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Ama
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I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side when the driver has its phase terminal connected to live power and its neutral terminal not connected to anything.

Detailed configuration under regular working wiring:

schematic

simulate this circuit – Schematic created using CircuitLab

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals,) disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.

schematic

simulate this circuit

  • Then, keeping one of the DC terminals disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, the LEDs are back glowing at 5%.

schematic

simulate this circuit

To sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if very little. When I measured the voltage, it was around 3 volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs lighting up when one end of the circuit is disconnected from the source of power cannot be explained if we consider DC power.

Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the voltage gets very high, and might be sufficient for some current to bypasstravel through the LEDsdiodes of the bridge rectifier, at counter current.

Is it possible that when this DC cable is disconnected, the voltage on the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals hence, propagating the capacitance effect from the AC side?

I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side when the driver has its phase terminal connected to live power and its neutral terminal not connected to anything.

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals,) disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.
  • Then, keeping one of the DC terminals disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, the LEDs are back glowing at 5%.

To sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if very little. When I measured the voltage, it was around 3 volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs lighting up when one end of the circuit is disconnected from the source of power cannot be explained if we consider DC power.

Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the voltage gets very high, and might be sufficient to bypass the LEDs at counter current.

Is it possible that when this DC cable is disconnected, the voltage on the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals hence, propagating the capacitance effect from the AC side?

I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side when the driver has its phase terminal connected to live power and its neutral terminal not connected to anything.

Detailed configuration under regular working wiring:

schematic

simulate this circuit – Schematic created using CircuitLab

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals,) disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.

schematic

simulate this circuit

  • Then, keeping one of the DC terminals disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, the LEDs are back glowing at 5%.

schematic

simulate this circuit

To sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if very little. When I measured the voltage, it was around 3 volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs lighting up when one end of the circuit is disconnected from the source of power cannot be explained if we consider DC power.

Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the voltage gets very high, and might be sufficient for some current to travel through the diodes of the bridge rectifier, at counter current.

Is it possible that when this DC cable is disconnected, the voltage on the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals hence, propagating the capacitance effect from the AC side?

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JRE
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I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side ifwhen the driver has its phase terminal connected to live power, and its neutral terminal not connected to anything.

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals),) disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.
  • Then, keeping one of the DC terminalterminals disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, then the LEDs are back at glowing at 5%.

So, toTo sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if littlevery little. When I measured the voltage, it was around 3 Voltsvolts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs littinglighting up when 1one end of the circuit is disconnected from the source of power cannot be explained if we consider DC power. 

Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the Voltagevoltage gets very high, and might be sufficient to bypass the LEDs at counter current.

My question: isIs it possible that when this DC cable is disconnected, the voltage aton the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals;terminals hence, propagating the capacitance effect from the AC side?

I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side if the driver has its phase terminal connected to live power, and its neutral terminal not connected to anything.

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals), disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.
  • Then, keeping one of the DC terminal disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, then the LEDs are back at glowing at 5%.

So, to sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if little. When I measured the voltage, it was around 3 Volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs litting up when 1 end of the circuit is disconnected from the source of power cannot be explained if we consider DC power. Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the Voltage gets very high, and might be sufficient to bypass the LEDs at counter current.

My question: is it possible that when this DC cable is disconnected, the voltage at the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals; hence, propagating the capacitance effect from the AC side?

I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side when the driver has its phase terminal connected to live power and its neutral terminal not connected to anything.

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals,) disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.
  • Then, keeping one of the DC terminals disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, the LEDs are back glowing at 5%.

To sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if very little. When I measured the voltage, it was around 3 volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs lighting up when one end of the circuit is disconnected from the source of power cannot be explained if we consider DC power. 

Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the voltage gets very high, and might be sufficient to bypass the LEDs at counter current.

Is it possible that when this DC cable is disconnected, the voltage on the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals hence, propagating the capacitance effect from the AC side?

Source Link
Ama
  • 173
  • 1
  • 7

Bridge rectifier and phantom voltage

I have a constant current LED driver which keeps the LEDs at about 5% glow when the AC side if the driver has its phase terminal connected to live power, and its neutral terminal not connected to anything.

Whilst trying to figure out a way to absorb the leaking energy so that the LEDs do get to 0% instead of 5%, I noticed some weird behaviours on the DC side of the driver:

  • Under the context described above (only one wire, live, on the AC terminals), disconnecting one of the terminals on the DC side results in the LEDs still glowing, at about 0.5%; this is very very weak, but still there.
  • Then, keeping one of the DC terminal disconnected, but going back to connecting the AC terminals under a regular phase + neutral scenario, then the LEDs are back at glowing at 5%.

So, to sum up, after disconnecting one of the AC and one of the DC terminals of the driver, the LEDs are still getting some power in, even if little. When I measured the voltage, it was around 3 Volts, but I suspect this measurement was altered by the multimeter.

In my humble opinion, LEDs litting up when 1 end of the circuit is disconnected from the source of power cannot be explained if we consider DC power. Since the driver is a constant current supplier, I suppose the following is happening when a DC terminal is disconnected: "Infinite" resistance, which means the Voltage gets very high, and might be sufficient to bypass the LEDs at counter current.

My question: is it possible that when this DC cable is disconnected, the voltage at the DC side of the bridge rectifier is so high that the diodes of this bridge cannot block the reverse voltage, and the DC terminals become AC terminals; hence, propagating the capacitance effect from the AC side?