R2 between base and emitter

Ian0

Joined Aug 7, 2020
3,280
This resistor increases the turn-on threshold of the transistor switch.
. . . in a potential divider with R1, so that it can be connected directly to a logic level output.
R2 also helps remove any stored charge and helps the transistor turn off more quickly.
 

Thread Starter

SpNw

Joined May 27, 2020
11
. . . in a potential divider with R1, so that it can be connected directly to a logic level output.
R2 also helps remove any stored charge and helps the transistor turn off more quickly.
Thanks, one more question, it can work like a pull down resistor too? To avoid that base start floating?
 

LvW

Joined Jun 13, 2013
1,300
This resistor increases the turn-on threshold of the transistor switch.
As we can see in the nice diagram from Bordodynov, it does NOT influence the thresholds but it improves the rise time at the output when the input goes down. This is because the time constant for discharging the BJTs capacitance is determined by the 47k resistor(s) - single vs. parallel.
 

Ian0

Joined Aug 7, 2020
3,280
As we can see in the nice diagram from Bordodynov, it does NOT influence the thresholds but it improves the rise time at the output when the input goes down. This is because the time constant for discharging the BJTs capacitance is determined by the 47k resistor(s) - single vs. parallel.
If R1 were absent, then it wouldn’t, but with R1 in circuit R2 sets the switch on threshold.
 

LvW

Joined Jun 13, 2013
1,300
If R1 were absent, then it wouldn’t, but with R1 in circuit R2 sets the switch on threshold.
R1 absent? Open circuit?
I think, for a fair comparison we have to consider two cases - both with the same input voltage:
* With one single resistor R1
* With a voltage divider R1-R2.
Now it is clear that (for the same switching threshold) the time constant of the second alternative can be much smaller than for the first one.

Example: Input signal (max) Vin=6V; required base current (saturation) Ib=0.1mA: Vbe=0.8V
* case 1: Series resistance Rs=(6-0.8)V/0.1mA=52kOhms
* case 2: Current through R2 (selected) I=0.1mA ;
R2=0.8/0.1=8kOhms and R1=(6-0.8)/(I+Ib)=5.2/0.2mA=26kOhms.
 
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