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The canonical breadboard layout consists of two strips of five contact points for every column, separated by a divider that allows you to accommodate DIP-style ICs. Is there a special reason why the strips are five contacts each?

Is wiring up five different components a common occurrence? My experience indicates this is actually quite rare. Is it perhaps related to the fact that five contact make up 0.5", allowing you to lay out 1"+ DIP packages?

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    \$\begingroup\$ partially to accommodate 600 mil wide devices I would think. \$\endgroup\$
    – kenny
    Commented Apr 17, 2011 at 19:43
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    \$\begingroup\$ In the wire wrapping days the pins were long enough for 3 connections and usually that was enough. 5 slots, 1 for the chip, 3 according to wire wrapping and +1 for good measure? I wouldn't really know, just a wild guess. \$\endgroup\$
    – Faken
    Commented Apr 17, 2011 at 21:02

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Considering the contact group size, unused points represent wasted real estate, so there's a reason to make the group size as small as possible. In some parts of a circuit, you'll only need to connect two component leads together, but just as commonly you'll need to connect three or four. Sometimes more. Only two points per group would not allow you to connect three or more leads. You could make the group size 3 points, which would be the minimum that would allow you to bridge groups and connect more leads, but to connect four or more components you'd be using 1/3 to 2/3 of your available contacts just to bridge groups. Obviously if the group is big enough, you don't have to bridge groups at all. So, five per group just seems like the best compromise between unused contact points and points tied up bridging groups.

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As kenny surmised, the main reason was to accommodate 0.600" DIP packages. Back when microprocessors first started appearing, many of them came in 40-pin DIP packages like this Z80 show below. Nowadays most microcontrollers that come in DIP packages, even those with 28-pins and above, are available in the narrower 0.300" form factor.

enter image description here

Even some 7400 series logic devices which had many pins (such as the 24-pin 74150 mux and 74154 demux) came in 0.600" DIP packages. These wide packages only allow for two connections on either side of the device.

Makers of small DIP modules, like the Gabotronics Xprotolab mini-scope shown below, are now making use of this extra width. Note that only one pin is available on either side of the module to connect to other parts of the circuit.

enter image description here

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  • \$\begingroup\$ Nowadays most microcontrollers come in surface-mount, not DIP, packages. \$\endgroup\$ Commented May 11, 2012 at 16:30
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    \$\begingroup\$ @BrianCarlton Yes, I realize that. I said "Nowadays most microcontrollers in DIP packages..", and have edited my post to make it more clear: "Nowadays most microcontrollers that come in DIP packages..." Microchip is one of the few vendors that still offers DIP packages of most of their newly-released microcontrollers, including a few of their PIC32s. Great for prototyping. \$\endgroup\$
    – tcrosley
    Commented May 11, 2012 at 16:38
  • \$\begingroup\$ @Brian - the solderless breadboards are still targeted at DIPs, not just discrete PTH components. \$\endgroup\$
    – stevenvh
    Commented May 11, 2012 at 16:42
  • \$\begingroup\$ "Nowadays most microcontrollers that come in DIP packages, even those with 28-pins and above, are available in the narrower 0.300" form factor." 0.3 inch 28 pin is fairly common but I don't think i've ever seen a 0.3 inch 40 pin. \$\endgroup\$ Commented Feb 13, 2016 at 3:49
  • \$\begingroup\$ @PeterGreen I don't know if anyone ever made one, but there are no 0.300" 40-pin µC's now -- I checked on Digi-Key. As you said, all 40 pin ones are 0.600". Some oddballs though, 0.400" 32-pin DIPs from both Freescale and STMicroelectronics, and a 0.750" 64-pin monster DIP from Intel -- dated 1994. \$\endgroup\$
    – tcrosley
    Commented Feb 13, 2016 at 4:11

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