Using BJT transistors as switches

Thread Starter

dcbingaman

Joined Jun 30, 2021
480
I have heard from certain people on different forums that a 'base to emitter' resistor is required for 'stability' of a BJT being used as a switch. And I have heard you do not need it as well. This example I found shows it with no base/emitter resistor:

1630027754571.png

Here switch on is simple. With switch off we have open base. We can source current into Rb but in this case we do not sink Rb to ground. I know that in this situation the turn off time will be a considerably longer with no base-emitter bias resistor due to base to collector capacitance will drive the base until charged. I have used this circuit configuration multiple times with no issues. But some EE's state it can cause 'issues'. Being that the base is open when the transistor is in cutoff, this is a high impedance situation the might be susceptible to capacitive noise coupling?

Here is what many people say is the 'correct' way to build the circuit. I will admit if you require fast turn off times this is fine, but if you do not require fast turn off time, what is the benefit?

I am interested in how the community in general feels about this. I have never had issues with the no base to emitter resistor circuit.

1630028340122.png
The change is shown in red but according to many websites I have looked at that is not necessarily required:

https://www.electronicshub.org/transistor-as-a-switch/

Looking for the communities opinion on this matter. And if Rbe is required a logical reason why besides the fact that it significantly improves the turn off time. I am also very much aware that Rbe can be used to increase the 'turn on voltage' of the transistor being that it creates a voltage divider. So I know about that benefit as well.

Thanks
 

MrSalts

Joined Apr 2, 2020
656
I have heard from certain people on different forums that a 'base to emitter' resistor is required for 'stability' of a BJT being used as a switch. And I have heard you do not need it as well. This example I found shows it with no base/emitter resistor:

View attachment 246657

Here switch on is simple. With switch off we have open base. We can source current into Rb but in this case we do not sink Rb to ground. I know that in this situation the turn off time will be a considerably longer with no base-emitter bias resistor due to base to collector capacitance will drive the base until charged. I have used this circuit configuration multiple times with no issues. But some EE's state it can cause 'issues'. Being that the base is open when the transistor is in cutoff, this is a high impedance situation the might be susceptible to capacitive noise coupling?

Here is what many people say is the 'correct' way to build the circuit. I will admit if you require fast turn off times this is fine, but if you do not require fast turn off time, what is the benefit?

I am interested in how the community in general feels about this. I have never had issues with the no base to emitter resistor circuit.

View attachment 246658
The change is shown in red but according to many websites I have looked at that is not necessarily required:

https://www.electronicshub.org/transistor-as-a-switch/

Looking for the communities opinion on this matter. And if Rbe is required a logical reason why besides the fact that it significantly improves the turn off time. I am also very much aware that Rbe can be used to increase the 'turn on voltage' of the transistor being that it creates a voltage divider. So I know about that benefit as well.

Thanks

Without the resistor, the base is floating - if your bjt is actually a Darlington, some capacitive, RF antenna effects, or human touching it can cause the LED to turn on/flicker occasionally. If you want to really insure it doesn't turn on and leak any current in a low power (extremely low power, a 22k to 100k is a good idea (for example if you put a PIC in sleep mode and turn all GPIO to inputs before shutting it down, the base will be floating but some static could occasionally cause the base to drift high enough to leak a bit of collector current - but that is button cell battery stuff and looking for every microamp.

Otherwise, I don't think it is too important for a standard BJT w Hfe < 250.

at least that is my opinion.
 

Thread Starter

dcbingaman

Joined Jun 30, 2021
480
Without the resistor, the base is floating - if your bjt is actually a Darlington, some capacitive, RF antenna effects, or human touching it can cause the LED to turn on/flicker occasionally. If you want to really insure it doesn't turn on and leak any current in a low power (extremely low power, a 22k to 100k is a good idea (for example if you put a PIC in sleep mode and turn all GPIO to inputs before shutting it down, the base will be floating but some static could occasionally cause the base to drift high enough to leak a bit of collector current - but that is button cell battery stuff and looking for every microamp.

Otherwise, I don't think it is too important for a standard BJT w Hfe < 250.

at least that is my opinion.
Thanks for sharing your opinion.
 

Thread Starter

dcbingaman

Joined Jun 30, 2021
480
Rbe will improve the turn off time. Does it matter for your application?
In this application the turn off time is not critical it could be 100ms and still no issue for this application. Of course I know it will be faster than that. Thanks for your input.
 

crutschow

Joined Mar 14, 2008
27,960
At the cost of one extra resistor, it's cheap insurance to minimize the chance of the transistor turning on due to interference or high temperature leakage.
 

Ian0

Joined Aug 7, 2020
3,533
Often, the base signal comes from a logic gate or something with push-pull outputs. In that case the logic gate provides the pull-down current, so no resistor required.
It could also come from an op amp which can’t get within a volt of ground. In that case Rbe is mandatory to make a potential divider, or it will never switch off.
The situation where the input signal is either V+ or open circuit doesn’t seem to occur very often.
If you want to save one resistor, then replace the BJT by a MOSFET. Then you will definitely need the pull-down resistor, but can eliminate the series resistor, unless V+ >15V.
 

DickCappels

Joined Aug 21, 2008
7,906
Behold! The venerable uL914, Dual RTL nor gate, one of the early logic chips:
1630057475865.png

The 7402 TTL/74L02 quad NOT gate (partial).
1630057663145.png

A lot of these were sold and used. I haven't heard any complaints about a missing B-E resistor.

I agree with crutschow in post #6. The additional resistor is cheap insurance, but if you use the part within the power supply and temperature limits spelled out in the datasheet you souldn't worry.
 

Ian0

Joined Aug 7, 2020
3,533
Behold! The venerable uL914, Dual RTL nor gate, one of the early logic chips:
View attachment 246668

The 7402 TTL/74L02 quad NOT gate (partial).
View attachment 246670

A lot of these were sold and used. I haven't heard any complaints about a missing B-E resistor.
But the base will get grounded every time it gets connected to another output. It was always said, but not recommended that TTL inputs could be left floating and they would read a logic high, but TTL inputs are common base and the initial question was about a common emitter circuit.
 

BobaMosfet

Joined Jul 1, 2009
1,883
I have heard from certain people on different forums that a 'base to emitter' resistor is required for 'stability' of a BJT being used as a switch. And I have heard you do not need it as well. This example I found shows it with no base/emitter resistor:

View attachment 246657

Here switch on is simple. With switch off we have open base. We can source current into Rb but in this case we do not sink Rb to ground. I know that in this situation the turn off time will be a considerably longer with no base-emitter bias resistor due to base to collector capacitance will drive the base until charged. I have used this circuit configuration multiple times with no issues. But some EE's state it can cause 'issues'. Being that the base is open when the transistor is in cutoff, this is a high impedance situation the might be susceptible to capacitive noise coupling?

Here is what many people say is the 'correct' way to build the circuit. I will admit if you require fast turn off times this is fine, but if you do not require fast turn off time, what is the benefit?

I am interested in how the community in general feels about this. I have never had issues with the no base to emitter resistor circuit.

View attachment 246658
The change is shown in red but according to many websites I have looked at that is not necessarily required:

https://www.electronicshub.org/transistor-as-a-switch/

Looking for the communities opinion on this matter. And if Rbe is required a logical reason why besides the fact that it significantly improves the turn off time. I am also very much aware that Rbe can be used to increase the 'turn on voltage' of the transistor being that it creates a voltage divider. So I know about that benefit as well.

Thanks
This will give you a really easy education on BJTs, plus example problems, answers, and good explanation:

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3
 
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