# Can you live with the burden?

May 6, 2020 | Electrical

By Chuck Newcombe

One of the realities of electrical testing is that you can't measure something without changing it in some way.

The usual 1 or 10 Megohm input resistances offered on most meters generally minimize circuit loading effects on the measured source, so we tend to forget about it. Of course, if the source resistance of the thing you're measuring is very low, such as a car battery, or an energized 20 amp receptacle, then these loading effects are not significant.

One exception that shouldn't be ignored occurs when measuring high frequency ac signals. In that case, the shunt capacitance at the dmm input creates a much lower impedance load than the 10 megohm resistive divider input. I wrote about this recently in the column titled 'Off the Wall Measurements.'

And, with dc volts, if you are looking at the output of a photocell light sensor, or similar electronic devices, the source resistance may be very high, and attaching your 10 megohm meter input can significantly change the reading you see, not to mention the function of the device itself.

But, not to worry! If you want to measure a high impedance dc source, and the anticipated voltage is less than 600 mV, the Fluke 87V has a power-up feature that can save the day. Just hold the 'Hz %' button while you turn the meter on to remove the 10 megohm divider from the circuit. Now, the dmm's loading effect is greatly reduced, with an input resistance of 1000 Megohms or more.

What is Burden Voltage?
One thing we rarely look at is the effect we have on a circuit when we attempt to measure the current flowing through it. In this case, the act of inserting the dmm 'A' or 'mA' input shunt in series with a current loop can significantly reduce the actual current flowing in the circuit. If the source voltage in the loop is very high, or the loop is current regulated, then the effect is minimized, but when you are operating in a circuit using a pair of AA batteries in series as a power source, you have only 3 volts to play with, and the shunt of a dmm's mA input can present a significant circuit loading error if you fail to account for it's effect.

So how much effect does the current shunt present? There is a little used specification in the user manual that can help you figure that out - it's called Burden Voltage, and once you know how to use the information, you can make corrections to your measurements to compensate for the effect.

In the case of the 87V which I mentioned earlier, the burden voltage for the 400 mA range is typically 1.8 mV/mA. What that means is that if we're reading 1 ma, there will be 1.8 mV drop across the terminals of the meter, and if we're reading 300 mA, there will be roughly 0.54 volts drop across the terminals.

What it all boils down to is that this burden voltage has been subtracted from the voltage available to the circuit being tested. So, if you had a circuit that was drawing 1 mA when connected to a 3.0000 V source such as 2 AA batteries, you would only have 2.9982 volts available to operate the circuit. The 1.8 mV drop represents only 0.06% of the original supply voltage, and since the dc current measuring specification of the 87V is 0.2%, you can probably ignore the burden voltage effect.

If, on the other hand, you measured 100 mA in this 3V circuit, the available voltage would be only 2.82 V, and the 0.18 V burden voltage now represents 6% of the original supply voltage - now you may have a significant error.

In this last case, if the circuit load is linear in nature, then you can estimate the current that would flow with the meter removed from the circuit by adding 6% to the reading, for an estimated circuit current of 106 mA.

So, what is the resistance of the meter's input that is being placed in series with the test circuit? We can complete the experiment using Ohm's Law.

0.18 V divided by 0.1 A equals 1.8 Ohms, and that is the approximate combined value of the mA shunt and its protection fuse. (The actual measuring shunt alone is 1.0 ohm)

Again using Ohm's Law, the effective resistance of the circuit load is 2.82 V divided by 0.1, or 28.2 Ohms, so the battery load is 28.2 + 1.8, or 30 Ohms. By removing the meter, the battery load resistance drops to the 28.2 Ohms, which when powered by 3V yields 106.3 mA.

The example shown is probably an extreme case. As mentioned above, if you're working with higher voltages, the potential error is less, and maybe negligible, but it's always a good idea to compare the estimated burden voltage with the available source voltage to get a better idea of the possible circuit loading error that might be present.