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UK diy (uk.d-i-y) For the discussion of all topics related to diy (do-it-yourself) in the UK. All levels of experience and proficency are welcome to join in to ask questions or offer solutions. |
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#1
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OK, we have a energy efficient lamp in the kitchen - normal ceiling
rose/pendant light fitting. The kitchen has two light switches for this light, one at each door. Strange thing was we noticed that the energy efficient lamp, once warmed up, would flash (dimly) for some time after being switched off. Stuck a multimeter into the pendant and there appears to be 50 volts across the pins when switched off! Checked in the living room, which also has two light switches, and it too had 50v across the pins! Checked upstairs in a bedroom (one switch) and no voltage at all. I take it this voltage presence is wrong? Anyone any pointer as to what could be causing this? Cheers, Colin. |
#2
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Stuck a multimeter into the pendant and there appears to be 50 volts
across the pins when switched off! A modern high impedence multimeter on a voltage range will give close to a random reading when attached to an open circuit that has any length of cabling. Any capacitive or inductive coupling will causing spurious voltage readings. Christian. |
#4
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John Rumm wrote:
wrote: I take it this voltage presence is wrong? Anyone any pointer as to what could be causing this? A digital multimeter I take it? No - it's a 'mini' analogue needle multimeter - a few years old now though. If so the chances are you are seeing something called capacitive coupling. Cable runs will have a small but detectable capacitance between the conductors. This allows a tiny amount of current to flow between the conductors even though when they are not connected to each other (and in fact separated by insulation!). This allows some of the disconnected sections of cable (i.e. switched off bits) to "charge up" toward some fraction of mains voltage. A digital multimeter will tend to show this since it places very little "load" on the cables when carrying out the measurement. (Conventional meters place a higher load on the circuit under test and this tends to discharge the cables). I assume that this doesn't rule out the analogue meter also picking it up though? You may well find the effect is more noticeable on two way switched circuits since they will have longer runs of cable. The flashing you can see with the CFL bulb is quite common, and seems to be an extension of the same problem. The small leakage current due to the capacitive coupling is also enough to charge capacitors in the input circuitry of the lamp. Once these has accumulated a sufficient level of charge, it is enough to momentarily "fire" the lamp. This discharges the circuit, the lamp goes out and the process starts over. Thanks for this - it certainly sounds like a reasonable explanation to me. (Don't need to call an electrician?) As long as we aren't 'seeping' electricity that we're having to pay for - or are we? I take it that the amount lost in this way is so small as to be negligible? :-) Cheers, Colin. |
#5
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In article .com,
" writes: John Rumm wrote: wrote: I take it this voltage presence is wrong? Anyone any pointer as to what could be causing this? A digital multimeter I take it? No - it's a 'mini' analogue needle multimeter - a few years old now though. If it's something like 20kohms/V (50µA movement), then on a 300V range, it's got nearly the same input impedance as many digitial multimeters. You will probably find the voltage reading is different when set to different ranges (lower on a lower voltage range, as the meter's input impedance drops). This would be an indication it's just leakage through a high impedance, most likely caused by capacitive coupling in the cables. -- Andrew Gabriel |
#6
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wrote:
A digital multimeter I take it? No - it's a 'mini' analogue needle multimeter - a few years old now though. I assume that this doesn't rule out the analogue meter also picking it up though? It is less common to see the effect with these since they load they place on the circuit is higher, but it is still possible. (The "load" of a meter is given by its "ohms per volt" spec. For AC voltage measurement this is typically some "thousands" of ohms per volt. A digital will usually be several million ohms/volt. So for example a 5k ohm/volt meter on a 250V range will place a load of 250 x 5000 = 1.25 MOhm load on the circuit. This will pass a current of 240/1.25M = approx 0.2mA). Have a look at your meter and see if you can see a rating for its ohms/volt. Usually the higher the number the better (since it means the less the meter is affecting the thing it is trying to measure), but it also means that you can end up seeing stray voltages even where they are of such a high impedance that they would quickly fade to nothing with the merest hint of a load. Thanks for this - it certainly sounds like a reasonable explanation to me. (Don't need to call an electrician?) I should not of thought so. If you stick an ordinary bulb in, the fact that it goes off when you turn the switch off is a good indication there is not really 50V there! (if there was it would still be glowing). As long as we aren't 'seeping' electricity that we're having to pay for - or are we? Well, every flash is coming from energy that you are paying for... I take it that the amount lost in this way is so small as to be negligible? :-) Chances are you are sinking micro amps here, so I would not be overly worried if I were you! -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
#7
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![]() "John Rumm" wrote in message ... wrote: I take it this voltage presence is wrong? Anyone any pointer as to what could be causing this? SNIP The flashing you can see with the CFL bulb is quite common, and seems to be an extension of the same problem. The small leakage current due to the capacitive coupling is also enough to charge capacitors in the input circuitry of the lamp. Once these has accumulated a sufficient level of charge, it is enough to momentarily "fire" the lamp. This discharges the circuit, the lamp goes out and the process starts over. -- I thought I had one of these yesterday but it turned out to be a duff nearly-new lamp :-( This may have been the cause of temporary massive interference I had on my AM radio .... Maybe I will surgically remove the lamp part and keep it until I have a blown lamp/good ballast available ... (This was a B&Q 18 watt "eco-light" - in future I may try to keep the receipts ...) |
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