7/28/2019 How to Measure Current Using a Shunt Resistor

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How To Measure Current Using A Shunt Resistor

You have a digital multimeter. It’s rated at 10 A max, but you would like to measure a current at about 30 A.How is that possible? Good news is that by using Ohm’s Law you can circumvent such limitation! Ohm’s Lawdescribes the relationship between voltage (V), current (I) and resistance (R) in the form: V = I x R.

This article will talk about how we can perform such measurement and also how toverify that our instrument is capable of handling such a task by using real examples.Some good practices and precautions will also be mentioned here. With just a slight manipulation of that equation, we can obtain I = V / R, which indicates to us that we canalso derive the value of the current measurement if we know the voltage and resistance. From this, weobserve that the current is inversely proportional to the resistance. So, if we could fix the resistance to somevalue using a resistor, we could then measure the voltage across that resistor to obtain the current. Thisresistor should be of a very small in order not to disrupt the circuit. This is termed as the shunt resistor and isplaced in series with the circuit load as shown in Figure 1. 

Figure 1 - Using a digital multimeter & shunt resistor to measure current

Let’s consider this real-life scenario - we are still interested to measure 30 A and the multimeter that you areusing is for example, the  Agilent U1253B 4.5 digit handheld digital multimeter, which has a maximum currentrating of 10 A. Then consider some precision current shunts from Ohm-Labs. From the selection of shuntresistors, we identify one which is within the right range (rated greater than 30 A). Hence we select the CS-50 shunt resistor  model with the following specs:

Model  Amps  Ohms  Output  Accuracy 

CS-20  20  0.05  1.0 V  0.1% 

CS-50  50  0.01  0.5 V  0.01% 

Table 1 - CS-50 shunt resistor specifications 

Figure 2 - Image of CS-50 shunt resistor from Ohm-Labs. 

CS-50 is a 0.01 Ohms shunt resistor with a 0.5 V output at maximum rated current of 50 A. Notice the fins in Figure 2? That's a heat sink to regulate the temperature onthe resistor as things can get pretty heated up during operation. Back to Ohm’s Law, if we are measuring current of 30 A with shunt resistance of 0.01Ohm, we should obtain: 

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Vmeasure = 30 A x 0.01 A = 0.3 V 

This 0.3 V is definitely within the safety rating of the  Agilent U1253B which is 1000 V. This proves that byusing shunt resistors, we can measure current (30 A) that are greater than the maximum rating (10 A)of the multimeter. 

 Although ensuring that our measurement stays within the instrument’s maximumratings is crucial, we also need to ensure that we have sufficient resolution tomeasure the smallest step change that interests us. For example, to detect every 0.1

 A of change in our measurement, our multimeter must be able to measure voltage assmall as:

∆Vmeasure = ∆0.1 A x 0.01 Ohm = ∆1 mV 

 As an illustration, Table 2 below shows some values of Vmeasure as we vary the Imeasure at a resolution of 0.1 Aabout our target current value of 30 A. Notice that with every 0.1 A step of change, the voltage also changewith a step of 1 mV or 0.001 V. This table is also useful to help convert the voltage measured to thecorresponding current value. 

Vmeasure  Imeasure  Rshunt 

0.295  29.5  0.01 

0.296  29.6  0.01 

0.297  29.7  0.01 

0.298  29.8  0.01 

0.299  29.9  0.01 

0.300  30.0  0.01 

0.301  30.1  0.01 

0.302  30.2  0.01 

0.303  30.3  0.01 

0.304  30.4  0.01 

0.305  30.5  0.01 

Table 2 - Imeasure vs Vmeasure 

Referring to the datasheet of the Agilent U1253B under ‘DC Specifications’ section (see Figure 3 below), wezoom in on the 500 mV or 0.5 V range as our measurement falls within this and find that the resolution is0.01 mV. This means that the  Agilent U1253B is more than sufficient to resolve a 1 mV change. 

Figure 3 - DC voltage specifications of the Agilent U1253B

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Example DIY Shunt Resistor

In this example we will look at making a 0-10 Amp shunt resistor with a voltage drop of 10mV at 10 Amps.

Therefore using a standard multimeter or 0-10 mV voltmeter it will be easy to read the numerical value of 

the current (and ignore the units displayed) - i.e. if the voltmeter reads 6 mV we know this corresponds to 6

Amps of current, 2.4 mV corresponds to 2.4 Amps and so on.

Therefore, using Ohm's Law it can be seen that since the voltage drop of 10mV divided by the current of 10

Amps give the required resistance of the shunt = 0.001 Ohms, we need to make a 0.001 Ohm resistor.