My war against dual rcd boards

If you were to do a Type A RCD test on the RCDs installed in the consumer unit, what test current does the installation tester deliver?

AJJewsbury said:

I thought that A type RCDs only needed to cope with 6mA of d.c.?

The DC component of the Type A test current is 6 mA DC.

- Ross

Chris Pearson said:

Hang on. BS 7671 643.8 specifies, "alternating current test at rated residual operating current (I_Δn)". No more, no less.

Recall however that:

• BS 7671 (643.7.1) and (643.8) require, as stated above, the application of an alternating test current, which is achieved with the instrument set to Type AC.

• BS EN IEC 61557-6, the standard for RCD test instruments, allows the alternating test current to be -0 % / +10 % relative to the RCD rated residual operating current.

- Ross

There is a good summary of the requirements for RCD testing in accordance with BS 7671 in Issue 91 (July 2022) of the IET Wiring Matters magazine:

https://electrical.theiet.org/wiring-matters/years/2022/91-july-2022/changes-to-rcd-testing-in-bs-76712018plusa22022/

- Ross

Ross C said:

A multiplier of 1.4 comes to mind, so with the instrument set as in the photo (30 mA × 1 × 1.4) = 42 mA ?

Why?

I'm not sure of the exact derivation, but the IET Wiring Matters article concurs:

'When the Type A setting is selected on the instrument, a half wave pulsating residual test current superimposed on a smooth direct  current of 6 mA is produced, which effectively applies a 1.4 multiplier to the rated residual current (I_Δn). For example, if the 30 mA setting is selected, the RCD will be subjected to a test current of 42 mA (30 x 1.4 = 42 mA)'

- Ross

This is messy. A half wave rectified AC is not there at all for half the time, and the relationship between its RMS value, peak value and average (ho yes, unlike an AC the average is not zero..) is not a simple thing.
1.4 sounds like the ratio of the peak to the RMS of the AC before it was rectified, but without the qualifier, it is far from clear. 

(

A Sine wave of 30mA RMS has 
42mA pk, 84mA p-p and zero average voltage

Remove say the -ve half cycles and this becomes

15mA RMS, (half the heating power, as off exactly half the time)
42mA pk and p-p are the same (as everything below the line removed)
and average of about 14mA (~33% of peak, derivation by integration is left to the student *) 

)

More importantly, I have no idea what they actually intended people to use in the spec.

42mA pk,

42mA RMS

or 42mA average ?

Mike.

* I'm not that mean, look here where it is demonstrated for you in some lecture notes from a college in Babylon, I think .... 

BS EN IEC 61557-6 doesn't appear to be specific on the matter, only giving special mention to the test current required for a Type B RCD, which is to be an 'increasing smooth direct residual current'.

The only mention of the other RCD types is a table in Annex A (Informative) of the standard, which states which of the three types of test current (AC sinusoidal / AC half-wave / DC) are relevant to each RCD type and which also includes a mysterious Type B+ RCD.

As discussed in the IET Wiring Matters article, we are probably in an not-well-defined area somewhere between manufacturers production testing and the in-service testing of RCDs, but for compliance with BS 7671 only a AC sinusoidal (Type AC) test current is required.

- Ross

I think that I have an explanation.

The rating of an RCD is the nominal RMS value of the alternating residual current which causes it to trip, I_Δn.

The RMS value of a sinusoidal alternating current is the peak value divided by √2, but for a half-wave pulsating current, it is half the peak value. The difference in RMS value is because the first step is to square all the instantaneous voltages. Rectification takes half of them away, so the mean square value is halved. When the square root is taken, the RMS value of the half-wave current is not the RMS value of the full sinusoidal current divided by 2, but divided by √2.

So for a 100 A peak, the RMS value of the alternating sinusoidal current is 71 A, and for the half-wave rectified current it is 50 A.

Table 1 of BS EN 61008-1:2012+A12:2017 gives the maximum values of break time for full a.c. for types AC and A RCDs at I_Δn, 2 I_Δn, and a choice of 5 I_Δn, or 0.25 A. Table 2 gives the maximum values of break time for type A at 1.4 I_Δn, 2.8 I_Δn, and 0.35 A (which ≈ 0.25 A x √2).

Unsurprisingly, the maximum values are the same because the mains voltage, and hence current, has not changed, it has simply been rectified. If the tables had used the peak current, they would have been identical.

I hope this makes sense.

mapj1 said:

I'm not that mean, look here where it is demonstrated for you in some lecture notes from a college in Babylon, I think

Three wise men came from the east, but I am not sure that he is one, because he has not really explained himself.

My apologies, I am not sure that that is entirely correct.

The explanation may be quite simple. I assume that a certain amount of power is required to activate the mechanism of an RCD. Half-wave rectification clearly reduces the power by a factor of 2 because no current flows for half of the time. So, given that power, P = I²R and R remains constant, if the same power is to be produced by the half-wave rectified current as the full a.c. one, I² must double and I must be increased by a factor of √2.

My apologies, I am not sure that that is entirely correct.

The explanation may be quite simple. I assume that a certain amount of power is required to activate the mechanism of an RCD. Half-wave rectification clearly reduces the power by a factor of 2 because no current flows for half of the time. So, given that power, P = I²R and R remains constant, if the same power is to be produced by the half-wave rectified current as the full a.c. one, I² must double and I must be increased by a factor of √2.