Measuring breadboard wiring resistance: how bad is bad?

This article is a brief record of my early breadboard wiring adventures, because learning things the hard way is fun! You may also read the related post if you wish:

NiMH vs Li-Ion voltage sag under small loads

Naive beginnings

As it may be the case with most people, I started my adventure with electronics by building simple circuits using passive components, low-power microcontrollers, and basic sensors. My first DIY kit was a bunch of breadboards and a pile of assorted pre-made 26 AWG stranded wires like these:

This worked well until I started standardizing my projects around ESP32 microcontroller boards, replacing low-power Arduinos. By that time, I've been mixing the pre-made wires with some 22 AWG solid core ones:

I haven't really spent much time measuring the actual impact of wiring quality until it was forced upon me by the ESP32 hiccup described in the linked article. Since then, I've moved away from the 26 AWG pre-made wires altogether. In this brief piece, I'm going to look at some wire resistance measurements to explain why.

Test setup

The test setup is a breadboard with 10 wires arranged a zig-zag pattern, plus 2 wires for measuring the resistance:

I'll test the following wire types:

• Pre-made 26 AWG, stranded copper wire. These are the same wires I've had since my first project.
• 22 AWG solid core copper wire cut from a spool.
• 26 AWG solid core copper wire cut from a spool. This is a very thin wire, not suitable for breadboarding, but I'll test it anyway.

I'm not sure where the 22 AWG wire came from, but it's been a great performer so far and I now use it for all my breadboarding setups. The 26 AWG wire comes from RadioShack and I absolutely love it for building permanent, soldered versions of my devices:

These 26 AWG wires are thick enough to support 1 A without issues, and they're flexible enough to be shaped as needed e.g. wrapped around MCU pins before soldering, or generally routed around a board using tweezers with very little effort.

Anyway, each of the twelve wires is ~4.5 inches long (~54 inches or ~4.5 feet total) and was tested for continuity i.e. there should be no bad wires.

Measurements will be taken using a bench-top multimeter in 4-wire resistance mode.

Here's the test rig using pre-made 26 AWG wires:

This is with 22 AWG custom-cut wires:

The test rig using 26 AWG custom-cut wires looks identical except for the wire color being different.

Static tests

First off, let's measure resistance across the entire path without touching the wires.

For each wire type, I took three measurements. For each measurement, I pulled all wires from the breadboard and re-inserted them into different holes within the same row.

Results:

• Pre-made 26 AWG (Ohm): 47, 27, 30
• Solid core 22 AWG (Ohm): 5, 4, 8
• Solid core 26 AWG (Ohm): 1100, 650, 930

Right off the bat, it's obvious that the 26 AWG solid core wires are not thick enough to make proper contact with the breadboard, resulting in abysmal results across all three measurements.

What was somewhat surprising to me is just how bad the pre-made wires are. In my particular case, resistance in the double-digit Ohms range are very likely to cause issues with ESP32 sensor modules that easily hit >400 mA at peak. Upon closer examination, it appears that the main reason for poor wire performance is the square-ish tip that's fairly thick but unable to make good contact with the breadboard due to sharp edges. 

By comparison, the 22 AWG solid core wires sit much more firmly in the breadboard and their resistance is much, much better.

Dynamic tests

This is just a fancy way of saying "let's touch some wires".

Here's the test with pre-made 26 AWG wires:

And here's the 22 AWG custom-cut version:

Quite the difference!

Soldered version

The scenario wouldn't be complete without a soldered version thrown in for comparison. I'll test the 22 AWG as well as 26 AWG solid core wires soldered onto a perforated board as shown below. The number and total length of wires in each set remains unchanged.

For the 26 AWG variant, the total path resistance is quite good indeed at ~0.17 Ohms whether you touch it or not. The measured resistance is roughly in line with the expected value for ~4.5 ft of 26 AWG wire (4.5 ft * ~0.04 Ohm/ft = ~0.18 Ohm). My multimeters and LCR meters don't have the resolution to accurately measure such resistances, so I used a milliohm meter instead:

For the 22 AWG variant, the measured resistance is again consistent with the calculated expected value (4.5 ft * ~0.016 Ohm/ft = ~0.07 Ohm):

What this means

It's all down to Ohm's law:

In simple terms relevant to this article:

• The amount of current passing through the circuit is inversely proportional to the circuit's resistance. More total resistance means less current.
• The same amount of current passes through each component in the circuit.
• The total voltage drop (supply voltage) is a sum of voltage drops across the individual components.

Adding resistance to a circuit, such as a resistor or a loose wire, will affect the current and voltage in the whole circuit.

Wiring in a circuit introduces additional resistance e.g. a bad wire or a loose contact in series with a microcontroller's power pins effectively acts as a resistor. Following Ohm's law (I=V/R), one can calculate the impact the additional resistance has on the circuit.

For example, consider this Tinkercad circuit with a 5 V supply and a light bulb to simulate electrical load:

The current through the light bulb is 104 mA. We can calculate the bulb's resistance to be R=V/I or R = 5 V / 0.104 A = ~48 Ohm. The bulb is operating at 5 V * 0.104 A = 0.52 W. There are no other components in the system and wiring resistance is assumed to be zero, so the entire supplied voltage is dropped across (or available to) the bulb (load).

Now, add a 10 Ohm resistor in series with the load:

The power supply is still only providing 5 V, so the added resistance is going to limit the voltage/current available to the bulb (load). Total circuit resistance is ~58 Ohm (~48 for the bulb, 10 for the resistor simulating a loose wire) which limits the current flowing through the entire circuit to I = V / R or I = 5 V / 58 Ohm = ~0.086 A. This is now the current that will flow through each individual component in the circuit.

Per the formula V = I * R, voltage drop across individual elements is proportional to the element's resistance, and so 10/58th of the available voltage is dropped across the resistor and won't be available to the load (bulb). The bulb is now at 4.14 V * 0.086 A = ~0.35 W, down ~30% from the earlier value.