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[Reply #45 on:]

Its all really the same, you can't separate voltage from current, one needs the other
However since you'd call the controller by what it was designed to focus on
a voltage regulator attempts to keep the voltage stable as a current regulator keeps the current stable
and the motor example isn't exactly the best relation because there is alot more than just "resistance" going on that affects its actual impedance and hence the current it draws, which would change dynamically with the conditions applied to the motor, the only example that works really is a regular resistor and even then that has a temperature coefficient that can be applied,

[Reply #46 on:]

Nope. The voltage is whatever comes out of the power supply. The driver has no control over that.

It achieves its function by varying its own resistance so that the resistance of the driver + the resistance of the string of LEDs allows a target current to pass. It's therefore controlling resistance, the voltage.

The voltage out of the power supply has no relevance here. What a (linear) led driver does is to change its resistance so that the voltage across the led string generates the right current level.

Switching mode drivers work similarly, except that they generate a voltage directly across the led string, by stepping up / down the supply voltage.

To my knowledge, there is no led drivers, linear or switching mode, that controls the current directly (aka shunt style), for obvious reasons.

[Reply #47 on:]

1 single 4.2v (Fully Charged) lithium battery will supply max brightness (in the flashlight world they say "direct drive", most would say a short i guess lol) ...

BUT, that 4.2v soon becomes 4.1...3.8 in less than a few seconds, your current as a consequent drops... the best way is to keep that current up is to build a driver, sadly most people in the electronics
think of super bright LED's as anything under 30ma lol... i'm dealing with 5watt and 10watt RGB and 15watt white LED's which all run fine at 4.2volts without a resistor (except for the 12/24vdc LED's)

but you need a resistor if you wish the arduino to power it (unless you want to damage the pin/processor). if you wish to power you need a suitable power supply, a power transistor or a mosfet and a big heatsink, also you may prefer to use a PWM pin and use analogWrite on it with say a power transistor and gradually dim/light it (as i do with a 12watt 4.2v LED/TIP31,for that, i run it at around 0.9amps or Full 1.3amps, but i use a 300 ohm resistor on base to protect the pin on the arduino, i could go a lot higher with another method)

your power supply you plug into the wall does not deliver like a battery, chances are give it 4.7volts and you will see smoke, batters can only supply 4.2v, this is why manufacture specifications are important.

[Reply #48 on:]

1 single 4.2v (Fully Charged) lithium battery will supply max brightness (in the flashlight world they say "direct drive", most would say a short i guess lol)

This is simply because the output impedance of the battery limits the current drawn. This would not necessarily apply to any voltage source of 4.2V.

I think this is what the OP is seeing when he says that the figures do not add up. By not taking into account the supplies output impedance you get a false idea of the current you can draw.
It is not that simple either. The output impedance changes with the current draw, that is it is not a linear impedance, which is why the graph of voltage against current is not a straight line.

The voltage out of the power supply has no relevance here.

Apart for setting the maximum voltage output of the constant current driver (minus a bit for the drop across the driver's internal components) .

Some of us remember the constant current output of mechanical teleprinters. They outputted a 20mA constant current into the next teleprinter. If this was a few feet away the voltage needed to drive this current was small. But they could be connected at a distance of a few miles apart, this required a much bigger voltage. The ultimate was when it was disconnected so it never stood a chance of driving 20mA then it could output 180V, enough for you to get a bit of a shock from, but connected up it had a very low voltage on the terminals.