Hi guys,

I have a MOSFET (n-type) which i want to use to control an led..

I have the drain connected to 5v

the source is connected to the LED (positve end)

and I will be using 5v to control the led at the gate...

however, having 5v at the gate gives me a very dim light...

i looked at the connections and calculated that I am in the Saturation region ( Vds is greater than Vgs - Vt )..(should I have it such that I operate in the linear if i want to use the FET as a switch? and if so , how do i connect it to properly bias the FET to work as a switch ? )

thanks guys

 

Hi guys,

I have a MOSFET (n-type) which i want to use to control an led..

I have the drain connected to 5v

the source is connected to the LED (positve end)

and I will be using 5v to control the led at the gate...

however, having 5v at the gate gives me a very dim light...

i looked at the connections and calculated that I am in the Saturation region ( Vds is greater than Vgs - Vt )..(should I have it such that I operate in the linear if i want to use the FET as a switch? and if so , how do i connect it to properly bias the FET to work as a switch ? )

thanks guys

You want to connect the drain of the FET to the cathode of the LED, the source
to ground and the gate to 5V. The anode of the LED should be connected to 5V.

The problem with wiring the source to the anode of the LED is that voltage drop across
the LED reduces your gate to source voltage which increases your drain to source
resistance which reduces your LED current.

(* jcl *)

 

You may also have a problem if the FET isn't a logic level type. Ordinary FET's don't go into full conduction until the gate voltage is close to 10 volts. VN10LP's are good logic level types in TO-92.

 

If i connect the LED's cathode to the drain , there will be a voltage drop from my power supply across the LED of around 0.7 (say..) that reduces the Vds, and if i were using a power supply of 5 volts and 5 volts to the gate, i have to ensure that Vgs - Vt is less that Vds to get it into saturation...so does that mean i would have to use a lower gate voltage?

 

Essentially, I think you want to do something as in the attached. You can ignore D1, and adjust R3 per your supply voltage vs. LED current requirements.

 

If i connect the LED's cathode to the drain , there will be a voltage drop from my power supply across the LED of around 0.7 (say..) that reduces the Vds, and if i were using a power supply of 5 volts and 5 volts to the gate, i have to ensure that Vgs - Vt is less that Vds to get it into saturation...so does that mean i would have to use a lower gate voltage?

Vt is the threshold voltage. To turn the LED on you want Vgs (the gate to source voltage) above the threshold voltage. Vds does not need to be above Vgs and will not be when the
LED is on in your application. When the FET is conducting the Vds will be close to zero.

Also the Vf of LEDs is much higher than the Vf of standard diodes. Typically a red or green LED is in the 2-2.6V range, a blue LED is in the 3-3.5V and a white led is around 4V.

(* jcl *)

 

Vds does not need to be above Vgs and will not be when the
LED is on in your application.
(* jcl *)

Don't I need Vds to be greater than Vgs - Vt ( overdrive voltage) to have the FET workign in Saturation region ?

Thanks

 

Don't I need Vds to be greater than Vgs - Vt ( overdrive voltage) to have the FET workign in Saturation region ?

Thanks

No. All you need is Vgs > Vt. If you have a logic level FET, then your Vt should be around one or two volts. If you are using a power FET by mistake, your Vt could be 4 volts or higher.