Create a circuit for the precise measurement of low resistances.
You already have a DMM to measure resistance, but it can be used with resistances lower than 1Ω? And if so, these low resistive readings are reliable?
This project will show you how to make your low resistance meter; it only uses a handful of components and can measure resistances up to one hundredth si Ω.
Resistance measurement can be performed using a variety of methods, but in this project the chosen method is to use the most fundamental equation in electronics:
V = I x R
A constant current source will establish the current through the resistance under test and measure the voltage drop produced by the resistance. This voltage drop will then be amplified and fed into a standard multimeter. The amplitude of the voltage will be equal to the resistance in ohms (the art. 1V = 1Ω).
You must select a current that produces reasonable voltages for the amplifier stages that follow the constant current stage.
An important consideration is the offset voltage of the op-amp input, which is modeled as a voltage source in series with the inverting or non-inverting input terminal of the operational amplifier. This voltage is multiplied by the non-inverting gain of the operational amplifier and is an error source because it can make the output voltage different from what we would expect from an ideal circuit. I have to design the circuit so that the effect of this offset voltage is low.
The op-amp used does not have offset-null functionality, I'm using the LM324, which does not include this pin. instead, I can easily reduce the influence of the offset voltage by ensuring that the input signal is much larger than the offset voltage, which is ± 2mV for the LM324.
My goal is to measure resistance up to 0.1Ω. This means that I have to choose a constant current source that creates a voltage significantly greater than 2 mV when the current passes through a resistance. This is a compromise, because higher currents have disadvantages and lower currents reduce the voltage drop across the resistance under test. The problems with higher current are as follows:
Greater energy consumption, while lower energy consumption helps with portability.
Lower currents produce less heat generated by the constant current source circuit.
Lower currents reduce power dissipation and therefore increase the temperature of the resistance under test; with a lower current, we can measure the resistance of the circuit elements that are most susceptible to heat damage (thin threads, eg).
The current chosen for this circuit is 200 mA. This amount of current is not too high but generates 20 mV through a resistor from 0,1 Ω e ± 2 mV equals ± 0.01 Ω. I consider it appropriate to be a value that falls within the intrinsic tolerance of the resistors used which it is usually 5%. With appropriate circuit arrangements I will significantly reduce this error
The constant current source consists of
U1A – LM324
Q1 - BD135 with thermal resistance of 10 ° C / W without fin, starting from 5V of power supply admitting to using the terminals short-circuited between them I will have 5V x 0.2A = 1W = 10 ° C of temperature increase in the transistor A small fin will guarantee me an increase within 5 degrees.
R1 – potentiometer for adjusting the reference voltage applied to the non-inverting terminal of the operational amplifier and consequently the current supplied
R8 – detection resistance (1Oh, tolerance 1%)
With a constant current of 200 mA through the sensing resistor from 1 Oh, the power dissipation is 40m W (hence the choice of 1/4 W)
The next stage after the constant current source is a differential amplifier with a gain of 1 and an adjustment of the offset voltage. Here we are using an amplifier “differential” because we want to detect the voltage drop through the resistance under test, or the difference between the voltage on one side of the resistance and the voltage on the other side of the resistance.
The differential amplifier consists of
U1B – the operational amplifier
R11, R12, R13 and R14: these resistors configure U1B as a differential amplifier
R15, R17 and R18 – offset adjustment
The circuit consisting of R15, R17 and R18 allows us to add an adjustable offset voltage to the differential amplifier output. To do this I need a negative voltage source.
I get it with this stage.
Q1 and Q2 form a classic square wave oscillator. R5, R6, C1 and C2 determine the operating frequency which is actually not critical at all. Q4 and Q5 are configured as an almost complementary push-pull symmetry amplifier. The use of PNP transistors is due to what I actually have at home during this period of forced isolation. At the push-pull output I have C4 which uncouples the DC voltage and the two D1 and D2 straighten the negative component of the signal and filter it with C5, by creating this negative voltage it is useful for correcting the intrinsic offset of the integrated. With this system I reduced both the input and output errors.
The last stage is an amplifier with gain of 5. This additional gain sets the overall measurement ratio to the convenient value of 1: 1, or 1Ω of resistance produces 1V at the output.
The use of components with standard tolerance would compromise the accuracy of the instrument, therefore I added a last calibration with R19 on the gain of this last stage to have the correct reading.
I used an old cell phone charger, 5V 600mA but half the current would have been enough. A good system to free the drawers of components that I didn't want to throw because they are functional but, which have now become unusable due to the programmed obsolescence of the devices they were supposed to charge. Nothing prevents you from using a USB cable and connecting it to the PC socket more and more an ubiquitous "tool" near the workbench.
The first part of the circuit to be calibrated is the constant current source. The simplest method is to use a multimeter (connected to J2) to measure the constant current.
Adjust the value of R1 until the measured current is 200 mA. Start with R1 set to its minimum resistance. This minimizes the setting of the initial constant current and therefore prevents potentially harmful quantities of current through Q1 and R8.
With the constant current set, we have to compensate for the error in the differential amplifier output. You can do this by measuring a known resistance and then adjusting R18 until the differential amplifier output corresponds to the known resistance (eg, a resistance of 1Ω should produce a differential output of 200mV), or you can measure the voltage through a small resistance using a precise voltmeter and then adjust RV2 so that the output of the differential amplifier is equal to the measured voltage.
The final calibration phase consists in adjusting R19 so that the gain of the U1C amplifier is equal to 5. Measure the non-inverted input voltage of U1C and adjust R19 until the output is 5 times the value of the entrance, or more simply give exact 1V with a 1Ω resistor.
Only one factor I could not solve. The output with inputs in short circuit give a resistive value that is not zero. It does not depend on design errors, it's just that the operational used internally has the transistor output stage.
From the internal equivalent scheme it can be deduced that in saturation the maximum output will be V + minus the voltage drop of about 1.4V caused by the darlington consisting of Q5 and Q6 while the minimum output voltage will never be zero, it is necessary to add the saturation voltage of Q13 which it is typically around 5mV. That is the minimum displayable value. Practically in short circuit it will signal me that the resistance will be 0,005 Ω ossia 5m Ω, if this value is exceeded, the displayed resistance will be the effective one.
https://www.elettroamici.org/wp-content/uploads/2020/04/icon-resistor.jpg275361Amilcarehttp://www.elettroamici.org/wp-content/uploads/2017/08/FAVICON-1-300x271.pngAmilcare2020-04-21 16:39:472020-04-22 12:29:04Low resistance meter