# Question about return current in high speed design

1. Mar 20, 2007

### NoName707

I have a question about return currents. Now, all current needs to travel in a loop. so any charge that leaves a source must return to it (correct me if i'm wrong).

usually on a high speed design the return current wants to travel right underneath the trace which is usually the ground plane. when a signal is launched onto a trace, it couples with the plane underneath essentially charging capacitors all it's way down. so electrons on the trace would push out electrons from the plane which return to the negative terminal of the source.

say i have a very very badly designed board and have no ground plane but a ground trace that is no where close to the signal trace. would happen if there is no plane and what would it couple with to obtain the return current?

Thanks!

2. Mar 20, 2007

### light_bulb

i thought the skin effect ment current would tend to stay all along the outside?

Last edited: Mar 20, 2007
3. Mar 20, 2007

### Staff: Mentor

In a transmission line, the characteristic impedance is basically the square root of the inductance per unit length divided by the capacitance per unit length.

$$Z_o = \sqrt{\frac{L'}{C'}}$$

Practical controlled PCB characteristic impedances range from about 50 to 100 Ohms. When you separate the ground return as you describe, it raises the inductance significantly, and you no longer have a controlled transmission line. You have a loop antenna, and more of your energy gets radiated away, plus you get lots of parasitic ringing, because the LC resonance of the loop is probably down in the frequency range of the signal waveform's spectrum.

BTW, one important (and often overlooked) aspect of using controlled microstrip transmission lines on PCBs -- You need to keep your signal trace on the side of the PCB that has the inner ground plane layer adjacent to it, and not feed through to the other side of the PCB. For example, with a typical 4-layer PCB stackup, you have the top signal layer, the inner ground plane layer, the inner Vdd plane layer, and the bottom signal layer. You need to route your transmission lines on the top layer only, or you need to add more layers to the PCB where you can run traces adjacent to other ground layers.

4. Mar 21, 2007

### NoName707

berkeman, thanks for the explanation ! just a couple of follow up questions to fill in the gaps in my understanding.

1. what exactly causes the inductance to rise when you move the return path away?

2. say you had a 2mA draw when the return path was under the trace. When you move it away do we have smaller current draw?

3. Is all charge leaving the source ultimately making it back to the source?

Thanks!

5. Mar 21, 2007

### Staff: Mentor

-1- The inductance ratios with the loop area. For a single turn loop of wire (or signal trace) in air, my crib sheet shows the inductance as:

$$L = \frac{\mu_0 \pi R}{2}$$

So the smaller loop area of a trace over ground plane is what keeps the L' in the range where you can get good transmission line behavior, and a consistent $$Z_0$$

-2- It depends on the source and the excitation frequency. If the source is a voltage source and the frequency is in the range where the inductive reactance becomes comparable to the source impedance of the voltage source, then you will get less AC current. If the source is an AC current source, then you will still get the 2mA AC. You always need to consider the real source characteristics when asking a question like this.

-3- Yes, the charges move in the loop of the completed circuit.

6. Mar 21, 2007

### light_bulb

where do you get those 'crib sheets'? do have to make them yourself?

7. Mar 21, 2007

### Staff: Mentor

Yes, for the most part they are crib sheets that I've done over the years. I keep them sorted by subject area in a 3-ring binder. (Actually one copy at home and one copy at work.) There are also a few pre-printed crib sheets included that I've purchased from book stores when I thought they were pretty good (a couple Calculus crib sheets, a C++ crib sheet, a couple periodic tables, etc.).

I got into the habit back in undergrad of making a good summary crib sheet for each of my technical classes, whether it was allowed on exams or not (generally they were not). It helped me to understand what the key concepts were, and helped me when it came time to study for exams. Particularly in studying for the final, since that covers material that you've learned months ago, and maybe have not applied recently.

8. Mar 22, 2007

### kendr_pind

one interesting question regarding reflections in a board... mathematically when there is a mismatch in impedance at the load side, the signal might reflect...the question is what physical phenomenon causes it to really alter the voltage levels at the load side...

9. Mar 22, 2007

### Staff: Mentor

I think the simplest way to picture it is that the voltage, current and load impedance do not obey Ohm's law when there is a mismatch. When an EM wave is travelling down the tranmission line, the voltage and current are consistent with the characteristic impedance. And when the transmission line is terminated with a real resistor equal to the characteristic impedance, the V, I and R all obey Ohm's law, so the current is correct for the voltage level, and all the energy of the wave is absorbed by the resistor and turned into heat.

But when the wave hits a load resistor that is mismatched (or encounters some other discontinuity in the transmission line, like a change in Zo), the current waveform is mismatched for the voltage waveform and the load impedance, but the energy has to go somewhere. The result is a partial reflection of energy, which propagates backwards up the transmission line.