Hey guys I was reading an earlyer thread on substituting a risitor for a ccfl lamp in a computer display.
Here was Plain Bill's example:" Ohms law. At a rough guess a CCFL runs at 650 - 1000 Volts and draws 3-5 mA.
E / I = R or 800/.004 = 200Kohms
I X E = P or .004 X 800 = 3.2 watts
So a 200K ohm, 5 watt resistor hooked to an appropriate connector should do it."
Ok so my measurements were somewhat different. One lamp leg was 260 Volts and the current draw was 7.5 mA.
E / I = R or 260/.0075 = 35Kohms it was acctually 34666.66667 so I rounded up.
I X E = P or .0075 X 260 = 1.95 or 2 watts
I have four ccfl lamps two on top two on bottom. I figured 260 fluctuating voltage time two lamps 520 volts for just that half of the inverter.
I originally tried 160K 2 watt resistors two in series and although they kept the ccfls on, the resitors got extreamly hot so I abandoned that set up. So then I placed them in paralell cutting my resistance in half but still extream heat on the resistors and so no go. I wish I could measure the heat. Does paralell with resistors double there watt ratting or in series? or does the watt ratting stay the same no matter the configuration.
So if my calculations are correct I would need 35K at 2 watts. Less resistance less heat. I think. So any ideas. I don't want to start a fire.
Here was Plain Bill's example:" Ohms law. At a rough guess a CCFL runs at 650 - 1000 Volts and draws 3-5 mA.
E / I = R or 800/.004 = 200Kohms
I X E = P or .004 X 800 = 3.2 watts
So a 200K ohm, 5 watt resistor hooked to an appropriate connector should do it."
Ok so my measurements were somewhat different. One lamp leg was 260 Volts and the current draw was 7.5 mA.
E / I = R or 260/.0075 = 35Kohms it was acctually 34666.66667 so I rounded up.
I X E = P or .0075 X 260 = 1.95 or 2 watts
I have four ccfl lamps two on top two on bottom. I figured 260 fluctuating voltage time two lamps 520 volts for just that half of the inverter.
I originally tried 160K 2 watt resistors two in series and although they kept the ccfls on, the resitors got extreamly hot so I abandoned that set up. So then I placed them in paralell cutting my resistance in half but still extream heat on the resistors and so no go. I wish I could measure the heat. Does paralell with resistors double there watt ratting or in series? or does the watt ratting stay the same no matter the configuration.
So if my calculations are correct I would need 35K at 2 watts. Less resistance less heat. I think. So any ideas. I don't want to start a fire.
Comment