Transistor not cutting off at 0V

Question

EDIT: I edited the question as it was my misunderstanding to think that the transistor was not getting into saturation.

I'm driving an LED with a microcontroller through a transistor with a PWM signal.

When the gate is grounded, no current flows, as expected. As soon as I start to drive the transistor (1% duty cycle), I get a voltage drop in TP8 of ~2V, with a 1% of the cycle at 0V. This turns the LED on at quite a high level of brightness, way more than 1% DC from the battery voltage would suggest.

As the duty cycle increases, the semiperiod at 0V at TP8 increases (as expected). The result is that VLED is something like (b), instead of seeing something like (a).

$$(a) Duty Cycle * 3.3V \\ (b) DC_{value} + ((3.3-DC_{value}) * Duty Cycle)$$

My PWM frequency is rather low, as I'm using a low frequency clock to generate it, so it's only 128Hz, and I have an 8 bit resolution. The positive rail for the LED is the battery voltage, so it will vary from 4.2V to 3.XV. The MPN of the LED and the transistor are in the schematic (Transistor datasheet, LED Datasheet). The threshold voltage of the transistor is 2.1 in the worst case, so I can't see why is not fully switching.

Can somebody shed some light?

edit: I am expecting the voltage at TP8 to go to 3.3V, but it's 1.6V. If the pin is technically floating, why can't I measure 3.3V at TP8?

Answer 1

When the voltage at the gate of the transistor is 3.3V, the transistor goes into saturation, and the voltage at TP8 is close to 0V.

When the voltage at the gate of the transistor is 0V, the transistor goes into cutoff, and TP8 is left floating. I was expecting it to read 3.3V, given that one pin of it is connected 3.3V via the current-limiting resistor, but that's not the case as there's a diode in the between, isolating TP8 from 3V3.

Answer 2

So many people confuse Vth or Vt or Vgs(th) with the required gate voltage to guarantee RdsOn.

If you read the specs carefully the conditions for Vth are VDS = VGS, ID = 250 µA

This means a 2.1V Vds/ 250µA = 8.4Kohm resistor the so-called threshold of conduction.

Normally you need at least 1.5x Vt max and typically 2x to 3x Vt but here they only guaranteed RdsOn at Vgs=4.5 (4.5/1.6Vtnom = ~3x Vt)

So you are seeing the effects of insufficient gate drive to achieve 1 Ohm RdsOn.

This means if low battery is 3.0V you need a Vgs(th)=1.1V max

e.g.DMG1012UW-7

Other points to consider

Also if you wish to regulate current, there are better ways to keep it constant, but you may compute \$I_f\$ as battery reduces from 3.7 to 3.0V using KVL with this formula. Note the curves are typical and not guaranteed so they have a tolerance of often +25%/-12% even on the best Samsung parts, where others might be +50%/-15% (ballpark)

If=V/R= (Vbat-2.7V)/(RdsOn+Rled+Rs) ( for If>200mA) If you choose my suggested part, RdsOn=0.5 Ohm @ 3V(Vgs)

\$R_s=\dfrac{(V_{bat}-2.7)}{I_f}-R_{dsOn}-R_{led}\$ e.g.

So if you wanted 200mA at 3.8V, Rs=1.1/0.2 -0.5-1= 4Ω @ 1/4W (not 27Ω) but at 3.0V If<0.3/5.5=55mA why "If<" because Rled is now ~2Ω so If~0.3<6.5= 46mA . A 1W LED needs at least 6cm² surface area heatsink.

^{simulate this circuit – Schematic created using CircuitLab}