Quote from: magnethead794 on Dec 22, 2011, 08:05 pmSo if it's dropping 2.9 volts, I is 19.3 mA ,Can you verify that this is the voltage being dropped, using a multimeter?QuotePassing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?The LEDs should drop about 2 volts but that may not be the case at such small currents. I don't know what LED you're using and even if I did the manufacturer may not test it for such low currents.Can you measure the voltage across one of the LEDs when it's in the dim condition and the voltage across "its" resistor?If you're actually getting 19.3mA through the single "dimming" resistor, this will be split between all 39 LEDs and their resistors, so it should be about 0.5mA each. This would give a volt drop across the 39 LED resistors of about 75mV. Again any actually voltages you can measure would help. Iain
So if it's dropping 2.9 volts, I is 19.3 mA ,
Passing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?
Vs = 5.050VVdimmer = 3.219VVprimary = 160.7mVVLED = 1.667V21.35 mA flowing between Vs and the dimmer resistor
Strangely, both sets of LED's were about the same brightness at both full brightness and reduced brightness. I found that somewhat peculiar since I used a 150 ohm resistor on the lower-voltage set of LED's.
OK. Given your selection of digits it may not be much of an issue.So are you happy with the power calculations?Iain
Once you said the second digit wasn't showing a 1 or 7 I didn't think there would be much of an issue. You won't see a 1 next to an 8.Iain