# LED driven by MOSFET is dim

I have this schematic:

I had to replace T1 transistor with a MOSFET (because D4 on an ESP8266 has to be high at boot time and the transistor reduces the pin voltage, so the ESP doesn't boot - see my previous question).

So I replaced it with a IRF3205 MOSFET datasheet (I know it is total overkill) base = gate; source = emitter; drain = collector.

The LED works (i.e. it turns off and on) but the LED in the "ON" state is really dim. What is wrong? It has to be something elementary, but I can't figure it out.

The LED is inside a button (12 V with internal resistor). D4 is the ESP8266 pin (3.3 V logic)

Thanks.

• This FET is not designed to be driven with 3.3V. – Hearth Apr 10 at 19:12
• Look at the Vgs in the "Conditions" column in the row for RDSon of the specs table. – DKNguyen Apr 10 at 19:14

This FET is not designed to be driven by 3.3V logic.

The threshold voltage is the gate-to-source voltage at which the FET is just barely on. This is not the point at which the FET is fully on, which would require quite a bit more than the threshold voltage.

• Ok, thanks. So I will have to order some mosfet with lower Vgs.. If I understand by datasheet Vgs voltage could be somewhere between 2-4V so if I drive it with 3.3 the threshold is not meet and mosfet is not fully open so it limits current D-S. It is right ? – Fires_CZ Apr 10 at 19:18
• Even if you do reach the threshold voltage, the current is proportional to $V_{GS} - V_{GS,th}$. You need to be substantially above the threshold voltage. Look for "logic-level" FETs. – Hearth Apr 10 at 19:22
• Yet this is the problem but not the solution – Tony Stewart Sunnyskyguy EE75 Apr 11 at 13:24
• @SunnyskyguyEE75 Yes. This is answering the question "What is wrong?" that the asker asked. – Hearth Apr 11 at 14:02
• As EE's we must think outside the Question only and give an answer that ALSO solves the problem rather than simply answer the incomplete question , what is wrong. But this is a systemic issue on this site. We constrain ourselves to just the question, when we need to think outside the little box. – Tony Stewart Sunnyskyguy EE75 Apr 11 at 14:05

When you see a Vgs(th) for a FET. You will need at least 2 to 3x Vgs(th) (threshold) conduct fully.

Change **to R = 500 Ohms or less depending on LED string current such that input current is 5~10% of switched output current.**

IRFxxx FETs are Vgs(th)=3V +/-1 tolerance This is the threshold for 500uA not Amps !!

So you need >4.5V for Vgs or this IRFxxx family of FETs and preferably 10V for rated RdsOn as shown in specs.

You ONLY get 10% of hFE = Beta as a saturated switch.

We want to ensure base current = >5% of LED current. Preferably 10%. (never use Vgs(th)2~4V FETS on 3.3V logic. It needs >5V.) All IRFxxx's I think are this way, Now they make Logic level FETS with a different prefix in P/N.

Your design was (3.3V-0.6Vbe)/5k=Ib= 540 uA.

Change Rb to >= 100 Ohms and it will work off any 3.3V logic and NPN open collector. Iout=(3.3-Vbe)/(Rcmos+Rs)= (3.3-0.7)/(33+100)=20 mA so it can drive 200mA to 350mA on collector with low Vce(sat)