The concept of Half Ohm low resistance meter

I watched EEVblog's video about debugging a short circuit with precise multimeter. He determined the direction of shorted place by comparing resistances in different places. I wanted to debug like that too and also measure resistances of wires and connectors, but all cheap multimeters measure only down to 0.1Ω. To get 10mΩ I have to buy 400€ multimeter. So I searched and found an article describing cheap and dirty way to measure low resistances. You need a known voltage source and known resistors and then you can form a voltage divider and measure the resistance. Awesome! But the form factor and all the math behind it sucks, as a plug and play device it would be perfect. I though about making it its own box, but was almost impossible to find cheap probe connectors for panel or PCB. So it will be a one PCB product.

First, the concept.

• It should measure resistances from 1Ω to 0.1mΩ
• General purpose - can be plugged in any multimeter.
• Output should be in mV, because most multimeters have mV display.
• No math, so 1mΩ is translated to 1mV and user doesn't have to calculate anything.
• Precise enough - 1% is nice number, but it's not very important, as usually we need the resolution, not the absolute precision.

Secondly the schematic. It is fairly simple: power supply -> voltage reference -> current limiting resistor -> connector for test probes -> voltage amplifier -> amplifier do to the math -> connector to multimeter.

It will be using small 3V coin battery as power supply. I though about the schematic and BR1225 coin cell would be perfect size and I already have 30 of those. Maximum urrent consumption will be about 10mA so I checked how much voltage I will get out of 3V BR1225 @10mA. The voltage started out from 2.5V and dropped to ~2.1V, then it stayed there. So our power supply gives out voltages between 2V and 3V. Edit: From there we can calculate that the internal resistance of the battery is (3V*330Ω)/2V - 330Ω = 165Ω.

The higher the reference voltage, the less error from amplifier, but too high voltages and currents can damage schematics, so we have to mind that. Less current means more error from amplifier, but higher current - more error from test leads and connectors. So I started to search cheap voltage reference below 2V. The highest voltage I found was 1.24V. I chose 1% shunt reference with up to 20mA current handling. The resistor between voltage reference and battery should limit the current so that with maximum voltage of 3V the current will be smaller than 20mA. So the value of the resistor has to be at least (3V-1.24V)/20mA = 88Ω. The nearest round value that I like is 100Ω, so I will go with that.

Next, the resistor. The shunt and current limiting resistor in parallel with it form a current divider. The value of measuring resistor should be chosen so that with minimum battery voltage, still at least 1mA of current flowing through the voltage reference. There is (2V-1.24V)/100Ω = 7.6mA of current flowing through the voltage reference and resistor. 1mA to voltage reference and we get that the resistor has to be 1.24V/6.6mA = 188Ω. I think that I will go with rounder 200Ω.

The amplifier will be configured in positive feedback  non-inverting negative feedback mode. When measuring 0.1Ω resistor, the output must be 0.1V. The gain must be: 0.1V / ((1.24V/(200Ω+0.1Ω))*0.1Ω) = 161. Since no common resistor values divide by 161 then the easiest way to get gain like this is to use 1k resistor and 160k resistor in parallel with another 1k resistor. Edit: Since I suck at math I wrote that wrong. The gain of the op-amp is (R1+R2)/R1. So I need 1k for R1 and 160k for R2.

Okay, all this seems pretty on paper, but will it work? Next post will be on error calculations to find out how precise the beast is.

Posted on 2012-01-09, 12:28 By

## 4 thoughts on “The concept of Half Ohm low resistance meter”

1. Iby K. says:

what do you mean by "too high voltages and currents can damage schematics"? how can schematics sustain damage? it is a drawing... 😉
btw. nice project...

2. Iby K. says:

btw, yur feedback (through R4) goes to inverting input, therefore it is negative (text says "positive feedback"). i think you meant to say 'non-inverting amplifier'.

3. jaanus says:

You are right, I don't know how I managed to write that wrong. Thank you.