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I'm trying to route 10 DIP ICs on a single-sided PCB, but because of my available pcb space, I cannot fit all 10 ICs in one single row. Instead, I lined them up in three rows.

It seems that if I connect the VCC pin on every IC to the positive through some sort of jumper wire instead of a direct connection on the PCB itself then the odds of me being able to route the whole board to 100% rise significantly. Same deal with GND pins.

If I'm not mistaken, adding jumper wires introduces either inductance or resistance and too much of that can create undesirable results. The ICs I used are of mixed types: some digital like the 74HC, an op-amp, sound chip (ISD1700), DAC, etc.

I was initially thinking connecting jumper wires from +ve of battery to VCC and using maybe a couple 0 ohm resistors in parallel to minimize bad effects, but I'm not sure. The point is, I want to be able to route the whole circuit to 100%.

Which is better:

Connecting the VCC pin of every chip (or at least the digital IC's) to the +ve of the battery through a jumper wire and connecting ground directly to the -ve of the battery?

OR

Connecting the ground pin of every chip (or at least the digital IC's) to the -ve of the battery through a jumper wire and connecting VCC directly to the +ve of the battery?

and why?

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  • \$\begingroup\$ What frequency range are you working with? \$\endgroup\$ – vofa Dec 13 '16 at 1:52
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    \$\begingroup\$ the highest frequency is about 350Mhz which is a separate radio component that I'll attach to the main board. For the main board in question, the frequency is approximately 20Mhz \$\endgroup\$ – user116345 Dec 13 '16 at 2:04
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Generally speaking, you want the best signal integrity on your ground network, because signals as well as power are referenced to it.

That said, it is not necessarily true that a narrow trace on a single-sided PCB is better than a wire jumper or zero-ohm resistor in terms of its impedance. A 1-mm trace on "1 oz." (35 µm) copper is roughly equivalent to AWG30 wire, whereas soldered jumpers are typically AWG22, which has 1/6 the resistance.

Of much greater significance is how well you bypass or decouple the power on each chip. You want to use the most direct path possible between the capacitor(s) and the power/ground pins of the chip.

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  • \$\begingroup\$ If i used ferrite beads around my jumper wires, would resistance be lowered? and I ask because I might have to connect the capacitor to each chip through jumper wires, and based on your answer, the material for the jumper wires seems to have importance. \$\endgroup\$ – user116345 Dec 13 '16 at 2:08
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    \$\begingroup\$ No, ferrite beads can only increase the impedance, primarily at very high frequencies (100s of MHz). This might be a good thing on the Vcc distribution, but would definitely be a bad thing for the decoupling caps. \$\endgroup\$ – Dave Tweed Dec 13 '16 at 2:11
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This is a case of best practices rather that a completely right or wrong way of routing power to parts on a PCB. The best way is to have a multi layer board and using power planes with thermal reliefs where the parts connect. But since you are confined to a single sided board this will be problematic. Regardless, routing power in the manner you describe should work for many cases. Especially ones that do not involve higher radio frequencies.

For more in depth reading this ti.com pdf looks like a good source.

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