If you want to know the resistance of low resistance components such as wires, switches, and coils, this milliohmmeter can be used. It is very simple and inexpensive to make. It even fits in your pocket. Most ohmmeters are accurate down to 1 ohm but this one is sensitive to low resistance in the range of milliohms or even microohms.

## Step 1: Materials

R1: ~220 ohm resistor

R2: Unknown resistance

2x thin wires (eg. mobile charger cords)

Rectangular shaped plastic box

5V source (eg. USB port, mobile chargers)

2x alligator clips

DC jack and connector (optional)

Solder

Hot glue

Multimeter with ohms and millivolts ranges (the lower the voltage range, the more sensitive the milliohmmeter)

Calculator

## Step 2: Drill Holes on Case

Drill holes for the wires and leads.

## Step 3: Soldering

Soldering can be done boardless. Just hot glue the parts to the box. If your power supply is bullky and you want it detachable, include the DC jack and connector.

## Step 4: Using the Milliohmmeter

Before testing the unknown resistance, measure the resistance of R1. It should be close to 220 ohms.

To measure the unknown resistance (R2), attach it to the milliohmmeter's test leads. Measure the voltage across R1 and R2. When measuring R2's voltage, measure it across R2 directly. Do not measure the voltage across the alligator clips because the contact resistance will add up the voltage drop and overestimate the resistance.

Based on Ohm's law, we know that R1 and R2 has equal current flowing through them. Because of this, we can use V2 and the current to calculate the unknown resistance.

R2 can be calculated as follows:

R2=V2/(V1/R1)

Where

V1=Voltage across R1

V2=Voltage across unknown resistor

R1=Measured value of R1 (~220 ohms)

In the second picture, I used an ammeter as an example.

This link has more details about the low resistance tester..

http://www.robotroom.com/Measuring-Low-Resistances.html

## Step 5: Measurements of Low Resistance Parts

Based on the calculations and expected values, this milliohmmeter was reasonably accurate.

Since the voltmeter has a range down to 0.1 mV, it can measure down to 0.01 ohm. To increase the sensitivity, you can purchase a more sensitive voltmeter or use a lower resistor value. Because resistors are sensitive to temperature changes, the power rating needs to be higher.

## 11 Discussions

1 year ago

another point of observation. when making very low ohm measurements where a high current is passed thru the low value resistor a Kelvin connection is necessary i.e. two wires at each terminal; one pair for the high current and the other for the voltage measurement at the point of contact, so this now becomes a more elaborate instrument and is not simple anymore. again thanks for sharing this idea.

1 year ago

if i need to increase accuracy i can add an x10 or even a 100x op-amp powered from the 5V.

there are many ways to skin a cat and this approach is very simple. and obviously one does not need two DVM's. take one reading and then the other. If as one of the readers was talking about measuring closed relay contact resistance in that case one needs to measure in the micro ohms range. i want to measure the resistance between two Nickle strips for spot welding. i was making assumptions about their contact resistance and did not know what is needed to design my battery terminal spot welder.

1 year ago

Thanks for sharing. this is an excellent idea and it is accurate enough for my needs

3 years ago on Introduction

3 years ago on Introduction

Field low resistance ohm-meters inject a relatively high current (5amp up to 20 amp) just measure those low resistances in the micro-ohm range. You need a high enough current to measure the voltage drop on stuff such as circuit breaker (closed) pole resistances, bolted connections etc. A low current application is not suitable for field work.

3 years ago on Introduction

=== the only suitable circuit to measure very small resistances is the Wheatstone bridge circuit. ===

With all due respect, a Wheatstone Bridge is not the _only_ suitable circuit for such measurements, but it is well suited for a lab where you have access to things like calibrated precision resistors and a rheostat that has a calibrated dial. The nice thing about the Wheatstone bridge is that it only depends on resistance ratios, and not on needing to measure voltages or currents precisely. However, most home experimenters will not have access to known precise low-value resistors, or a rheostat with a calibrated dial. But, digital multi-meters with relatively accurate millivolt measuring capability are very common and inexpensive, and the method described in this Instructable takes advantage of this measurement capability. Both this method and the Wheatstone one have limitations, but both are theoretically sound methods. Use whichever one you have the equipment for.

3 years ago on Introduction

god job.

6 years ago on Introduction

Sorry, the only suitable circuit to measure very small resistances is the Wheatstone bridge circuit. It is not much difficult than your diagram. You even do not need both voltmeters to measure the resistivity correctly :-)

Reply 6 years ago on Introduction

In this milliohmmeter, you only need one voltmeter to test the voltage across R1 and the unknown resistance. I've never used the Wheatsone bridge circuit before but I will check it out. Thanks for sharing.

Reply 6 years ago on Introduction

In my time as a physics student we had a laboratory work where we used a wire-rheostat, known resistance (to minimize the error it should be close to the unknown one) and the lamp. The rheostat in your case you can substitute with a high-resistant wire (I have seen such wires with the resistivity of 1 Ohm/cm and 100 Ohm/cm).

The idea of the method is: R3 is your known resistivity, R1 and R2 together is the wire-resistivity, voltmeter is replaced by the lamp. So, if you will find the point on the wire-resistivity which correspond to the condition of no current through the lamp (lamp is off), that will mean that the bridge is balanced. And for the balanced bridge: Rx=R1*R3/R2, where R1/R2 can be replaced by the ratio of their lengths.

No two voltneters, no Excel, only wire, lamp and resistor :-)

Reply 6 years ago on Introduction

An absent picture of the bridge :-)