Need help to calculate values of BJT switch function

[Sebastie]

Hello
I have to calculate the value of resistor of a simple switching stage. My question is what is the method to calculate the value of resistor ?
Since the value of the beta is dependant of the VCE voltage and collector current (and as the voltage of my application doesn't match with the example in the datasheet), how can I do to evaluate the current of the base ? and the base voltage ?

How can I calculate RB1 and RB2 for worst case analysis ?

I hope that you could help me

thanks a lot

**broken link removed**

 

[alexan_e]

Please upload your images here, your links are not convenient to download the images and have time restrictions.

Alex

 

[Sebastie]

I forgot to tell you that my application need a IC=5mA.

 

[ducvilla]

I think Rb1 and Rb2 play the role only bias for transistor. The Ic is given, then you can easily calculate Ib. Then applying Kirchoff law at the node B of transistor, note that when transistor is ON, the Vbe ~ 0.7V depend on the transistor you use. See two char. you can estimate the value of Rb1 and Rb2.

Hope this help!
Duc.

 

[jony130]

Normally the only information we have is that if we "force" the base current to be equal Ib = Ic/10 then the BJT for sure will be in saturation region.
Most small-signal BJTs saturation specs are defined saturation when Ic/Ib (called forced beta) = 10.
https://elenota.pl/pdf/ON_Semiconductor/p2n2222a-d.pdf (figure 11)
But in practice, most small-signal transistors will be in saturation when Ib > Ic / Hfe_min

Rb1 = 10*Rc or
Rb1 = (Vin - Vbe)/ (0.1*Ic) = (3.3V - 0.6V)/ 500uA = 4.7KΩ
Rb2 = 10*Rb1 = 47K


Or we use

\[ Rb \leq * Rc * \frac{Hfe_{min}}{K} * \frac{Vin - Vbe}{Vcc - Vce_{sat}}\]

K - overdrive coefficient ( typical from 2 to 5)

 

[Sebastie]

Hello Jony130,
Thanks a lot for you reply.
RB2 doesn't seem to be very important, I suppose the relation RB2=10*Rb1 that is a convention ...
I don't understand the relation ib > Hfe_min*Ic, do you mean Ib > Ic/Hfe_min ?
why do you take Ic/Ib=10 ? (since the beta min is specified at 35 in the datasheet ?) does it include the overdrive coef ?

Could you please explain in more details how can you deduce your last equation between RB and RC ?

Thanks a lot for your help.

 

[jony130]

RB2 doesn't seem to be very important, I suppose the relation RB2=10*Rb1 that is a convention ...

In this case RB2 is simply pull-down resistor.
But in some cases the low input voltage is greater than 0.6V so we need voltage divider to ensure BJT cut-off.

I don't understand the relation ib > Hfe_min*Ic, do you mean Ib > Ic/Hfe_min ?

Yes, I made mistake.
This is what I mean Ib > Ic/Hfe_min

why do you take Ic/Ib=10 ? (since the beta min is specified at 35 in the datasheet ?) does it include the overdrive coef ?

The most datasheet show Vce_sat for Ic/Ib = 10
See the example
https://elenota.pl/pdf/ON_Semiconductor/p2n2222a-d.pdf (figure 11)
So to be one hundred percent sure that your BJT will be in saturation you must use this so-called forced beta technique.
For european BJT Ib/Ib = 20
https://elenota.pl/pdf/Philips/bc846_bc847_bc848_5.pdf

Also notice that Ic/Ib = Rb/Rc
So if Rb/Rc > Hfe_min the BJT is in active region

 

[Sebastie]

Hello Jony130, Thanks a lot, it's more clear for me now

 

[jony130]

Could you please explain in more details how can you deduce your last equation between RB and RC ?

This equation come form this

Ib = (Vin - Vbe)/Rb

Vce = Vcc - Ic*Rc = Vcc - hfe*Ib*Rc =Vcc - Hfe*(Vin - Vbe)/Rb*Rc = Vcc - hfe*Rc/Rb * (Vin - Vbe) --> Solve for Rb

Rb = (Vin - Vbe)/(Vcc - Vce) * hfe * Rc