Query about single-ended LED tubes

[Steve4Physics]

TL;DR Summary

Single-ended LED tube has around 2 ohms resistance between the non-input pins. Why?

Not sure if this is more DIY than Electrical Engineering. However...

I have some single-ended LED tubes, i.e. the live (‘hot’) and neutral connections are at one end – pins A and B on the diagram:

Pins C and D are internally connected with a resistance of around 2 ohms between them.

The circuit used is essentially this:

This is to allow the tube to be fitted either way round. But in this case I don’t understand why there is a resistance of around 2 ohms between pins C and D. Surely the pins should be directly connected inside the tube so there is negligible resistance between them. Can anyone explain this?

Thanks

 

[berkeman]

Can you link to the datasheet? If this is a line-operated LED string, it likely has control circuitry inside and those extra pins could be for dimming or similar.

 

[Baluncore]

Surely the pins should be directly connected inside the tube so there is negligible resistance between them.

Maybe the LED tube is presenting a fake starter filament at each end of the tube, as expected with legacy tubes and starters. Two ohms will not drop more than 0.2 V, with the 100 mA LED current, so it is not a problem.
I suspect that the external connection, you show between the ends, is not actually present in the batten circuit, as that link would prevent a normal discharge tube from operating.

 

[Steve4Physics]

Can you link to the datasheet? If this is a line-operated LED string, it likely has control circuitry inside and those extra pins could be for dimming or similar.

Non-dimmable AFAIK.

They are Philips CorePro 14.5W 1200mm LED tubes. I’ve had them a few (about 4) years but Philips don’t now list the exact model. But this one seems almost identical:

Product leaflet PDF:
https://www.lighting.philips.co.uk/...zed_commercial_leaflet_929003519702_en_GB.pdf

Installation instructions PDF:
https://www.lighting.philips.co.uk/...line_QIG_for_324166183191_Corepro_LEDtube.pdf

I should add that the tube work fine in the Post #1 circuit. It’s just that I’m puzzled about the reason for the resistance between the non-input pins.

It’s not that important – please don’t spend any significant amount of time on this.

 

[Steve4Physics]

Two ohms will not drop more than 0.2 V, with the 100 mA LED current, so it is not a problem.

Yes - in operation the end with the non-input pins doesn't even get warm.

I suspect that the external connection, you show between the ends, is not actually present in the batten circuit, as that link would prevent a normal discharge tube from operating.

The external connection in between the ends is definitely there!

There used to be a fluorescent tube + ballast in the fitment. The ballast has been completely disconnected and the wiring changed to what is shown in Post #1.)

The old starter has been replaced by one that contains just a single thick wire for continuity. (I discovered this after accidentally breaking one!

I've now have discovered that this particular model of LED tube can run with a ballast (EM) or directly from the mains with no ballast. It has been wired to run directly from the mains as shown in Post #1. Maybe the small resistance between the non-input pins is something that is needed if used with an EM ballast.

EDIT - typo's.

 

[Baluncore]

Something does not add up. With single ended operation, if the tube is turned end-for-end, then the AC supply will be connected across the hypothesised internal 2 ohm resistor. I would expect symmetry, so that 2 ohm measure should be at both ends of the tube.

 

[Steve4Physics]

Something does not add up. With single ended operation, if the tube is turned end-for-end, then the AC supply will be connected across the hypothesised internal 2 ohm resistor. I would expect symmetry, so that 2 ohm measure should be at both ends of the tube.

Maybe I haven’t explained what’s going on very well. This might clarify things...

The tube is meant for 230V operation and has some sort of PS circuit inside. When measured in isolation with a multimeter:
Resistance between input pins: $R_{AB} = R_{BA} =$ very high.
Resistance between non-input pins: $R_{CD} = R_{DC} = 1.3 ~\Omega$.

The tube can be wired like this (with C and D unconnected):

I’ve tried it and it works fine.

But the problem with this is that if a replacement tube is accidentally fitted the wrong way round, live and neutral are connected to C and D - effectively a short.

However with the following circuit the problem of fitting the tube the wrong way round is avoided:

Suppose the tube is inserted with A and B on left:
Circuit path is: L-A-[internal circuit]-B-D-[$1.3 ~\Omega$. resistance]-C-N

If the tube is inserted with C and D on left:
Circuit path is: L-C-[$1.3 ~\Omega$. resistance]-D-B-[internal circuit]-A-N

It doesn't matter which way round the tube is inserted. However it would be better if there were negligible resistance between C and D, not $1.3 \Omega$. So the $1.3 \Omega$ is present for some unknown reason.

Edit - typo's
Edit 2 - It looks like I changed $R_{CD}$ from 2 $\Omega$ (in Post #1) to 1.3 $\Omega$ (above). I have three of these tubes, $R_{CD}$ for one the them is 1.3 $\Omega$ but the average value is 2 $\Omega$. I should have stuck with the average value to avoid confusion. Apol's.

 

[Baluncore]

But the problem with this is that if a replacement tube is accidentally fitted the wrong way round, live and neutral are connected to C and D - effectively a short.

Which explains why: "This method is not permitted in Switzerland".

This design is a compromise. Maybe the 2 ohms you measure, only at the CD end, is an internal fusible link, or a protection circuit. Two wires from CD run the full length of the tube to supply circuitry at the AB end of the tube.

 

[berkeman]

However with the following circuit the problem of fitting the tube the wrong way round is avoided:

Thanks for the clarification, it is helpful.

The tube is meant for 230V operation and has some sort of PS circuit inside. When measured in isolation with a multimeter:
Resistance between input pins: $R_{AB} = R_{BA} =$ very high.

Do you know what test voltage your DVM uses for that resistance measurement? Maybe try using the Diode Test setting instead, to get a source voltage high enough to detect the input diode bridge, if that's the first stage of the power supply.

 

[Steve4Physics]

Which explains why: "This method is not permitted in Switzerland".

Well spotted! For anyone reading this and puzzled, the quote is from a small diagram in the installation instructions (link in Post #4). Apparently the regulations in Switzerland, do not permit the option to connect to one end of the tube only. I’m surprised that the Swiss approach is not more widely used.

This design is a compromise. Maybe the 2 ohms you measure, only at the CD end, is an internal fusible link, or a protection circuit.

Yes, that seems to be the case.

 

[Steve4Physics]

Do you know what test voltage your DVM uses for that resistance measurement? Maybe try using the Diode Test setting instead, to get a source voltage high enough to detect the input diode bridge, if that's the first stage of the power supply.

I don’t know the test voltage the DVM uses for resistance measurement. The DVM is only a cheap one (max. resistance measurement 20M$\Omega$) but it does have a diode-test option. The diode-test gives open circuit both ways round for pins A and B. However, the tube works just fine when power is applied to pins A and B, so there's no issue with that. It's just the reason for 2$\Omega$ between pins C and D that puzzles me.