# BJT NPN transistor as low side switch: Use voltage divider or in series resistor?

I want to use a PMBT3904,215 transistor as a switch on low side. My question is, which technique is the best and why? To drive the base of the transistor using a voltage divider like so:

simulate this circuit – Schematic created using CircuitLab

Which I think ignores R2 when the base is 5V, since all the current flows from the base to the emitter of the transistor.

Or a resistor in series:

simulate this circuit

I think a voltage divider would be a better idea to avoid my transistor's base "floating," but would there be another reason? After all the signal from the MCU can be 0V and it is not actually floating, so I can "save" some money throwing away that extra resistor (when we are talking about many products being produced.)

• The divider is not always necessary, but won't hurt, and can help turn off when MCU is reset. But adjust R2,R3 to higher value, 100Ω will most likely pull too much current ... you need roughly I_load / β (give it a generous margin) Commented Feb 23, 2021 at 15:14
• R2 is not ignored, but the voltage across it is limited to Vbe, and this determines the current it can steal from the base (it's not a gate!)
– user16324
Commented Feb 23, 2021 at 15:22

You, as the designer, should be considering two cases:

1. Am I driving the base sufficiently to drive the transistor into saturation while at the same time making sure that I don't either overdrive the base or my MCU's GPIO output specification?

2. Is the transistor going to be fully into cutoff when the output from the MCU is LOW or undefined, such as on power-up before the pin has been initialized as an output.

So the answer to 1 is usually just an Ohm's Law calculation based on the required current to switch the transistor on.

Number 2 can be a bit more complex due to the difference between a LOW on the output vs. whatever your specific MCU does at power-up. You'll need to understand that and design your interface circuit between the MCU and the transistor accordingly.

What are the two levels (on/off) provided by the MCU? As far as I know, the most popular solution is to use just a series resistor R3. The minimum value of R3 must ensure that the transistor can be driven in deep saturation.

As a reliable indication for this mode, the base current (calculated based on VBE=0.7 V and the MCU on-level) should be at least app. 1/10 of the selected collector current. Such a relatively large base current indicates that the base-collector diode is now open (saturation condition).