MOSFET to source power to LEDs

[StudentSA]

Good Day,

I would like to know how to setup a mosfet to take a 5V/0V (from Arduino) output and use this to switch Vcc to an LED array.

The Hi Power LED array has 4 LEDs in common anode configuration. the cathodes of the LEDs are fed into an LED driver IC.

So basically I would like to switch the string on by connecting the LED strings common anode to Vcc when the arduino IO goes high(5V) then back to ground when low(0V)

I put together this simple representation online hope it help to explain

 

[MrAl]

In a word:
Logic level mosfet

 

[Tony Stewart]

LEDs ( and all other diodes) have an ESR inversely proportional to power rating.

Such that a 65mW 5mm LED is ~ 16 Ohms and 1W LED = 0.5~1 Ohm such that W*Ω~1

A mosfet has an ESR called RdsOn.

Then use the lowest voltage available e.g. 3.3 or 5V , not 12V unless stringing all 4 in series)

A 3V White LED has a threshold Vth = 2.7 and rises to 3 or higher due to ESR at If from bulk resistance.

THus use Ohm's Law to compute what Series R is needed for the drop voltage and desired current no matter if in series, parallel or an array like 3S3P from 12V

Ensure the driver Resistance is << 10% of the ESR & Rseries for cool operation. lower the better.

Common Anode is preferred as you have chosen with N channel Mosfets preferred or BJT's with current gain only of 10 min when saturated

 

[alec_t]

Which LED driver IC are you using? Does it have an 'enable' input?

 

[Tony Stewart]

Keep in mind Common Anode is negative logic so active low turns On, so we call this a Low Side switch which sinks Current limited by a series R and all voltage drops in the KVL loop.

Also Active 1 stage Switches are inverting, thus you now have positive logic and they offer Logic level sensitive N-channel MOSFETs by distributor selection. ( thousands of types)

So choose RdsOn to be << ESR of power LEDs for low dissipation.

3V 5mm Blue or White LEDs are not power type and can be driven from low voltage logic levels such as 3.3V directly or 5V with 100 Ohms in series.
(5-3=)2V/20mA=100 Ohms.... but since 5V logic CMOS driver may have an ESR be 25 or 50 Ohm's (Vol/Iol).. check... , the series R can be reduced to 50 Ohms from 5V. CD4xxx logic has 200~300 Ohms ESR,

 

[StudentSA]

Thanks for the advice, So far I have come up with the following circuit to address my issue:

However I am not happy with the need for Ra, this implies that current for the LEDs are drawn through this resistor. also when the mosfet is "ON" current is again drawn through here.

Ideally I would like point A to be switched between Vcc and ground (or even hanging).

Also ideally I would like the 5V to the MOSFET Gate to tie A to Vcc not as present where it is reversed.

 

[MikeMl]

Look at the circuit of post #8 and post #13 in this recent thread. It is called a "high-side-driver".

 

[StudentSA]

Thanks MikeML,

So something like this? maybe with a 100k resister between A and GND?

 

[MikeMl]

What is Vcc (Volts)?

What is the source of the signal that drives the left end of Rin?

 

[StudentSA]

Both are 5v MikeML. Rin driven by arduino digital pin

 

[AnalogKid]

Thanks MikeML,

So something like this? maybe with a 100k resister between A and GND?
View attachment 94018

Not quite. Rgs goes from Gate to Source; in your drawing, that's from the Gate to Vcc. It is a pull up resistor. The Arduino output goes LOW to turn on the LED's, HIGH to turn off.

ak

 

[Tony Stewart]

he means For bidirectional ports , All your FETs will be turned on at the same time, when uC is reset to all inputs.
If this is not intended, Change Rgs to pullup. Right now its Rgd.

Otherwise Rgs is redundant in operation with a CMOS driver.

Your individual Driver box is now upside down as a Low Side driver to gnd.
Ensure your High Side Driver can carry all low side currents with low drop ( low RdsOn)