High Voltage LED Solution for 4.2A Circuit

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Discussion Overview

The discussion revolves around the challenges of integrating an LED indicator into a high-voltage circuit (680V) that operates at a current of 4.22A. Participants explore various methods for sensing voltage and current, particularly focusing on the use of voltage dividers and the limitations of components available for such high power levels.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant describes the need for an LED to indicate when a capacitor is draining in a high-voltage circuit, expressing difficulty in finding suitable components.
  • Another participant suggests using larger resistor values in the voltage divider to reduce power dissipation, proposing R1 as 600 KOhms and R2 as 22 KOhms.
  • A participant questions the understanding of power dissipation in resistors, explaining how to calculate it using the power equation and providing an example with a 1MΩ resistor.
  • Concerns are raised about the safety of working with high voltages by another participant, who questions the original poster's understanding of basic electronics.
  • The original poster clarifies that this is their first experience with high voltage, typically working with embedded systems.
  • Participants inquire about the purpose of the system and the reason for adding components, seeking to understand the context better.
  • Another participant asks for clarification on what is meant by "drain" in the context of the capacitor's voltage.

Areas of Agreement / Disagreement

There is no consensus on the best approach to integrate the LED indicator, and participants express differing opinions on the safety and feasibility of the proposed solutions. The discussion remains unresolved with multiple competing views on the use of voltage dividers and component ratings.

Contextual Notes

Participants express uncertainty regarding the appropriate resistor ratings for high power levels and the implications of using voltage dividers in high-voltage applications. There are also concerns about the safety of handling such high voltages without adequate understanding.

btb4198
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I have a circuit and it has a cap that has around 680V and the high current that is pull it 4.22A. I need to add an LED to let me know when the Cap is drain. The problem I am having is that since the voltage and current are so high, I can't use a Hall effect sensor . I can't find one rated for that much voltage. I could do a voltage divider and have R1 be 600 ohms and R2 needs be 22 ohms but I can't find a resistor rated for 2911.8 Watts. Can someone help me ?
 
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Why can't you make the voltage divider resistors much larger? If R1 is 600 KOhms and R2 is 22KOhms the power will be less than 1 Watt.
 
phyzguy said:
Why can't you make the voltage divider resistors much larger? If R1 is 600 KOhms and R2 is 22KOhms the power will be less than 1 Watt.
ok to i use the power equation P = V*I to get my power and I measure my current to be at a max of 4.22 and my voltage around 680V, This is why no resistors in my resistors in my circuit, so now you are saying if I add the voltage divider to the cap, it will not burn up the resistors ? I thought any resistor i added to the circuit would have to be rated for the max watts. If I am wrong can you please explain to me ?
 
btb4198 said:
ok to i use the power equation P = V*I to get my power and I measure my current to be at a max of 4.22 and my voltage around 680V, This is why no resistors in my resistors in my circuit, so now you are saying if I add the voltage divider to the cap, it will not burn up the resistors ? I thought any resistor i added to the circuit would have to be rated for the max watts. If I am wrong can you please explain to me ?
Power dissipated in a resistor: R⋅I2. Since I = V/R in this case, you get P = V2/R. Since V is 680V, assume R = 1MΩ and calculate the power dissipation: P = (6802/1000 000)W = 0.46W. The current through the resistor is (680/1000 000) A = 0.68mA. This current will make a LED glow, but not very brightly.
 
btb4198 said:
I have a circuit and it has a cap that has around 680V and the high current that is pull it 4.22A. I need to add an LED to let me know when the Cap is drain. The problem I am having is that since the voltage and current are so high, I can't use a Hall effect sensor . I can't find one rated for that much voltage. I could do a voltage divider and have R1 be 600 ohms and R2 needs be 22 ohms but I can't find a resistor rated for 2911.8 Watts. Can someone help me ?
Um, why are you working with a 680V circuit with so little understanding of basic electronics? Sounds like kind of a dangerous situation to me...
 
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berkeman said:
Um, why are you working with a 680V circuit with so little understanding of basic electronics? Sounds like kind of a dangerous situation to me...

It is my 1st time with voltage this high. I normal work with embedded systems.
 
btb4198 said:
It is my 1st time with voltage this high. I normal work with embedded systems.
What is the system/device? What does it do, and why are you adding things to it? Please help us understand the situation.
 
berkeman said:
What is the system/device? What does it do, and why are you adding things to it? Please help us understand the situation.
it is a system and it give power to a VFD. I have 120 Vac coming in from a UPS and I convert that to 680V DC and that goes to the DC bus on the VFD.
 
What do you mean by "drain" in your post #1? Drained below a set voltage?
 

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