I am budding engineer working on RE testing. I know about series termination. We are using 50 ohm as trace impedance. We are getting large reflection and emission in SD clock. I used series resistor values upto 33 ohm as suggested in datasheet. But If I put 100 ohm emission is reducing. Why????

  1. Trace impedance value is 50 ohm. If we match the output impedance of input driver to that value, then I will get minimum reflection.
  2. But If I increase the value higher (approx. 100 ohm), then I am getting less reflection. Why????
  • \$\begingroup\$ Note that "emission" and "reflection" are two very different things. Optimizing one does not necessarily optimize the other. \$\endgroup\$
    – Dave Tweed
    Commented Mar 27, 2018 at 11:57

1 Answer 1


Impedance matching is something that makes sense when you have a transmission line and load of the same impedance. That is not what you have here. A series resistor on a short clock net is there for damping as much as impedance matching. Because the load is not anywhere near 50 Ohms (or 75 or whatever) there will ALWAYS be reflections back from the load. A larger series resistor simply attenuates the reflections more.

A typical CMOS input is like a few pF capacitor as far as loading goes. Any resistive component will be in the MOhms.

A large series resistor, along with the input capacitance of the load, forms an RC network. So the waveform you will observe at the load (and you definitely should look at it with an oscilloscope) is a classic rounded RC low-pass response. If the R gets too big, you will start to see that the waveform never reaches GND or VCC. Also, setup and hold timing may no longer be met for the overall SD bus. So you have to be careful with large resistors. I think I did ship a product once with 82 Ohms in series on the SD clock, but we carefully checked setup and hold times, and tested it while over-clocked and under-volted and hot and cold before we agreed to ship it.

Most of the time, the IC which sends out the SD clock will have internal registers that allow you to adjust drive strength. Before you use large series resistors, you should experiment with reducing the drive strength to improve radiated emission levels. The reason this works better is that the drive strength registers work more like a fixed current limit rather than just a series R. So instead of getting a rounded RC type of waveform at the load, you will see a trapezoidal waveform.

Ultimately, as the voltage slew rate at the load is reduced, emissions will also be reduced, and EMI will improve. Reducing the drive strength typically gives you less degradation of signal integrity (setup and hold times) for the same reduction in radiated emissions. This is based on experience. Perhaps it could be demonstrated with theory or simulation, but hopefully you will trust me enough to check it out for yourself, assuming you have the option of reducing drive strength.

  • \$\begingroup\$ Hi Sir,Thanks for your reply.But I have a small doubt sir. The series termination resistor is larger than trace impedance, so there itself (at source side) should be some reflection due to imp.mismatch, right??If wave reaches the source side after reflecting from the load. It will further cause ringing at the waveform. So there must be high reflection there. Also, I am seeing the waveform is damping due to RC. But the ringing voltage is very less. Why??. Whether I am understanding something wrong??. Please help me if my understanding is wrong. Thanks!!! \$\endgroup\$ Commented Mar 28, 2018 at 10:13
  • \$\begingroup\$ Normally the series resistor is placed very close to the source. If it is placed that way, you will not see ringing at the source due to the series resistor. In order for ringing to occur because of the series resistor, the rise time of the signal must be faster than the round trip flight time to the series resistor. \$\endgroup\$
    – user57037
    Commented Mar 30, 2018 at 20:13

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