bjt switch

Thread Starter

jerome89

Joined Jan 14, 2013
5
hi,

i want the bjt to act as a switch with the goal of getting
when bjt turned off - Vout = 20V
when bjt turned on - Vout = ~0V

Vin is 5v

what is the proper step to get the desired resistance on collector and base?
from the bjt datasheet, what should i take note of?

your help is much appreciated


 

Jony130

Joined Feb 17, 2009
5,230
When we use a BJT as a switch (saturation / cut-off) we use forced beta Ic/IB = 10 --->Rb/Rc = 10 --- > Rb = 10 * Rc

Or Ib = (K * Ic)/βmin
where
K = 3...10 - overdrive factor
 
Last edited:

anhnha

Joined Apr 19, 2012
881
When we use a BJT as a switch (saturation / cut-off) we use forced beta Ic/IB = 10 --->Rb/Rc = 10 --- > Rb = 10 * Rc

Or Ib = Ic/(βmin * K )
where
K = 3...10 - overdrive factor
Jony, are you assuming that RB and Rc use the same voltage supply? In the OP, Vin is 5V and Vcc = 20V.
 

Jony130

Joined Feb 17, 2009
5,230
Jony, are you assuming that RB and Rc use the same voltage supply? In the OP, Vin is 5V and Vcc = 20V.
Yes.
the most datasheet show Vce_sat for Ic/Ib = 10
See the example
http://www.solarbotics.net/library/datasheets/2N2222.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 Ic/Ib = 20
http://hep.fi.infn.it/PAMELA/pdf/bc847.pdf

If Vin is not the same as Vcc we can use this equation.
Ib/Ic = 10

Rb = (Vin - Vbe)/(0.1*Ic)

Rc = (Vcc - Vce_sat)/Ic

Or
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_min - Vbe)/(Vcc - Vce_sat) * hfe_min/K * Rc

K = 3...10 - overdrive factor
 
Last edited:

WBahn

Joined Mar 31, 2012
26,398
Basically, if you set K=1 then you are assuming that β=hfemin. But if, at the same time, you use the datasheet values for Vcesat you have introduced a problem because Vcesat is usually measured under conditions that force β=10, which is considerably lower than hfemin for most transistors. Also, hfemin is a parameter that is valid for a specific range of operating conditions. In many applications this will probably work out just fine, but in others it could lead to marginal or non-operational results. If, on the other hand, you set K equal to a small value, say at least three, then you are providing protection for when you would otherwise be in those marginal regions without having to do the detailed circuit analysis that you really should be doing if you are using K=1. By the time you have gotten to about K=10 you have probably reached (or are reasonably close to reaching) the forced β=10 case since hfemin is seldom much greater than 100.
 
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