??? 03/18/07 04:20 Read: times |
#135186 - Critical length Responding to: ???'s previous message |
Hi Mike,
First, stop thinking about the frequency of the clock signal. It doesn't matter. What matters is the dv/dt of the signal. How fast does the signal go from 0 to 1, or 1 back to 0. Transition time matters, not clock frequency. You can be running a 1 Hz clock, and if it transition time is 1 ns you're going to have a boatload of hi speed digital effects. How can this be? Consider that the voltage presented on the trace does not propogate instantly. It propogates along the length of the trace at something close to the speed of light. In most circuit boards you can use an assumed speed of 6 to 8 inches per nano second. Now, if your signal transitions from 0V to 5V in 1 ns, and your trace is 12 inches long, then there is going to be a finite time during which the same trace is going to have 5V at one end and 0V at the other. And now you have a wave front propogating down your trace. So what happens when that wavefront hits the other end? Take a rope and tie one end to a door knob. Now hold the other end and whip it to send a wave up the rope. When the wave hits the door knob it will reflect right back up the rope to your hand. That signal will do the same thing. If the circuit is not impedance matched the wavefront is going to reflect back onto itself. This is why you use matched impedance traces and loads at the end of the trace. Of course, this is assuming that the trace is longer than the critical length for your device. And the critical trace length is determined by the transition time, not the clock frequency. Now if your traces are less than the critical length, none of this will really matter. The voltage propogates fast enough that both ends are essentially always at the same voltage and you don't have this problem. It becomes like trying to do the rope-whip-on-a-door-knob thing with a six inch rope. There have been instances of mature products, that were produced and performed reliably for years, suddenly becoming unreliable because a component provider improved their product to run faster. Since the parts were specified with just a minimum transition time, the newer faster parts easily met spec. But since the existing traces were suddenly well over the new critical length the circuits just weren't reliable anymore. So either make your traces controlled impedance, and add matching impedances to the ends, or make them shorter than the critical length. Either way your circuit should work. Joe |
Topic | Author | Date |
track calculation? | 01/01/70 00:00 | |
Critical length | 01/01/70 00:00 | |
seen that one![]() | 01/01/70 00:00 | |
try that | 01/01/70 00:00 |