How can I identify transistors with a low Vce-sat (collector-emitter saturation voltage)?

Question

I want minimum voltage drop between collector and emitter as I have very small Vcc voltage to supply the load.

My Vcc is 3.3 volts. I want to use this switch to turn on/off 3 LEDs connected in parallel with 1K ohms resistor in series at the collector. Is it possible to find transistor with low collector voltage?

Answer 1

What you want to do is saturate the transistor, which means you want a base current (very approximately) 0.1* the collector current - around 0.3mA, or higher currents won't hurt. Assuming you're switching from a 3.3V MCU output, you'll have about 2.6V across the base resistor after subtracting Vbe, so Rb should be less than 2.6/0.3 = 8.67kilohms (choose 8k2 or even reduce to 4k7).

What voltage does that give between collector and emitter (Vce)? Check the datasheet for your chosen transistor; it'll probably guarantee "below 0.2V". e.g. for the BC847, Table 7 shows "VCEsat : IC=10mA; IB=0.5mA 90(typ) 200(mak) mV" So Vce would be under 0.1V at currents below 10mA - and note the base current is only 0.05* the collector current.

Also see the graph in Figure 3 which shows measured performance of a sample transistor, where Vce (at 25C) is only around 50mv under these conditions, rising to 100mv at 30mA.

If that's not enough, the higher gain BC847 reduces Vce to about 30mv under the same conditions (see Fig.11)

Most small signal NPN transistors should have similar info in the datasheets, I've just used the BC847 (aka BC107 for old timers) as an example.

Answer 2

You are intending to use a 1 kohm resistor and this means that the maximum current (into a short) will be 3.3 mA (from a 3V3 supply). With the LED in circuit this might drop to about 1 mA.

If you have three such LED circuits in parallel then, you would need the transistor to supply about 3 mA. However, if you are intending to drive the transistor's base from a GPIO (or other similar source), it makes sense to throw the transistor away and directly drive the 3 mA load.

Answer 3

You say you want to switch 3 LEDs in parallel from a 3.3 V supply, and that you intend to have a single 1 kΩ resistor in series with the parallel LEDs. There are several issues here:

1. Putting the LEDs directly in parallel is not a good idea. Instead of one resistor for the whole bunch, use a separate resistor in series with each LED.

   You didn't say what kind of LEDs, so I'll use typical green as example. These drop about 2.1 V. Even if the switch drops no voltage at all, that leaves 1.2 V across the resistor, which means you intend the bunch of LEDs to be driven with 1.2 mA total. That will be quite dim, but I'll have to take on face value that this is what you want for some reason. This also means the parallel LEDs aren't for getting more light, but because you want them in physically different places.

   Your ideal average per LED is then 400 µA. (1.2 V)/(400 µA) = 3 kΩ, which is the resistance you should use per LED. This also makes sense just by seeing that 1/3 the current should be thru each resistor compared to the single resistor, so it should be 3x larger.

2. As Andy said, with only 1.2 mA total, you should be able to use the digital output directly. Check the datasheet, but 1.2 mA is quite likely within its capabilities.

3. In this case, the saturation voltage of a low side NPN transistor isn't a big deal. With only 1.2 mA collector current, you can easily run it well into saturation. 200 mV is a typical value of saturation voltage in a case like that. Even if it is as high as 500 mV, you can easily design for that just be lowering the resistor values.

   For example, if the switch drops 500 mV, and the LEDs 2.1 V, then that leaves 700 mV across the resistors. To get 400 µA per LED, you need (700 mV)/(400 µA) = 1.8 kΩ.

4. If you really needed very low on state voltage for some reason (none is apparent here), then use a FET instead. The IRLML2502, for example, is guaranteed to be 80 mΩ or less with 2.5 V gate drive. At 1.2 mA, this will only drop 96 µV. The slop in the forward voltages of the LEDs will be more than that.