Burden calculating for large current transformer (400A/1A)

Hi, I want to measure high currents up to 400A installed on rail like this:

I choose this transformer: https://www.wago.com/global/current-transformers-and-voltage-taps/plug-in-current-transformer/p/855-501_400-1001

This transformer is rated 400A/1A, so I would like to ask you if my calculations of needed burden resistor are correct. I will use 3.3V as reference:

burden_resistor = (system_voltage / 2.0) / ((I_RMS * 1.414) / ct_turns)

So I calculate that I need 1 ohm burden resistor, is this true? Also how I can calculate the needed power raiting of the resistor?

Thank you in advance!

Welcome, Georgi, to the OEM forum.

That formula is OK, but you must remember that current is not necessarily a nice “text book” sine wave, therefore your “1.414” might be wrong - and by a very large amount. It might be like this:
(Washing machine brushless d.c. motor on spin by @dBC Phase measurement and correction in IoTaWatt - #17 by dBC)

and in this case, the peak value - which is what limits the measurement - is much more than 1.414 × Irms. Only you know the type of load that is connected - and that will determine the shape of the current wave. It is very unlikely to be as bad as that image, but it is something you must allow for.

You must also allow for all the components to have a tolerance, so your 3.3 V might be low by a few percent, the resistors or regulator that biases to input to half the supply voltage might make that wrong also by a percent or two, and then there is the tolerance of the c.t. itself and your burden.

I usually say that, for a 3.3 V ADC input range, your rms voltage at the input must be approximately 1 V, and not more than 1.1 V, and that allows for some distortion of the current away from a nice sine wave, and in most cases enough allowance for component tolerances.

The power rating of the burden resistor is the same as any other resistor: P = V I = V² / R = I² R.
If the secondary current is 1 A, and the burden voltage is 1.1 V, the power rating is 1.1 W minimum. You must rate the resistor much higher than that for two reasons - you will extend its life significantly if it is working well below maximum power and therefore running much cooler, and the change of resistance due to temperature rise will be much less, therefore the overall accuracy will be better. I would suggest at least 5 W rating.

Thank you very much for your answer

Hi Robert, again I would like to thank you for your help in the other topic that I started (reagarding the calcIrms calculations).
Now I have a question regarding the choice of current transformer. I want to measure a building and I know its consumption profile for last year. The peak of the current is 360A so I decided to buy 400A current transformer. The problem is that all current transformer that I found on the market are 400A/5A and I calculate the burden resistor value of 0 ohms (my system is at 3.3V):

burden_resistor = (system_voltage / 2.0) / ((I_RMS * 1.414) / ct_turns)

Thanks in advance!

If you stop to think, that clearly won’t work – a zero ohm resistor will drop zero volts at any current, so you will always read zero! You need to be a little more precise in your maths.

I ignore those equations. Although they are correct, it’s the hard way to do the sums. Here’s my method:

  1. Calculate the rms value of a pure sine wave that will fit inside the ADC input range.
  2. Subtract a percentage for the worst case component tolerances.
  3. Subtract a few more percentage points for “headroom” (that’s some spare capacity for anything you forgot).

When I do that for an emonTx, I get 1.1 V rms. For the emonTx Shield, I get 1.6 V. These are the only numbers you need to remember for the emonPi, emonTx or emonTx Shield, or most other systems working on 3.3 or 5 V. Then

  1. Use Ohm’s Law with that voltage and your 5 A secondary current to calculate the burden resistor value.

I get 0.22 Ω. I can buy one of those:


The rating of 20 W is more than adequate (you only need about 5.5 W) and will give very little variation in resistance (i.e loss of accuracy) with temperature rise.

Thank you very much!