BJT as a switch: why the third resistor?

bjt switch circuit

I have an IC device which provides a pin in order to decide if turn the device ON or OFF. If the pin is HIGH, the device is OFF, otherwise is ON. Actually, the device pin is internally pulled-up at Vcc (equal to 3.8V) via a resistor R2 (equal to 47K) and if you would like to turn ON the device you have only to connect that pin to ground. I would like to remote control the turning ON/OFF, thus a BJT switch is needed.

Assume to not consider R3 (R3 = inf) in the figure below.
To decide how large should be R1, I made in this way:
The current which passes in the transistor from C to E (Ice) is about Vcc/R2 since Vce is about zero when the BJT is closed. By reading from the BJT datasheet the range of hfe (hybrid parameter), I considered the hfe minimum in order to estimate the maximum base current (Ib) needed to let BJT working at the desired Ice. In practice, to be further sure, I considered an high value for Ib: two or three times that calculated. Then, R1 is given by Vin/Ib.

Now, why does the IC device HW guide suggest me to use a BJT switch putting another resistor (R3) between C and E as weel? In a practical example, the guide says both R1 and R3 are equal to 47K in case of Vin=3 to 5 Volt. Let consider hfe = 200 to 450 and Vbe = 0.77 Volt.


Thx
N.

bjt switching circuit

R3 is used to turn ensure the BJT is turned off and to help reduce the likelihood of the transistor being incorrectly triggered by noise or similar event.

Furthermore, it is often useful to have a Base emitter resistor included on discrete component layouts, that way you can use the voltage divider input to further influence turn on / turn off characteristics of the transistor.

Experience tells me to not omit R3 in common emitter designs. This concept also applies to PNP transistors - where I find a Base Emitter resistor often a good idea.

oddbudman

application of bjt as a switch

if your circuit work fully as switch, i think R3 not needed. May be R3 used for limitting the base current and the feature is optional.

bipolar transistor switch

Thank you for your reply.
First of all, I wanna say R3 should be useful in case of using BJT as a switch as well, since lots of application notes suggest it and even many BJTs switch transistors are sold with an embedded R3 resistor.
That said, I have not understood yet the question very deeply:
1) If I consider R2=R3 and make the Thevenin equivalent at node B, it results Vin_Thevenin = Vin/2 and R_Thevenin = R2/2=R3/2. This means the Ib current should be the same. So, I have done some simulations but I have seen the Ib current in case of presence of R3 is smaller than in case of omitting R3! Why is it?
2) You say the BJT with presence of R3 is more robust in term of noise. Do you mean about noise on gnd? I only can imagine that in case of noise, the presence of R3 let the base voltage increase faster than not, and then avoid the turn-off of the transistor. Is it right? if not, in which terms did you mean that?
3) Still I cannot understand how the turn-on/off of BJT could be influenced by R3...
By simulating, I only have seen that in case of R2=47K and R3=inf the times taken by BJT to turn-on and to turn off are quite different; while in case of R2=R3=47K the times become more similar to each other.
4) Why is generally used R2=R3?

As you can see, I still have a lot of issue to address!
Thank you in advance.


N.

switch+bjt+circuit

1) With R3, Base current is lower because the total current through R1 now needs to travel through both R3 and the BE junction of the BJT.

2) Sometimes when you design a PCB, you will have some problems with voltage levels - often unexpectedly.

Say for instance you have a switching device such as a DCC regulator nearby your example transistor circuit. Such a device emits disruption onto the pcb (caused by high current switching). This coupled with high impedance terminations to transistor circuits like your example can often cause problems, as voltage can often appear across the high impedance terminations, causing transistors to turn on. Without R3 noise current through R1 can only make it to GND through the transistors base. This current can turn on the transistor

By putting R3 on the circuit you can have more control over the transistor. Sure it will work without it. But on all of my designs I have learned to put one there, its good practice. The PCB environment is much more complex than a simulator.

3) As i have explained in (1) R3 changes the base current, hence turn on characteristics.

4) I am not entirely sure, however, my guess would be a 50-50 configuration would be the most appealing combination for a range of designs. Hence its configuration popularity.

Hope this helps.

Regards,
oddbudman

bjt transistor switch circuit

Thank you very much, really kind and useful reply. At last, I think to have understood how noise can influence circuit..
Only a doubt still remains:
First of all, I made a mistake and when I mentioned R2 in the previous message I actually meant R1. My doubt concerns the Thevenin equivalent of R1, R3 and Vin made at node B with a R_thevenin = R1//R3 and Vin_Thevenin = Vin*R3/(R1+R3). In practice, if I chose R1=R3, results R_thevenin = R1/2 and V_thevenin = Vin/2: this means Ib still equal to Vin/R1, but in simulations it is not (consider no noise applyed). Why Ib is different in the two cases if the Thevenin equivalent assures the same value of current? What is my mistake...?
Thank you once more.

N.

bjt design as a switch

OK! I found my error... there was an error during the simulation. The circuit and the Thevenin equivalent are exactly the same. Apologies for my mistake.
Everything is ok now.

Thank you all.

N.

bjt switching high impendance

third reisistor is to limit the voltage drive across the transistor and provide discharge path for base emitter capacitor........

bjt as switch circuit configuration

dischge path for base capacitance........

transistor + base-emitter resistor

simple voltage divider ......... the switching is depend on the values of the resistors R3... if it is too small then may be transistor not work as a switch