# Trouble driving NPN into saturation [duplicate]

I can't saturate my NPN transistor!

Here is my circuit:

simulate this circuit – Schematic created using CircuitLab

The Vce is around 1.7 V, which is way above the saturation Vce of 0.3, meaning that I must be in linear mode right?

No matter what value resistor I try for Rb, the Vce doesn't change.

My LEDs are super-bright, 2.4v, 20 mA

• And the voltage across Rled? – Ignacio Vazquez-Abrams May 27 '14 at 3:14
• Smaller base resistors for more drive. || Higher beta transistors. eg BC337-40 = cheap and superb and high beta (current gain). – Russell McMahon May 27 '14 at 12:35

A few basic calculations:

The current through Rled, assuming the transistor is fully saturated: $$I_R = \frac{3.7V - 2.4V - 0.3V}{39 \Omega} = 25.6 mA$$

Looking at a 2N3904 Datasheet, they define saturation as the point hfe=10.

Thus: $$I_b = 2.56 mA$$

This means your micro controller needs a control signal of: $$V_m = 0.65V + I_b \cdot 3.9 k\Omega = 10.65V$$

You didn't specify what your microcontroller voltage supply is, but I'm willing to bet it's 5V or lower. To fix this, either lower the value of Rb or increase the value of Rled.

Assume $V_m = 3.7V$ and you don't want to change Rled: $$R_b = \frac{V_m - 0.65V}{I_b} = 1189.5 \Omega$$ So pick $R_b < 1.19 k\Omega$ to saturate the transistor.

One other worrying issue:

Having the 2 LED's in parallel as you have implies that the LED's have a very well matched voltage drop. This is usually not the case! In practice you'll get an asymmetric current flow, where one LED could have a significantly higher current than the other. You really should have separate diode current limit resistors if you intend to wire D1 and D2 in parallel.

simulate this circuit – Schematic created using CircuitLab

• Thanks for the response! Good point about the LEDs being unlikely to have the exact same forward voltage... that could definitely lead to some problems! I'm going back up to my shop this evening and I'm going to try connecting Vss directly to the base resistor at 1 kOhm. The Vce should then go to saturation, which I expect will be <0.5 V. – macdonaldtomw May 27 '14 at 21:36

If the LED's you are using have a forward voltage drop of 2.4 volts at a current of 20mA, then you will not saturate that NPN in the first place, although that is OK in this situation. You do have another issue elsewhere as the voltage across your transistor is 1.7 volts, but there is not information provided to definitively say what it is. Also, try to get your circuit working with one LED first.

Regarding saturation: Say you have the setup below. Say you are driving Q1 such that Ic = 20mA. The voltage across Q1 will be Vce = Vin - R1*Ic - Vd1 = 3.7 - 39*.02 - 2.4 = 0.52 V. To figure out how to make Ic = 20mA you need to find out the DC current gain. According to the datasheet the minimum is 100 with Ic at 10mA and 60 with Ic at 50mA. Let's assume the current gain is 60. So, Ib = Ic / hfe = 20 mA / 60 = .33 mA. Say your micro outputs 3.3V when the logic level is high. You want to figure out R2, so use Ib = (Vin - Vbe) / R2 to get R2 = (Vin - Vbe) / Ib. The datasheet says the max Vbe will be around 0.85V since the Ib will be less than 1 mA. Thus, R2 = (3.3 - .85) / 0.00033 = 7.4kohms. The power dissipated in Q1 will be P = Vbe * Ib + Vce * Ic = .85 * .00033 + .52 * .02 = 11 mW, which is way less than the absolute max 625mW.

simulate this circuit – Schematic created using CircuitLab

Try disconnecting the input and put the 3.7v to the base resistor to see if that turns it on. Make sure the leds work without the transistor first. I don't see why it wouldn't work. It must be something simple.

• Telling someone to short out a supply across a B-E junction is usually not a good first step. – Ignacio Vazquez-Abrams May 27 '14 at 3:25
• Not what I meant. I'll change it so that it implies connecting through the resistor. Thanks for the heads up. – Cameron May 27 '14 at 3:30
• Check to see if you have the collector and emitter switched. A transistor will still work that way, but the beta will be much lower, meaning you may see operation like you're describing. If that's not it, you might have a bad transistor, or something connected incorrectly. – John D May 27 '14 at 3:34
• @JohnD: Did you mean to leave that as an answer? – Ignacio Vazquez-Abrams May 27 '14 at 3:49
• @IgnacioVazquez-Abrams No, I don't know if it's the answer or not, just a comment on something to look for. – John D May 27 '14 at 17:34