Why does the base-emitter resistor affect the switching threshold voltage of a BJT inverter? What's the math behind it?

Thread Starter


Joined Mar 16, 2023
Welcome to AAC.
As this is homework, have you been taught that the base-emitter voltage is a fixed value?

This is regarding a practical we have to do. I've already chosen the values for the resistors such that the switching threshold voltage is shifted to Vcc/2 with R2, but don't know why it does so. I suspect it has something to do with R2 sinking current that would otherwise have gone into the base, but am uncertain of the mathematics involved.

We've been taught that Vbe(on) varies slightly with change in Ib (or the biasing of the transistor), as in the datasheet.


Joined Jan 29, 2010
Hi PH,
Check out this image of a silicon diodes Voltage versus Current.
The Base to Emitter junction is a Diode, forward biassed.
ScreenHunter 31.gif


Joined Mar 14, 2008
I suspect it has something to do with R2 sinking current that would otherwise have gone into the base
Yes, R8 and R2 form a voltage divider so the transistor doesn't start to conduct significantly until that divider voltage rises to the Vbe voltage and provides the base current shown in the data sheet for the desired collector current.

Do you understand how a two resistor voltage divider works?


Joined Mar 31, 2012
In order to explore the math behind the question, you need to decide on a mathematical model for your transistor. You have some options here and which one you use depends on how detailed you want the answer.

Probably the simplest model that is reasonable for this question is the model that has the Vbe of the transistor being a fixed value (say 600 mV) in the active region and the beta of the transistor also being fixed (say 100) in this region.

Now define the switching point. A reasonable choice for this might be when the output voltage is half the supply voltage. How much base current is needed to make this happen?

Now you are in a position to determine the mathematical relationship between the input voltage and the base current as a function of the base-emitter resistance.


Joined Jun 17, 2014

The simplest way to think about this is to assume a turn on threshold for the transistor which as an approximation can be taken as something around 0.7 volts,, and that is because the base emitter voltage conduction voltage is around 0.7 volts. It is just then a matter of doing the voltage divider formula with R8 and R2 and the input voltage.

So at first, assume a base emitter voltage turn on of 0.7 volts, then do the voltage divider formula. That's an approximation that will allow you to see about how this works. You can then later look deeper. If you dont know the voltage divider formula you can ask here too.

Note that as stated in this thread before this post the transistor models vary so depending on what model you use you will get different results, but the above approximation will help you understand roughly how this works. Spice model base emitter diodes can be as low as 0.4 volts and 0.65 volts typical, but 0.7 volts is a good starting point.