My Arduino is connected to the first breadboard. From there, I chained the positive and negative to the other boards. I have 6 breadboards connected right now. In the first board I see (with a meter) that I have 4.5V difference, and in the next board, 3.5V and in the last, I even get 2V.

Why am I getting this? I partially solved this (getting 3V to the last board,) by wiring every rail to every rail in the other boards, but it makes my design cluttered.

How do I debug this? What's causing it?

enter image description here

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    \$\begingroup\$ You have a lot of breadboards, and probably some combination of high power consumption and thin leads. \$\endgroup\$
    – pjc50
    Commented Oct 16, 2013 at 14:37
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    \$\begingroup\$ Just a thought when you are rearranging the power supply wires. At least with my cheap breadboards and cheap tinned-copper 0.6mm bell wire, I have found the resistance in power supply rails like these was dominated not by the length of the wire but by the number of wire-to-breadboard contacts. \$\endgroup\$ Commented Oct 16, 2013 at 16:45
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    \$\begingroup\$ Gil drives into a small town in a red sports car. Soon after, police lights flash in his mirror. He pulls over, and the officer walks up. "I'm sorry, was I speeding?" Gil asks. "No, sir. It's against the law to drive red cars in this town," the officer says. "Really? I didn't know that," Gil replies, surprised. "YOU THINK YOU ARE ABOVE THE LAW?!" the officer shouts. Moral of the story: Just because someone doesn't know the laws doesn't mean they think automatically they're exempt to them. (Also, the question revision history shows no indication whatsoever of an "attitude" from the OP.) \$\endgroup\$
    – JYelton
    Commented Oct 16, 2013 at 17:20
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    \$\begingroup\$ Well. I wasn't planning on a flame war. But just so you all know, my Arduino is really powering all of these ICs. There are 29 ICs in my circuit, and it works well. It's just that I was trying to understand if there is a way around the single-point of contract. This circuit is a VGA adapter, syncing at 640x480x60hz using only 74xx ICs. Certainly not above any law :) \$\endgroup\$ Commented Oct 16, 2013 at 18:35
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    \$\begingroup\$ @Gil I think your question is totally valid. I remember experimenting with assortments of electronics and having similar questions. I don't think this is a flame war, just sometimes we have to point out that the wise old master is sometimes a grouch. :) \$\endgroup\$
    – JYelton
    Commented Oct 16, 2013 at 20:05

4 Answers 4


You should have one single point of contact for the power supply power and ground. Wire that point to each board individually. If your power supply is the arduino, be aware that it can only supply a small amount of current, so you may want an external power supply. Also use the heaviest gauge wire you can fit in the breadboard (probably 22gauge) to reduce wiring losses.

The power supply could still be the Arduino - but honestly with that many TTL chips you really should be using a separate power supply if possible.

The following edit to your picture should give you the idea. I didn't pay enough attention to know if I mapped red to positive correctly, so ignore that aspect if I've got it wrong. The point is that you need an individual wire from your power source to each rail, rather than chaining them in any way. While it'll be a bigger rat's nest of wiring, daisy-chaining just isn't going to work for you, electrically, due to the losses inside the breadboard.

enter image description here

If you can't stand this solution, though, you might be able to rough up the wires between the boards, and shove them in and out of the holes in the rails a few times to clean off the contacts in the board. This will buy you some additional overhead, but it won't solve the base problem.

  • \$\begingroup\$ Thank you for your answer. I did what you suggested. I now get 4.5V to all rails. I was not having a problem with my circuit, was just trying to understand the reason why voltage drops so bad. I assume it's the resistance in the crappy breadboards. Anyway, you rock! \$\endgroup\$ Commented Oct 16, 2013 at 18:37
  • \$\begingroup\$ @Gil I think the lesson here is that you can think of wires, breadboard conductors, and other connectors as low value resistors. In long serial chains they can definitely add up to create a voltage drop. \$\endgroup\$
    – JYelton
    Commented Oct 16, 2013 at 20:06
  • \$\begingroup\$ Thank you very much! Solved my problem here! \$\endgroup\$ Commented Feb 18, 2023 at 6:59

You state you're suffering voltage drop between breadboards. "Star" from your power supply to each breadboard ensuring good mechanical and electrical contact. (As posted by Adam Davis and Passerby)

According to your picture, you don't have ANY electrolytic capacitors or any ceramic capacitors, that's where I'd start - not seeing ANY makes me cringe and is the point of this post.

Filter capacitors (the electrolytics) are polarized and go across the power supply rails to "condition" the power supply, keeping the voltage throughout the circuit relatively constant.

The ceramics also connect across the power supply rails and one should be connected across the V+ and 0V pins of each IC. The generic purpose of ceramics is to shunt high(er) frequency "noise" to ground, polarity is not an issue.

Since you're using 5V power supply, a 220uF 16V electrolytic (or two) on each breadboard would be an excellent start, they are common place (Radio Shack, etc.) and inexpensive.

Ceramic caps are very inexpensive and worth every penny. A .1uF 50V located at each IC and one next to each electrolytic will do wonders for clamping spurious noise.

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    \$\begingroup\$ Great point. Each chip should have a small bypass cap across its power supply pins, in addition to a larger cap on the rails where power comes in. Reading the voltage with a DC meter might actually show a low voltage when the reality is there's probably a ton of noise on the power line by the end of the daisy chain. \$\endgroup\$
    – Adam Davis
    Commented Oct 16, 2013 at 15:33
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    \$\begingroup\$ I feel like I'm missing something here. I can see how caps can stabilize local DC by mitigating effects from transients causing AC signals on your DC. But I don't see how it would help in this case, where the probable problem is voltage drops due to thin, long leads. Caps can't help with that; or am I missing something? \$\endgroup\$
    – Bob
    Commented Oct 16, 2013 at 16:18
  • \$\begingroup\$ @Bob the caps are a secondary concern of their setup. As they said in the first sentence, OP needs to star their power setup. \$\endgroup\$
    – Passerby
    Commented Oct 16, 2013 at 16:24
  • \$\begingroup\$ @Passerby, good call, I missed the first sentence because the two distinct ideas bled together. An edit would be nice to distinguish the two ideas: the answer is the first sentence, and the rest is all (very good) recommendations. But that's a style thing, and the answer is correct and complete. Thanks! \$\endgroup\$
    – Bob
    Commented Oct 16, 2013 at 16:56
  • \$\begingroup\$ @Bob My post includes "not seeing ANY [capacitors] makes me cringe and is_the_point_of_this_post" and is about the best formatting I could come up with, after all, the shock of not seeing any caps 'bout knocked me out of my chair! Help yourself to an edit if you wish. :) \$\endgroup\$ Commented Oct 16, 2013 at 17:37

Breadboards have a variety of problems, including high capacitance, inductance, and resistance. From your picture, your breadboard power rails are two per side, are split in the middle, and you have 6 boards, so that's a total of 2 rails * 2 sides * 2 splits * 6 boards = 48 different segments that you bridge with some jumper wire. And from the looks of it, you have a single power input point at the very end of the segments. All the resistance of the segments AND the jumper wire add up, and then there depends on how much current you are pulling.

Easiest Solution, make sure each board is in parallel. Run a pair of wires from your power supply to each board instead of board to board to board. Or two pairs, since you have positive and negative rails on each side of the boards. Or simplify your layout so only one set of rails are used (or positive always on the left side, negative on the right side).


The star power and ground distribution as posted by Adam Davies and Passerby is only a partial solution you must also:

Ensure the current capacity of the power supply is adequate for the peak power consumption of each bread board in use and preferably allow some margin of overhead.

Use decoupling capacitors as suggested by Joe from Ozarks

Use substantially heavier gauge wire for power and ground distribution as the other posters have suggested.

Finally if you get a sound 5v supply onto each of the breadboards and the circuit still doesn't work consider that there still may be one or more of the following to consider:

Clock distribution is not satisfactory between the boards due to the capacitive and inductive effects of the breadboard contacts.

Race hazards between signals introduced by the long leads and again breadboard contacts.

Depending on the 74XX logic family running the circuits at well below design voltage may have destroyed some of the gates. The latter tends to be more of a problem with CMOS but you don't give details of which one so refer to the data sheet.

Finally you state you're trying to run it at 640x480x60 = approx 18.4Mhz so there's is a pretty good chance of the circuit not working at all using those breadboards. If so run the circuit at lower frequency to debug the logic then solder a more compact version onto strip board.

  • \$\begingroup\$ Thanks for the explanation! The circuit works fine, there's a video here: facebook.com/photo.php?v=10151801133034947 It works well with only two caps (around high freq ICs.) I was just inquiring why the low voltage. I did the star power distribution and now getting 4.5V to all ICs. Just to be on the safe side with these TTL. \$\endgroup\$ Commented Oct 18, 2013 at 9:43

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