Do you think a 90° bend in a trace will affect a signal and cause problems? Of the thousands of engineers I've taught, about half admit to harboring this concern. I think I know why.
It is absolutely correct that in sensitive analog microwave circuits built on thick Teflon substrates, a 90° bend in a trace can sometimes affect circuit performance. I suspect that a number of early board designers were exploring the design of serpentine delay lines and discovered the huge problem a pair of 90° corners could create. But what is a 90° corner's impact on high-speed digital circuits?
As part of a series of demo boards for my public classes, I designed a simple serpentine line, which, if unwound, would be 21" long (Figure 1). This yields a delay of about 3.5 nsec. The line has a characteristic impedance of roughly 50 Ω. Each leg is 3" long and there are seven segments. The time delay of each segment is about 0.5 nsec. At the end of each segment are two 90° bends, very close together.
If we send a very fast rising signal into this serpentine, what would we expect to see as the reflected signal? Of course, the reflected signal displayed on a TDR is caused by the incident edge reflecting off of impedance discontinuities. As it propagates down the serpentine, the signal will see a uniform 50 Ω line and no reflection for the first 3 inches. Then, the fast edge will see the first pair of corners.
How will a corner affect a signal? Some believe that electrons zipping around the bend at nearly the speed of light will slow down, cause excessive radiation and crosstalk, which will cause an impedance discontinuity and losses in the transmitted signal.
This is totally wrong.
A corner will affect a signal by changing the instantaneous impedance the signal sees. Of course, if the line width were constant, there would be no change in the instantaneous impedance and no reflection. It's the presence of the extra metal associated with the corner that adds capacitance at the corner, causing a drop in impedance and a negative reflected voltage. The two corners at each 3" interval will have twice the capacitance of one bend and should give rise to a negative reflected signal. If corners affect signals, we should see a reflection from the first bend, after about 1 nsec. This is half a nanosecond for the edge to go down, and half a nanosecond for the reflection to make it back to the TDR.
Figure 2 is the measured TDR response from this serpentine trace. Sure enough, there is a negative dip after 1 nsec, a pretty large reflection. The reflected voltage is about 40 mV, with an incident signal of 200 mV. This is a 20% reflection, with a rise time of about 50 psec.
We would expect that each of the double bends should show a similar reflection as the fast edge hits it. Sure enough, every 1 nsec there is an additional reflected signal 6 in all. But if the reflections we see are from corners, why does the peak value drop off with each corner?
The magnitude of the reflection will depend on the rise time of the signal and the amount of capacitance in each pair of corners. If the geometry is the same for each bend, the capacitance is the same. What could cause the rise time to increase with each corner? One possibility is the losses in the FR-4. As the edge propagates down this 21" interconnect, the rise time will get longer, and the magnitude of the reflection from each capacitor will get lower.
The measured response from this serpentine is exactly like what we would expect if the reflections were caused by the corners. And the magnitude is huge: 20% with a 50 psec edge. This certainly seems to demonstrate that corners cause reflections, and large ones at that.
I think this observation is what leads many engineers to believe corners cause reflections and should be avoided. But is this the right explanation? Here is your challenge: Can you come up with another possible explanation for these dips in the reflected signal?
E-mail me your answers and in two months I will report some of your responses in this column. I look forward to hearing from you. PCD&M