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Chromoflex - Manual - English V1.07
Important: If the selected supply voltage is too low, the LEDs might not operate with maximum
brightness or some flickering can be noticed. Unnecessary heat must be dissipated if the supply
voltage is too high. Sets coming from our stock are already matched. When using other switching
supplies there is a small potentiometer on the supply, to adjust the output voltage within several
percent.
We recommend selecting the lowest voltage on which all LEDs perform at full brightness without any
disturbance.
The power supply must provide 1.2 Ampere maximum to each „I350"module (all channels on full).
Important note: The optimal operating voltage of the „I350" module is 12 Volts using 3 industrial
standard 3*1 Watt RGB-LEDs in serial connection. The following considerations have been made with
regard to design:
At least 2 channels drive LEDs with higher forward voltage (green, blue)
The total continuous power dissipation in the module is <= 2.5 Watt, The module is not designed
to run continuously at full load (even mixed „white" usually requires only 66% full load).
Good air ventilation is needed, environment air temperature should be <= 30°C
We recommend reducing the dissipating heat in the module by using an external power resistor to
lead off excess power if these limits are exceeded for longer periods, especially when using more
than 3 LEDs in serial connection:
Model „STRIPE"
This model does not have any internal current regulator. Supply voltage is passed through and the
voltage of the power supply must match the voltage of the LED stripe.
Regarding any possible voltage drop, please see the previous paragraph. Cable lengths up to 5
metres are allowed.
Note: LED stripes of 10 Volts, 12 Volts and 24 Volts are offered on the market.
Note: We also deliver flexible LED stripes on wheels in length of up to 4 metres. Each colour requires
up to 2 Amperes at 12 Volts.
- www.barthelme.de
In the picture a low cost 5 Watt power resistor mounted
on an aluminium plate was used. Its value can be
calculated with the formula:
R = U / 0,35
and
(U is the surplus Voltage in Volts, R the resistance, P the
load the resistor will take). An example: If using 6 red
LEDs in series with a forward voltage of 6 * 2.8 Volts and
operating the module with 24 Volts, the resistor shall
catch 6 volts so that the electronics has about 2 volts
headroom for regulation. Hence the resistor should have
18 Ohms and 5 Watt maximum load.
- 5 -
Rev. 19.08.2010
P = U * 0,35
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