PCB track width for MHz frequency

Well, I was hoping that someone else would answer this question, but apparently nobody wants to touch it.

@BarsMonster, the reason why nobody (except the fool that I am) wants to answer this one is because you're crazy. And I don't mean that in a negative way! But you are wanting to do something that very few people would try to do, and the odds of failure are very high.

Much of this answer is going to sound like I'm saying, "Don't even try this", but let me assure you that this is absolutely NOT what I'm saying. If you do this, I want you to have the best chances for success possible and that means knowing what you're getting yourself into.

Let me recap what your project entails, and then I'll get you answering your question.

You're doing a CPU in TTL logic (a.k.a. 74xxx type chips). This would normally require maybe 100+ chips. If you use some larger chips (like an ALU chip, etc.) you might get it down to 20+ chips. The PCB for this will be around 5"x5" on the small end, and maybe 12"x12" on the large end. With 20 chips, each with 14 signal pins, you will have a minimum of 280 connections. More likely there will be 800+ connections and way more than 20 chips.

You're estimating that the clock frequency of this will be around 100 MHz. Having hundreds of signals on a 2 layer PCB, all switching with a 100 MHz clock and 1-3 ns edge rates is going to be, well, crazy. Many of these signals are going to be on long traces with many branches and forks. Without a solid power/gnd plane underneath. Routing of power and signals is going to be super difficult, and managing signal integrity will be almost impossible.

For starters, you really need to read the book, High Speed Digital Design: A Handbook of Black-Magic, by Howard W. Johnson. It is an expensive book, US$100, but is worth every penny. On this project alone it will save you $100 in time and PCB mistakes. It talks about all the issues of routing a PCB like yours.

1) What is optimal FR4 thickness for such high-speed schemes? I have both 1.5mm and 0.1mm FR4. I guess 0.1mm FR4-based PCB would allow some distributed capacitance between VCC & GND...

The optimal thickness would be less than 0.010" (0.25mm) between your signal layers and your power/gnd planes. But since you have a 2 layer PCB there is no point in worrying about PCB thickness. The thickness will be whatever is required to give the board the mechanical strength required to hold all of your parts, about 0.0625" (1.5mm).

You also don't have power or ground planes. Filling in the unused space on the 2-layer PCB with power and ground does not constitute power and ground planes. Those "planes" will not function well (if at all) as a signal return path. That's because they are so broken up into smaller shapes due to the signals being routed on those same planes. The only thing those "planes" will do is make it slightly easier to route your power/gnd traces-- and I do mean only slightly easier.

"Distributed capacitance" is something that is approaching "Urban Legend" status. Everyone talks about it like it is some sort of holy grail, but few have gone through the math. So here it is: if you have SOLID power/gnd planes separated by 0.25mm with a dielectic constant of 4.5 then you will achieve 16 pF of capacitance per cm2. With a separation of 0.125mm it goes up to 32 pF/cm2.
Now let's assume that each chip on the PCB requires 2.0 cm2 of space and you do have a 4 layer PCB with 0.125mm separating the power/gnd planes then you'll get 64 pF per chip! Clearly that is not enough, so this distributed capacitance is not enough to remove actual caps from the PCB!

The above example (4 layer PCB) assumes that there isn't a lot of space between layers. If that were the case then to make the PCB 1.5mm thick then there will be a lot of space between the signal and plane layers. This has the effect of increasing the trace impedance to a point where it's useless. You really want the trace impedance to be in the 50-100 ohm range for high speed signals. If your signal and plane are 0.5mm apart then your trace width would have to be impossibly wide to get the impedance down (and a wide trace has enough capacitance to mess other things up). The point is, with a 4 layer PCB there is a trade-off between distributed capacitance and trace impedance. And since the dist-cap effect is so small you're much better off going in the direction of lower trace impedance.

The dist-cap of a 1.5mm thick 2 layer PCB is 2.66 pF/cm2.

2) What is the optimal trace width for high-speed digital signals? As I don't know input/output impedance of all TTL devices I have, I can't do impedance matching. What is the optimal clearance between trace & surrounding GND? I guess it's hard to get even 1pf trace to GND capacitance...

The optimal trace width is "as thin as is reasonable". More than 0.25mm is too thick. You're trying to reduce the capacitance of the trace by making the traces thinner.

The optimal space (clearance) between traces, or between traces or planes, is not super important. There is not much coupling when signals/planes are "edge on". So make it as small as possible, if only because it will make routing easier.

Sometimes putting two signals side by side for a long distance (several inches) can cause crosstalk. But honestly, crosstalk caused by that will be minor compared to the other signal integrity problems caused by huge AC signal return paths and impedance mismatches.

3) Do I need any termination for 5-10cm traces with 1-3ns fronts? If so, what it should be (220 ohm to GND?)?

Read that book. About 1/4 of that book is devoted to this exact subject, and I cannot hope to give you all of that info in only a couple of paragraphs. But I'll try! :)

You will need termination for any trace longer than about 5 mm. Also, I don't believe your figures of 1-3 ns edges. They will either have to get much faster to meet your 100 MHz estimate, or your clock rate will have to drop. But more about that in a moment.

The termination resistor should match the trace impedance on your PCB. But because your traces are not over a solid power/gnd place you have no control over your trace impedance. Worse than having no control, your trace impedance is going to be changing as the trace is routed over partial planes and other signals. The point is, with a 2 layer PCB you cannot do proper signal termination.

The exact type of termination you use (even with a 4 layer PCB) is going to depend on the signal topology. By topology, I mean does the signal only go from one driver to one receiver? One to Many with the trace branching out? One to many with everything daisy-chained? One to one, bidirectional? Many to Many, with tri-state buffers? Each of these will require a different method of terminating the signal.

One to One: A series termination resistor at the driver.

One to Many, w/branches: Also called a "star". There are lots of difficulties dealing with this one, and most people advise to not do it at all.

One to Many, daisy chained: A single termination at the end of the line. Usually a resistor to +1.65 volts (assuming 3.3V logic signaling). The +1.65v can't use a normal voltage regulator (there are special regulators for termination). You can also use a voltage divider between +3.3v and GND instead, but that takes much more power. For signals with a 50/50 duty cycle you can use a resistor and cap in series to GND (called AC termination).

Many to Many: Good luck. This is why all the high speed busses have gone serial point to point. Read the book on this one.

What you absolutely won't do is run a resistor of any value to GND. That just doesn't work with TTL signals. This only works when the signal goes both + and -, and averages out to 0 volts.

As I mentioned earlier, I do not believe that you'll get to 100 MHz. Actually, I don't think you'll get even close. A 100 MHz clock has a 10 ns period. So with 10 ns as your timing budget, let's look at what will eat into that budget:

1. Clock Skew: Clock skew is the difference in clock arrival times for the different flip-flops in the system. In your case, this is going to be mostly due to the difference in clock trace lengths. There are ways to route a clock signal to minimize the differences, but since you only have a 2 layer PCB you are restricted in what you can do. If your PCB is 6" x 6" then I am going to guess that you could have a 6 inch difference in clock trace length from any one flip-flop to any other. Since light travels at 6"/ns you have used up 1 ns of your 10 ns timing budget due to clock skew. 9ns remaining.

2. Clock Jitter: For this PCB, most of the clock jitter is going to come from signal integrity issues as well as the edge rates of the signal. This is similar in effect to "clock skew", but the source of the problem is different. For your clock signal, signal integrity is going to be much worse than your other signals simply because you are driving a lot of loads from a single source. There are ways to minimize the problems, of course, but you are still going to have issues simply because this is a big problem. My best guess is that your clock jitter is going to be somewhere around 3 ns. It could be as little a 1 ns if you use drivers with fast edges and do a superb job with routing and termination (which you can't do with a 2 layer PCB). 9ns - 3ns = 6ns remaining.

3. Gate propagation delay: Most of the modern 74xxx type chips have a max Tpd of 3ns or greater. If you have 1 level of logic (I.E., your signal goes through 1 gate between flip-flops) then you will use up 3 ns. If you have 3 levels of logic then you'll need 3 * 3ns, or 9 ns. An 8-bit full-adder could have 17 logic levels, or 51 ns. 6ns - 51ns = less than nothing. Of course you can use a larger chip to do the adder, but there are still plenty of other things that require more than 2 levels of logic. So either way you're out of time.

4. Signal routing delay: The delay of the signal on the PCB. Again, PCB routing and signal integrity matter most here. Without going into a huge analysis, I would estimate 1-3 ns per route. Most of this time is taken up by the edge rate of the driver. So if you have 1 logic level then you'll have 2 routes (one to the gate, one from the gate).

5. Setup time of the Flip-Flops: Expect about 2 ns for modern 74xxx parts.

6. Clock to out time of the flip-flops: The Flip-Flop that I looked at was a 54LVTH574, one of the faster ones, and it needed 4.5 ns.

Add all of this up and you need at least 15 ns per clock, and more likely 50+ ns. So what was 100 MHz is now 20 MHz, and even that is pushing it. Honestly, I would predict that 1 MHz is closer to what you'll achieve.

To conclude: You can do it. It is possible. Many of the things you'll learn doing this project are very valuable. But think very hard about if you want to tackle it all at the same time. Normally you'd learn logic design, PCB design, high speed signal layout, etc each separately.

That's actually two separate questions. It's the circuit voltages that determine the clearance requirements, while the current levels determine the width (and thickness) requirements.

Trace Width

Dealing with the latter first, it's the width and thickness of a copper trace on a PCB that determine its cross-sectional area, in the same way that the diameter does for an ordinary wire. The cross-sectional area determines its resistance per unit length, at which point, it's up to you to decide two things:

• How much voltage drop (ΔV = I × R) can you tolerate from one end of the trace to the other?

• How much heating of the trace (Power = I²R) can you tolerate?

One or the other of these will be the limiting factor for each trace.

For example, you might have "1 oz." copper on your PCB. This is shorthand notation for "1 ounce of copper per square foot", which translates to a thickness of 1.38 mils, or 0.035 mm. A trace that's 10 mils (0.254 mm) wide, then, has a cross-sectional area of 13.8 mil² which is roughly equivalent to an AWG38 wire. It will have a resistance of about 0.75 Ω/ft. and the current capacity is on the order of 10s of mA.

To handle higher currents, you might select "2 oz." copper (0.070 mm thick) and use traces that are, say, 100 mils (2.54 mm) wide. This gives you a cross-sectional area of 276 mil² which is roughly equivalent to an AWG24 wire.

Note that since the traces on a PCB are very flat and wide, they're actually much better at getting rid of heat to the enviroment than the equivalent circuilar wire is. So as far as I²R losses are concerned, you can put a lot more current through a PCB trace — but you still need to pay attention to the temperature rise and the associated thermal management.

Clearance

The required spacing between conductors is determined by the voltage difference between them and the amount of leakage current you can tolerate. Leakage current is primarily associated with surface contamination of the PCB (e.g., residual flux, as well as accumulated dust, moisture, etc.).

One guideline comes from safety testing services such as UL, which requires a creepage distance of 5mm per kilovolt for circuits that are supposed to be "isolated" from each other (material group I, pollution degree 2 from UL840).

Obviously, this guideline gives very small values for low voltages (0.05 mm or 0.002 in. at 10 V), so the limiting factor actually becomes the line/space widths that your PCB fab house is capable of.