10th. Since most PCB's are in the 10's of cms range it's not a big deal. If you were designing a PCB that was in the meters range, then you may have to worry about wavelength\speed at lower frequencies. If you look at most digital comm protocols, after roughly 50Mhz they start requiring matched transmission lines and drivers. \$\endgroup\$
    – Voltage Spike
    Aug 7, 2019 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.