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I have a question regarding bipolar transistor and it's operation as a switch.

I know that, in order to make transistor work as a switch, we need to avoid active region, meaning, we need to make transistor to work only in saturation and cut-off region.

This basically, cannot be implemented perfectly, but we can make transition period from saturation to cut off region as fast as possible, but I am not quite sure how to make transistor to work in cut off region, meaning, I want to know how to turn the transistor off correctly.

Let's take NPN transistor as an example:

The transistor is off if it's base-emitter and base-collector PN junctions are reverse-biased, however, that means that we would need significant negative voltage on base in order to do so, but I am wondering, how would this affect the circuit and transistor?

Another approach I've heard of is that we only need to make base-collector reverse biased and then we need to add a resistor between base and emitter, how is this any better? Do I need a resistor with huge resistance or with low resistance or it actually doesn't matter, and why?

Any help appreciated!

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    \$\begingroup\$ To cut-off, the bipolar transistor all you need to worry about is about the base-emitter voltage and the base current. Short the base together with the emitter and the BJT will be cut-off. Or set Ib = 0A. \$\endgroup\$
    – G36
    Nov 4, 2018 at 10:48
  • \$\begingroup\$ Ok, but, why that works? I mean, why is that a correct way to turn off transistor? What would happen if we tried to turn it off by negative voltage on the base? Why is that bad thing to do? Why is shorting the base and emitter better solution? \$\endgroup\$
    – cdummie
    Nov 4, 2018 at 10:53
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    \$\begingroup\$ It works because without the base current (Vbe <0.4V) transistor is cut-off. And transistor action only occurs if the base current is flowing. And sometimes to speed-up the BJT turn-off time we apply a negative voltage between base-emitter junction. electronics.stackexchange.com/questions/367750/… and about saturation electronics.stackexchange.com/questions/276146/… \$\endgroup\$
    – G36
    Nov 4, 2018 at 10:59
  • \$\begingroup\$ Why cant you use the transistor in active region? \$\endgroup\$ Jan 2, 2021 at 12:11
  • \$\begingroup\$ @the force awakens because then it's not a switch. For a switch, you want as much IC current as needed by your load. In active mode, your current is limited by IB, while in saturation it's limited by your load. \$\endgroup\$
    – KD9PDP
    May 23, 2021 at 16:08

2 Answers 2

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Since there is a lot written and explained about this subject, I will make it more practical - in the form of a series of experiments implemented through CircuitLab.

"Floating" base

Basically, if we leave the base of a transistor unconnected, its base-emitter voltage Vbe and base current Ib should be zero; so it should be cut off.

schematic

simulate this circuit – Schematic created using CircuitLab

Leaks occur

The problem, however, is that there are always some leaks (paths with resistance or capacitance) between the base and supply rails. Let's simulate such a leakage by connecting a high resistance (1 MΩ), Rleak resistor between the base and the positive terminal of the supply voltage Vcc.

schematic

simulate this circuit

As a result, a base current flows through the base-emitter junction and the transistor is saturated.

Increasing the leakage resistance

A straightforward way to solve the problem is to increase the leakage resistance (for example, by cleaning the board). For the circuit shown, if it reaches 100 GΩ, the transistor will be cut off again.

schematic

simulate this circuit

Grounding the base

If the leakage is significant (e.g., 1 kΩ), the best solution is to ground the base (short the base-emitter junction)... of course if this is possible in the particular application.

schematic

simulate this circuit

This is the preferred way of controlling the transistor switch when the turn-off time is important. The reason is that there is an accumulated charge in the base of the saturated transistor that must be dissipated through some path.

Shunting resistor

We usually do not have to be so brutal and can solve the problem by connecting a resistor Rshunt of moderate resistance (e.g., 10 kΩ) between base and ground.

schematic

simulate this circuit

The two resistors Rleak and Rb form a voltage divider. Its voltage is less than 0.7 V so the transistor is cut off.

schematic

simulate this circuit

Significant leakage

But if the leakage resistance is relatively low (e.g., 100 kΩ)...

schematic

simulate this circuit

... the divider's output voltage is more than 0.7 V, and the transistor is saturated.

schematic

simulate this circuit

Negative base voltage

Then the remedy is to "pull down" the base below ground by another but negative voltage source Vee.

schematic

simulate this circuit

Now the two resistors Rleak and Rb form a bipolar voltage divider whose output voltage is negative so the transistor is reliably cut off.

schematic

simulate this circuit

Positive emitter voltage

Instead to "pull down" the base below ground, with the same success we can "pull up" the emitter above ground by another positive voltage source Ve. The result is the same - the base-emitter voltage Vbe is negative.

schematic

simulate this circuit

An advantage of this method is that there is no need for a second source because the positive voltage can be obtained from the power supply, for example with an Rd-D network.

schematic

simulate this circuit

A disadvantage is that when the transistor is saturated, its collector voltage will be raised by 0.7 V.

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In addition to what others wrote in the comments: A bipolar transistor is basically a current amplifier from which follows that no current flowing into the base means no current flowing through collector/emitter.

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