A Guide about How to Build Your Own Low-Resistance Meter
A digital multimeter can definitely measure resistance, but not as low as below 0.1 Ohm. The project below is a comprehensive guide to how you can make your very own Low-Resistance ohmmeter to estimate low resistances.
The Theory of Low-Resistance Meter:
Generally, the Wheatstone bridge and the RC calculation method is used to calculate resistance. The method for this particular project is the most fundamental equation in electronics:
A constant current will be incorporated which will not only provide the current through the resistance under consideration, but will also estimate the voltage drop. Furthermore, the voltage drop will be attached to a standard multimeter to multiple the effect. It is important to note that the voltage will be equal to the resistance in ohms i.e. 1V=1 Ohm. By using the above mentioned equation and plugging in the estimated values of R (less than few ohms), we can select a sufficient current which would be able to generate voltages for the amplifier stages that follow the constant current stage.
In this project the operational amplifier’s input offsets voltage, which is replicated as a voltage source in series with the operational amplifier's inverting or non-inverting input terminals. The operational amplifier’s non inverting gain magnifies the voltage and thus as compared to an ideal circuit it can generate an error because of higher chances of fluctuation. Hence, the circuit needs to be designed in a way so as to minimize the chances of errors. By making sure that the signal of interest is greater than that of the offset voltage i.e. ±2mV for the LM358.
The project’s objective is to calculate resistance as low as 0.1 Ohm. This implies in order to flow the current through a resistance of 0.1 ohm, we need to use a constant current source which is able to generate a voltage i.e. larger than 2mV. A tradeoff has to be done as a higher amount of current has some setbacks and lower currents tend to decrease the voltage drop across the resistance under consideration.
Disadvantages of the higher current:
Following are some of the disadvantages of the higher current:
- Power consumption is higher, comparatively.
- A larger amount of heat is generated by the constant current source.
- High current won’t allow us to measure resistance of circuit components which are prone to heat damage (for instance: thin wires).
The amount of current in this circuit will be 100 mA. Even on this much amount of current, 100 mV will be produced across a 0.1 Ohm resistor.
The constant current source constitutes of the following:
- U1A - LM358
- Q1 - 2N3055 (TO- 3 package)
- RV1 - To adjust the reference voltage attached to the operational amplifier’s noninverting terminal
- R1 and R2 - divider
- R3 - sense resistors
- P2 - to measure the resistance
For the power dissipation to be 0.1 W, a constant current of 100 mA will be passed through the 1 Ohm resistor. When the resistor is connected to P2, Q1 will conduct 100mA and a TO -3 package is attached in the circuit to prevent the transistor to overheat.
A differential amplifier is used in this circuit to measure the difference between the voltage on one terminal of the resistance and the voltage on the other terminal of the resistance.
In this case the differential amplifier constitutes of the following:
- U1B – the operational amplifier
- R4, R5, R6, and R7 – these resistors configure U1B as a differential amplifier
- R8, R9, and RV2 – offset adjust
The addition of R8, R9 and RV2 in the circuit makes it possible for us to connect an adjustable offset voltage to the output of the differential amplifier. Hence, the error can be minimized. The Calibration section below will explain how to go about the compensation circuit.
At the end, a gain of 10 is produced by the amplifier. The extra gain allows us to fix the measurement ratio to 1:1 i.e. 1 Ohm of resistance is able to generate 1V of the output.
- U2A, RV3, and R10 – non-inverting amplifier with gain of 10 (RV3 set to 90K)
- U2B – output buffer
Bill of Materials:
|U1, U2||LM358 – DIP 8|
|R1, R4, R5, R6, R7||100K resistor|
|R2, R10||10K resistor|
|R3, R8||1R 1W metal film resistor, 1% tolerance|
|RV1, RV2, RV3||100K linear potentiometer|
|Q1||2N3055 BJT, TO-3|
|C1, C2||100nF decoupling capacitors|
Following are some suggestions to construct the circuit:
- Project Box - To keep the circuit within a box and to avoid any mess, an internal 9V battery and external connectors can be used.
- Multimeter attachment - Banana plugs can help you to connect a multimeter to your circuit.
- A meter - You may choose to construct your very own measurement device by adding a small voltage meter.
A split power supply will be required for full functionality. A negative rail would not be needed if your requirements are fulfilled by without the compensation circuit. If you are not incorporating a LM358, then you need to know that the input common mode voltage range of the operational amplifier must produce 0V, as we have used an input voltage of approximately 100 mV.
First, you need to calibrate the constant current source by using a multimeter to estimate the constant current. Then adjust the value of RV1 till the point the current is 100 mA. Start by adjusting the RV1 to its minimum resistance to avoid any potential harmful current flow through Q1 and R3.
The next step is to minimize the potential error that can be caused in the differential amplifier. This can be done by first estimating a known resistance and then adjusting RV2 accordingly, till the point the output of the differential amplifier corresponds to it.
In the very last step, you need to adjust RV3 in such a way that it results in a gain of 10 for the U2A amplifier. Calculate the non-inverting input voltage of U2A and adjust RV3 such that the output is 10 times the measure of output.
Test your circuit to check the functioning. If it’s functioning well, then you’re good to go. This circuit is very cheap to build and can save a lot of money and provide you an opportunity to practice over and over again.