Z
z_heychris
Guest
All,
I could really use some help with a project I'm working on. I decided to try a harbor freight LED worklight. Unfortunately they aren't very bright. I figured I might as well fix that by re-engineering it. I purchased some bright LED's on Ebay. 5mm 60° 5-Chips 0.5 Watt High Power WHITE LED's 3.4V, 100mA
The circuit used is similar to :
http://www.candlepowerforums.com/vb/...31#post3273531
Unfortunately I need more power for these LED's. There are 2 circuits one with 23 LED's and one with 22 LED's
My calculations voltage drop of 74.8 (22leds) and 78.2 (23LEDs), I can shoot for Maybe 81 Volts instead of 100?
the 81Volts needed +6 for the bridge rectifer loss =87 V. 118 volts of the line make 31 volts I could drop at the capacitor.
31V/.1A = 310 ohms. C = 1/(2*pi*f*Xc) = 1/(2*3.14*60*310)=1.77e-5 =8.56 uF
I tried a 8uF Capacitor and ended up with 200mA going through my LED's.. oops.. too much. Also the Voltage came out at 65 V. Which is too low.
I realized 331 ? as the impedance is correct for an 8 µF capacitor at 60 Hz. However, as that is a capacitive impedance, voltages on there do not add arithmetically to voltages across resistors or LEDs.
The average rectified current is about 1.11 times the RMS current, so you want about 90 mA RMS
The phases will be at 90º, so the voltage after a drop of 331 ? * 90 mA = 29.8 V is
sqrt(120² - 29.8²) = 116 V So the Capacitor isn't really doing much at all.
Second try 2uF Capacitor
Xc = (1/c)/2*pi*f = (1/.000002)/(2*3.14*60) = 1327 ohms
at 90ma -V=1327*.09 =119V
sqrt(120^2-119^2)=15.5
Result: 50mA and 56V
This math is driving me crazy. My gut tells me to pick something in between like 4uF and use a resistor on the DC side to limit the current.
Can anybody help?
I could really use some help with a project I'm working on. I decided to try a harbor freight LED worklight. Unfortunately they aren't very bright. I figured I might as well fix that by re-engineering it. I purchased some bright LED's on Ebay. 5mm 60° 5-Chips 0.5 Watt High Power WHITE LED's 3.4V, 100mA
The circuit used is similar to :
http://www.candlepowerforums.com/vb/...31#post3273531
Unfortunately I need more power for these LED's. There are 2 circuits one with 23 LED's and one with 22 LED's
My calculations voltage drop of 74.8 (22leds) and 78.2 (23LEDs), I can shoot for Maybe 81 Volts instead of 100?
the 81Volts needed +6 for the bridge rectifer loss =87 V. 118 volts of the line make 31 volts I could drop at the capacitor.
31V/.1A = 310 ohms. C = 1/(2*pi*f*Xc) = 1/(2*3.14*60*310)=1.77e-5 =8.56 uF
I tried a 8uF Capacitor and ended up with 200mA going through my LED's.. oops.. too much. Also the Voltage came out at 65 V. Which is too low.
I realized 331 ? as the impedance is correct for an 8 µF capacitor at 60 Hz. However, as that is a capacitive impedance, voltages on there do not add arithmetically to voltages across resistors or LEDs.
The average rectified current is about 1.11 times the RMS current, so you want about 90 mA RMS
The phases will be at 90º, so the voltage after a drop of 331 ? * 90 mA = 29.8 V is
sqrt(120² - 29.8²) = 116 V So the Capacitor isn't really doing much at all.
Second try 2uF Capacitor
Xc = (1/c)/2*pi*f = (1/.000002)/(2*3.14*60) = 1327 ohms
at 90ma -V=1327*.09 =119V
sqrt(120^2-119^2)=15.5
Result: 50mA and 56V
This math is driving me crazy. My gut tells me to pick something in between like 4uF and use a resistor on the DC side to limit the current.
Can anybody help?
