# Different printed circuit layout for a transistor

The below figure shows a schematic for a transistor where there exists an external magnetic field :

This schematic has two layouts to choose from. The first one is shown below :

The second layout is also shown below :

The second layout is better than the first because this will cancel the unwanted induced voltage in the transistor. I don't understand how this will happen, I am trying to figure this out using the right-hand rule but I always fail. Please, someone, explain this in details and how to figure out exactly the polarity of the induced voltage in this case.

• I don't see how the second one is any better. Just run the traces on top of each other or beside each other does the same thing. – DKNguyen Mar 9 '19 at 22:04
• @Toor This problem is captured from electromagnetic compatibility engineering by Henry Ott and according to him the answer is : The second layout is better because The magnetic field pickup is of opposite polarity on either side of where theinput traces crossover, which thereby cancels the noise voltage. – John adams Mar 9 '19 at 22:09
• Oh, I know that name. The stuff on his website has always made sense to me and was well explained. Seems odd this one would be so vague. Lemme think. – DKNguyen Mar 9 '19 at 22:14
• @Toor Ok waiting for your explaniation – John adams Mar 9 '19 at 22:27
• If I come up with anything at all... – DKNguyen Mar 9 '19 at 22:32

It seems that the author is suggesting that a PCB layout with built-in "twisted-pair" layout of the traces will result in lower EM pickup by the circuit board itself.

Figure 1. Untwisted and twisted PCB layout.

Shielded pair twisted pair cables reduce interference by nature of the twisting. Each twist of the cable picks up the EM interference in the opposite polarity to its neighbour with the result that the sum of all the interference tends towards zero due to cancellation.

To see how this happens, think of an external magnetic field noise source orientated directly out of the page (for example). Point your right hand thumb towards you so it points in the same direction as the magnetic field. Now curl your right hand fingers naturally - they'll circle anticlockwise. Trace around the two loops in the Figure above anticlockwise - notice how current induced in this way will be coming out of pin 1 (into pin 2) in the left loop, but out of pin 2 (into pin 1) in the right loop. The two loops mean stray magnetic fields (which direction they point) will induce noise currents that cancel each other out.

With the PCB layout suggested I would imagine that the crossover point should be set between the end of the infeed cable and the transistor such that the areas of the two opposing loops created are equal. Don't forget that the parallel pins of the connector will probably need to be considered as part of the first loop if one is going to this level of interference reduction.

• Thank you very much for your effort. Can you please explain how did you figure out the directions of EM interference in each cable? because this is my only problem – John adams Mar 9 '19 at 22:40
• I didn't! EM is alternating so the voltages on each of my loops will alternate. Note that we don't have any idea of the direction of the EM fields anyway so we can't work it out. If the PCB is oriented edge-on to the interference field that there will be no pickup. – Transistor Mar 9 '19 at 22:44
• I understand this but what I mean is how the nature of the twisted cable cancel the interference? – John adams Mar 9 '19 at 22:58
• Think of an external magnetic field noise source orientated directly out of the page (for example). Point your right hand thumb towards you so it points in the same direction as the magnetic field. Now curl your right hand fingers naturally - they'll circle anticlockwise. Trace around the two loops in Fig. 1b anticlockwise - notice how current induced in this way will be coming out of pin 1 (into pin 2) in the left loop, but out of pin 2 (into pin 1) in the right loop. The two loops means stray magnetic fields will induce noice currents that cancel each out out. – Heath Raftery Mar 9 '19 at 23:04
• @Transistor I hope you don't mind, I submitted an edit to your answer that includes my comment above, since it seems to have helped understanding. It's undergoing peer review. – Heath Raftery Mar 9 '19 at 23:58