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 ?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.
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: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 ?
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...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 ?
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...
What is the system/device? What does it do, and why are you adding things to it? Please help us understand the situation.btb4198 said:It is my 1st time with voltage this high. I normal work with embedded systems.
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.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.
A high voltage LED solution for a 4.2A circuit is a type of lighting system that uses LEDs (light-emitting diodes) and can handle a current of up to 4.2 amps. This allows for brighter and more efficient lighting compared to traditional LED solutions.
A high voltage LED solution works by converting the AC power supply into a DC (direct current) power supply using a driver. The driver then regulates the current and voltage to power the LEDs. The high voltage aspect refers to the higher voltage range (up to 100V) that these LEDs can handle, which allows for more LEDs to be connected in series.
There are several benefits to using a high voltage LED solution for a 4.2A circuit. These include higher efficiency, longer lifespan, and brighter light output. The higher voltage range also allows for more LEDs to be connected in series, reducing the number of drivers needed and simplifying the circuit design.
One potential risk of using a high voltage LED solution is the risk of electric shock if not handled properly. It is important to follow safety guidelines and use proper insulation when working with high voltage circuits. Additionally, the higher voltage range may not be suitable for all applications and may require additional components for voltage regulation.
While a high voltage LED solution for a 4.2A circuit can be used in a variety of lighting applications, it may not be suitable for all. For example, applications that require dimming or color-changing capabilities may require a different type of LED solution. It is important to consider the specific needs of the application before choosing a lighting solution.