The current handling capability of a breadboard is related to the breadboard’s finger-to-component resistance as well as the internal resistance of the metal strip connecting the fingers together. Recall that a solderless breadboard is composed of many terminal strips including 5 position strips and longer strips for the power rails.

Before we continue, we need to recognize that breadboard resistance is also a function of the breadboard’s age and quality. For this post a new, never used, breadboard was chosen.

## Methodology

Resistance measurements were conducted using a 4-wire method. A known current was passed through the breadboard (22 AWG wire) and the voltage was measured at the points shown below. For example, 0.5 A was passed through the breadboard’s power rail. The voltage measured as closely to the breadboard as possible was 40 mV. Using Ohm’s law, the calculated resistance is 80 mΩ. This measurement was repeated for each of the breadboard’s rails and the results averaged.

On average, this 5.75 inch section of the breadboard measured 80 mΩ as shown in the lower part of the picture. A second set of measurements were conducted using the right-hand side of the breadboard’s power rail. On average this 2.75 inch section yielded 48 mΩ.

The delta (difference) between the resistance measurements is 32 mΩ as shown in the red section of the picture. Given this 3-inch section and the 32 mΩ resistance, we can model the breadboard’s internal wire resistance as 11 mΩ per inch. With the assumption that the finger contact resistance is the same for all contacts on this new breadboard.

Continuing on, we can assume the right-hand side of the breadboard has a resistance of approximately 30 mΩ calculated as 2.75 inch x 11 mΩ per inch. The 18 mΩ difference between this 30 mΩ calculation and the measured 48 mΩ is the finger contact resistance.

## Breadboard Model

Taken together, we have modeled the breadboard as a device with 10 mΩ per contact and 11 mΩ per inch. With this information we can determine the voltage drop and power dissipated for any given current.

Once again, this is the model for a new, unused breadboard. I did conduct experiments using older but still serviceable breadboards with significant degradation. These experiments suggest a contact resistance of 50 mΩ may be more appropriate. That’s five times higher than our ideal model.

## Calculations at 1 A

Let’s perform a calculation assuming a 1 A continuous current. To be on the safe side, we will use the 50 mΩ approximation and voltage will pass through ½ of the rail.

R_{Total} = 50\ mΩ + (2.75 *11\ mΩ) + 50\ mΩ = 130\ mΩ

V_{drop} = 130\ mΩ * 1\ A = 130\ mV

P_{loss} = \dfrac{(130 \ mV)^2}{130 \ mΩ} = 130 \ mW

In my opinion, this appears to be a reasonable design. The heat will be distributed over the 2.75 inch length of the voltage rail.

## Calculations at 5 A

Let’s try again, by increasing the current to 5A. Again, we will use the lesser 50 mΩ approximation and voltage will pass through ½ of the rail.

R_{Total} = 50\ mΩ + (2.75 *11\ mΩ) + 50 \ mΩ = 130 \ mΩ

V_{drop} = 130 \ mΩ * 5 = 650 \ mV

P_{loss} = \dfrac{(650 \ mV)^2}{130\ mΩ} = 3.25\ W

This is an unacceptable situation. Recall that power increases as the square of current. Since we increased the current by a factor of 5, the 3.25 W dissipation is 25 times higher than the original 130 mW. This 3.25 W of heat inside the breadboard’s insulated plastic cavity is not good. It will likely melt the plastic. It will also burn you if you happen to touch the wires exiting this section of the breadboard.

## Closing Comments

The calculations in this post suggest that a breadboard can handle about 1 A. This is a good starting point but is certainly not a definitive answer. As stated in the opening, the condition of the breadboard’s fingers can have a significant impact on performance. Also, we need to consider what is happening on the top of the breadboard. As an extreme example, suppose we installed a 7805-voltage regulator or some other moderately high-power device. We can expect the heat from this device to wick into the breadboard’s fingers. This heat only adds to the I^2R losses. The heat will oxidize the fingers increasing resistance. It may also weaken the spring tension leading to an ever-worsening spiral.

It’s no wonder so many breadboards have melted sections!

Best Wishes,

APDahlen