Finding Unknown Resistor Value using Voltage

In the world of sensing, there are many sensors that change their resistance value based on the environment. Knowing the sensor resistance provides a measurement of the environment being measured. These variable resistor sensors include:

  • Thermistor – a variable resistor that changes value with the surrounding temperature changes. There are two types: negative temperature coefficient (NTC) and positive temperature coefficient (PTC). The NTC thermistor decreases in value when the temperature increases the the PTC thermistor increases in value when the temperature increases.
  • Magneto Resistor – a variable resistor that changes value when a magnetic field is applied. When the magnetic fields increases, the resistance increased. When the magnetic field decreases, the resistance decreases.
  • Photoresistor – a variable resistor that changes value based on light energy. The photoresistor resistance decreases when light energy is increased and increases when light energy is decreased.
  • Humistor – a variable resistor that changes value based on humidity.
  • Force Sensitive Resistor – a variable resistor that changes values based on the force that is applied.

Thermistors are variable resistors that are more sensitive to temperature changes then a standard circuit resistor. The simple first order thermistor relationship between resistance and temperature is:

ΔR = kΔT, where
ΔR is change in resistance (in ohms)
Δ is change in temperature (in kelvin)
k is first-order temperature coefficient (in ohms/kelvin)

In general the first order approximation is only accurate over a limited temperature range. The Steinhart-Hart equation is a widely used third order approximation that improves accuracy to less than 0.02 oC over a much wider temperature range.

T is absolute temperature (in kelvins)
R is the resistance (in ohms)
a, b, and c are coefficients

NTC thermistors can also be characterized with the Β (beta) parameter equation, which is just a specialized case of the Steinhart-Hart equation.

T is absolute temperature in kelvins
T0 is 298.15 K (25 oC)
R0 is resistance at T0
R is the resistance

Having the B parameter and measuring the thermistor resistance, the temperature can be determined. But most embedded systems don’t measure the resistance directly. So the question is how do we measure the thermistor resistance. The answer is to use a voltage divider. Measuring the voltage divider voltage, which is common using analog to digital converters, gives us a way to get the thermistor resistance.

Remember that a voltage divider is two series resistors in this case connecting power and ground. The voltage between the two resistors is given by:

NTC Thermistor Voltage Divider

VOUT = VIN x (R1 / (R1 + R2))
R2 = R1 x (VIN/VOUT – 1), where
R2 = unknown thermistor resistance (in ohms)
R1 = known resistance (in ohms)
VIN = known input voltage (in V)
VOUT = measured voltage between resistors (in V)

Generally NTC thermistors have a nominal resistance at 25oC. Most common is either 10K or 100K ohms. When picking the known resistor R1, the value should match the nominal thermistor resistance, e.g. for a 100K thermistor, R1 should be 100K.

Using an embedded controller like an Arduino UNO or Nano, the code is very simple to convert the sensed input voltage to the thermistor resistance, to a temperature as shown in following code segment.

  // in setup, one time calculation
  Rinf = NTCRESISTOR * exp(-1*BETA/298.15);

  // in Arduino loop
  tempIn = analogRead(TEMPPIN);  // 0 to 1023 values

  // find thermistor R
  // SERIESRESISTOR = R1, 1023.0 = VIN, tempIn = VOUT
  Rth = SERIESRESISTOR * ((1023.0/tempIn) - 1);
  // calculate temp in K and convert to C
  tempC = BETA/(log(Rth/Rinf)) - 273.15;

One final item to consider with a voltage divider is the input impedance of the measuring device. To limit the impact of input impedance on circuit performance, generally you want the input impedance to be at least 10 times the value of R1 in the circuit above. The input impedance is in parallel with R1 so if the input impedance is only 100K, then the effective value of R1 in our circuit is only 50K, which greatly affects the measurement and calculations.

Op Amp Voltage Follower

One way to solve this problem is to use a voltage follower op amp circuit. This circuit provides unity gain (voltage divider Vout equal op amp Vout), has a low output impedance, and very large input impedance. It is important to select an op amp that has stable unity gain.

The Node Method

In the next update to my book, AC/DC Principles and Applications, I plan to add a small section on the Node Method to Chapter Nine – Complex Network Analysis Techniques. Discussed in this chapter are Kirchoff’s voltage and current laws, superposition, Thevenin’s and Norton’s theorems. This post is an advance installment of the update.

The Node Method is a circuit analysis approach that uses the circuit element properties with Kirchoff’s voltage and current laws. This method helps to reduce analysis complexity as seen in the example.

Each element connection point is a circuit node. In using the Node Method, each node voltage is label with one node being selected as ground or 0 volts. Each node also has the current labeled showing the currents entering and leaving each node.

After all the node voltages and currents are labeled, KCL is written for each node. Based on Ohm’s Law, the current through a resistor is the voltage across the resistor divided by the resistance. The voltage across a resistor is the voltage difference between the terminals.

Using the set of KCL equations, solve for the missing information.

Node Method Steps

  1. Select a reference node to be 0 volts (ground). A good choice for this reference node is one that has the most connections.
  2. Label voltages and currents.
  3. Write KCL equations for each node that has an unknown voltage.
  4. Solve for missing values using the equations developed in step 3.


Using Figure 9-3, what is the voltage drop for a 2Ω load resistor that is connected between a 12V closed-loop circuit (loop A) with a resistance of 4Ω and a 6V closed-loop circuit (loop B) with a resistance of 3Ω?

Write KCL
IL = I1 + I2
e/2Ω = (12V – e)/4Ω + (6V – e)/3Ω

Solve for node voltage e
6e/12Ω = 3(12V – e)/12Ω + 4(6V – e)/12Ω
6e/12Ω = (60V – 7e)/12Ω
13e/12Ω = 60V/12Ω
e = 60V/13
e = 4.6V
IL = 4.6V/2Ω
IL = 2.3A

This solution was simpler than creating the KVL for each loop, solving for I1, I2, and then IL, and then finding the node voltage e.