# Casambi Bluetooth Milli-Amp drivers

Below a large collection of Casambi-ready drivers that serve to control LED spots that you can sometimes put in series again depending on the total power. There are various drivers that accept very long wires on the output side, such as the SSW60WCM from Vadsbo. As a result, the wiring can be concealed very easily because it can for example be designed as 4x0.8 (thin wire). It is based on a misunderstanding that because of the thinner wire (voltage drop) the LED will give less light; the driver will want to keep a set value of, for example, 350mA and thus simply slightly increase the output voltage. Below the pictograms we give some instruction on P = UxI or; 'how effective is your driver in relation to the chosen LED spot'.

P = U x I either; Power = voltage x current (in this explanation I do not include cosine Phi (power factor)). The explanation below actually applies to all drivers and you will often see a sticker on the back of drivers where you can read that lower 'mA' (current) also shows a lower 'Watt' (power).

Suppose you decide to work with four 250mA (= 0.25A) LEDs connected in series that each have a capacity of 3Watt (4x3 = 12W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 12W / 0.25A = 48V.

Suppose you decide to work with four 250mA (= 0.25A) LEDs connected in series that each have a capacity of 10Watt (= 10W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 40W / 0.25A = 160V. Drivers that can provide an output voltage of 160V are very exceptional (ELT is pretty close by the way), but you have a problem.

Suppose you decide to work with four 700mA (= 0.7A) LEDs connected in series that each have a capacity of 10Watt (= 10W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 40W / 0.7A = 58V.

So note that the larger the LED current passage (mA), the more efficiency (Watt) your driver has, or written differently; you buy fewer and fewer drivers with a higher LED current. In case you really want to work with a lot of 350mA LED spots, you might prefer to opt for drivers that can work with phase-cutting dimmers.

Suppose you decide to work with four 250mA (= 0.25A) LEDs connected in series that each have a capacity of 3Watt (4x3 = 12W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 12W / 0.25A = 48V.

Suppose you decide to work with four 250mA (= 0.25A) LEDs connected in series that each have a capacity of 10Watt (= 10W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 40W / 0.25A = 160V. Drivers that can provide an output voltage of 160V are very exceptional (ELT is pretty close by the way), but you have a problem.

Suppose you decide to work with four 700mA (= 0.7A) LEDs connected in series that each have a capacity of 10Watt (= 10W in total). The voltage that the driver must be able to offer is therefore; U = P / I either; 40W / 0.7A = 58V.

So note that the larger the LED current passage (mA), the more efficiency (Watt) your driver has, or written differently; you buy fewer and fewer drivers with a higher LED current. In case you really want to work with a lot of 350mA LED spots, you might prefer to opt for drivers that can work with phase-cutting dimmers.