[SOLVED] BJT inverter with high base to emitter resistance

[Y.li]

Hi,

I was making a BJT inverter when I came across this web page. Here they have mentioned that a high value of resistance (R2), 10 times of base resistance is to be connected b/w base and emitter of bjt to avoid base going into -ive region of operation during switching. Can any pls. give the explanation of this resistance. The link is as below:

http://techhouse.brown.edu/~dmorris/projects/tutorials/transistor.switches.pdf

Thanks,
Y Li

 

[FvM]

The explanation doesn't suggest much understanding of transistor operation principle.

R2 is not that important here; it has to do with stabilizing the base and preventing it from going slightly negative when you turn the device on and off. The web says it should be about 100*R1, which is good enough for me.

A value of 100*R1 is almost meaningless in most cases, except when the driving source is a single ended switch with leakage current.

A base-to-emitter resistance (with considerable lower resistance) can serve several purposes
- speed-up transistor switch-off ("drain" charge carriers)
- keep the transistor safely off in presence of a high Vce respectively Vcb voltage.

 

[Y.li]

Hi FvM,

Thank you for replying. So do you suggest that I skip the R2 resistance in this case.

Regards,

Yang Li

 

[godfreyl]

I suggest throwing the PDF away. As FvM suggested, whoever wrote it doesn't have a good understanding of how to use transistors as switches.

If you tell us a bit more about your circuit or what you want to do, we can help you with that.

 

[Y.li]

Well, I have simply done away with the inverter stage .